\subsection{Resolutions having a DGCA structure}
\label{sec:DGCA}

\subsubsection{An explicit arborescent Koszul-Tate resolution}

Several $ \CO$-module resolutions $(\FM_\bullet,d)$ of $\CO/\CI $  admit a structure of a differential graded commutative algebra  (DGCA). Still, they may not be Koszul-Tate resolutions, because they may not be graded symmetric algebras. However,  such $ \CO$-module resolutions $(\FM_\bullet,d)$ allow the construction of an explicit arborescent Koszul-Tate resolution in quite a straightforward manner.
Before stating our result, let us introduce a convention: for all $n \geq 2$, denote by $t_n$ the tree with $n$ leaves and one root:
\begin{equation}
\label{eq:onerootnleaves}\begin{array}{c}t_n = \\ \vspace{.4cm} \end{array} \overbrace{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[, tier =1]
[, tier =1]
[,edge= dashed,  tier =1]
[,edge= dashed,label={[mystyle2] {$...$}}, tier = 1]
\path[fill=black] (.parent anchor) circle[radius=2pt];
\end{forest}}^{\hbox{$n$ leaves}}
. \end{equation}
Notice that $t_2$ and $ t_3$ already appeared as $ \vee$  and $ \scalebox{1.5}{\wtree}$ in the expressions $ \psi_\vee$ and $ \psi_{\wtree}$.

\begin{proposition}
\label{prop:psi.dga}
Let $(\FM_{\bullet}, d)$ be a free $\CO$-module resolution of $ \CO/\CI$ such that $ \FM_\bullet \oplus \CO$ is equipped with a graded commutative associative multiplication $\star$ turning it into a DGCA. Here the multiplication $\star$ is assumed to extend the given $\CO$-module structure:
$F * a =a * F=Fa $ for all $ F \in \CO$, $a\in \FM_\bullet \oplus \CO$. Choose the arborescent operations $\psi \colon t \mapsto \psi_t$  as follows
$$ \psi_{t_n} \left(    a_1 \otimes \dots \otimes a_n\right) = a_1 \star \cdots  \star a_n \hbox{ for all $ a_1, \dots, a_n \in \FM_\bullet$}$$
and  $\psi_t=0 $  for any tree not of the previous form.
%\in \CO \oplus \FM_\bullet $
Then the pair $(S(\CT[\FM_{\bullet}]),\delta_\psi)$
is an arborescent Koszul-Tate resolution of $ \CO/\CI$.
\end{proposition}
\begin{proof}
%We leave it to the reader to check that the arborescent operations above satisfy Equation \eqref{eq:deltaPsiSquare}.
It suffices to check Equation \eqref{eq:deltaPsiSquare} for \emph{i)} the trees $t_n$ with $n \in \mathbb N$ and  for \emph{ii)} the trees obtained by plugging $ t_n$ on some leaf of $ t_m$ for  $ n,m \geq 2$.
\emph{i)} holds true because $d$ is a derivation of $*$ and \emph{ii)} holds true because $*$ is associative.
\end{proof}
\vspace{1mm}
\noindent We now have two DGCAs: $(\FM_\bullet \oplus \CO,\star, d) $ and $(S(\CT[\FM_{\bullet}]),\odot,\delta_\psi)$. Proposition \ref{prop:Homotopy} provides a projection ${\mathrm{Proj}}_\psi $ from the second complex to the first complex which is a chain map.

\begin{proposition}
\label{prop:backtoKoszul}
For an arborescent Koszul-Tate resolution as in Proposition \ref{prop:psi.dga}, the projection $${\mathrm{Proj}}_\psi \colon S(\CT[\FM_{\bullet}]) \longrightarrow \FM_\bullet \oplus \CO$$ is a morphism of differential graded  commutative algebras.
\end{proposition}
\begin{proof}
%${\mathrm{Proj}}_\psi $ is a chain map
%by Proposition \ref{prop:Homotopy}.
Let us check that ${\mathrm{Proj}}_\psi $  is a  morphism of graded commutative algebras,
\begin{equation} \label{eq:morphismneeded} {\mathrm{Proj}}_\psi( T_1  \odot  \cdots \odot T_k) =  {\mathrm{Proj}}_\psi( T_1  ) \star \cdots \star   {\mathrm{Proj}}_\psi( T_k  ) \end{equation}
%Let $S \in S(\CT[\FM_{\bullet}])$ be of the form
% $$  S= T_1 \odot T_2. $$
for all $ T_1, \dots, T_k$ in $\CT[\FM_{\bullet}] $.
If  $ T_1=|\otimesk a_1 $, $\ldots$,  $T_k = |\otimesk a_k$ are all trivial trees, then, by the definition of the projection in Equation \eqref{eq:matricesInclProj}:
$$ {\mathrm{Proj}}_\psi \left(\begin{array}{c} \\ \scalebox{0.5}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{2}{$a_1$}, tier =1]
\end{forest}} \begin{array}{c}  \odot \dots \odot \\ \vspace{.2cm}  \end{array} \scalebox{0.5}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{2}{$a_k$}, tier =1]
\end{forest}} \\ \end{array}\right) \begin{array}{c} \\= \psi_{t_k}(a_1,\cdots,a_k)=  \\ \\ \end{array}  \begin{array}{c} \\  a_1\star \cdots \star a_k \, .\vspace{.5cm} \end{array}$$
If, on the other hand, one of them is not a trivial tree,
then ${\mathrm{Proj}}_\psi( T_1 \odot \dots \odot T_k)=(\psi \circ \r) (T_1 \odot \cdots  \odot T_k)=0$ since $ \r (T_1 \odot \cdots  \odot T_k)$ is not a tree of the form of Equation \eqref{eq:onerootnleaves}.

\end{proof}

\begin{remark}
\normalfont
Note that the kernel of ${\mathrm{Proj}}_\psi $ satisfies the requirements on the ideal $\mathfrak J $ in Proposition~\ref{prop:quotient}.
\end{remark}

\vspace{0.1cm}

\noindent
Examples \ref{ex:4.1} and \ref{ex:4.2} in Section \ref{sec:basicExamples} above provided two examples of  $ \CO$-module resolutions coming with a graded commutative associative product. We conclude this discussion here with a last and more exotic example.

\begin{example}
\normalfont
Consider $ \CO= \mathbb K[x,y]/\langle xy \rangle$ with the two generators $\overline{x}$ and $\overline{y} $. Let $\CI $ be the ideal generated by $\overline{x}$.
Let $\FM_i  $ be the free $ \CO$-module with one generator $ e_i$ for all $i \geq 1$.
Then $ \FM_\bullet$ becomes an $ \CO$-module resolution by choosing the boundary map $d$ according to
$$d(e_{i} )=\left\{\begin{array}{ll} \overline{y} \, e_{i-1} & \hbox{ for $i$ even} \\  \overline{x} \, e_{i-1} & \hbox{ for $i$ odd.} \end{array}\right.  $$
%for even integer $i $ and $ d(e_i) = x e_{i-1} $ for any odd integer $ i $.
$(\FM_\bullet \oplus \CO,d) $ is turned into a DGCA by equipping it with the product $\star$ defined by means of
$ e_i \star e_j = \mu_{i,j} \, e_{i+j}$ with
$$ \mu_{i,j} =
\left\{\begin{array}{cl} 0 & \hbox{if $i,j $ odd}  \\[10pt]
\begin{pmatrix} \tfrac{1}{2}(i+j) \\ \tfrac{1}{2}j \end{pmatrix} & \hbox{ if $i,j $ are even} \\[10pt]
\begin{pmatrix} \tfrac{1}{2}(i+j-1) \\ \tfrac{1}{2}(j-1) \end{pmatrix}& \hbox{ if $i $ is even and $j$ is odd} \end{array}\right.
where $\begin{pmatrix} n \\ k
\end{pmatrix}$ denotes  binomial coefficients.
This can be verified by a direct calculation.

\vspace{2mm}
\noindent
Let us remark that this DGCA is a Koszul-Tate resolution, since it is isomorphic, as a graded algebra, to the symmetric algebra of the free $ \CO$-module with one generator of  degree $1$ and one of  degree~$2$.

\end{example}

\subsubsection{Koszul complex}
\label{sec:Koszul}

Assume that an ideal $\CI \subset \CO$ is generated by $r$ elements $ \varphi_1, \dots, \varphi_r$.
Let $V$ be a vector space of dimension $k$ with basis $e_1, \dots, e_k $.
The pair $(K_\bullet(k),\delta)$ given by $K_\bullet(k) := \mathcal O \otimes_{\mathbb K} \wedge V^* $,  and $ \delta = \sum_{i=1}^r \varphi_i \otimes \mathfrak i_{e_i}  $ is a differential graded commutative algebra called \emph{Koszul complex}.
The following result, attributed to Jean-Louis Koszul, is well-known.
Recall that  $\varphi$  is said to be a \emph{regular sequence} if  $\phi_1 \hbox{ is not a zero divisor in } \CO$, and for all $i\in \{2, \dots, k\}$, $\phi_i $ is not a zero divisor in $ \CO/\langle \phi_1, \dots \phi_{i-1} \rangle $.
%and $(\varphi_1, \dots, \varphi_r)\CO \neq \CO$.

\begin{proposition} [13, 29] Let $\CO$ be a Noetherian ring. The Koszul complex $(K_\bullet(k),\delta)$ is a Koszul-Tate resolution of $\CO/\CI $ if and only if  $\varphi_1, \dots, \varphi_k \in \CO $ is  a regular sequence.
\end{proposition}

\begin{example}
\normalfont
\label{ex:K1}
Consider $\CI$ an ideal generated by a non-zero divisor $ \varphi_1 \in \CO$. Then $\varphi_1 $ is a regular sequence of length one.  As an $ \CO$-module, $K_\bullet(1) $  has therefore two generators: the first one is the unit $1 $ and we denote the second one by $\theta $.
The differential is given by $ \delta(1)=0$ and $ \delta (\theta)=\varphi_1$.
\end{example}

\begin{example}
\normalfont
Consider a polynomial $\psi \in \mathbb C[x_1, \dots, x_n]$ which is homogeneous of degree $> 0 $. If $\psi $ has an isolated singularity at $0$, then  it is a classical result [33] that the sequence
$ \frac{\partial \psi}{\partial x_1}, \dots,  \frac{\psi}{\partial x_n} $ is a regular sequence. Its Koszul resolution is canonically isomorphic to the graded algebra of polynomial polyvector fields on $\mathbb C^n$ equipped with the contraction by the exact $1$-form $d\psi $.
\end{example}

\vspace{0.1cm}

\noindent
Since the Koszul resolution is an $\CO $-module resolution, one can construct its arborescent Kozul-Tate resolution as in Proposition \ref{prop:psi.dga}.

\begin{example}
\normalfont
\label{ex:K1_2}
%Consider the context .
The Koszul resolution of Example \ref{ex:K1} is an $ \CO$-module resolution for which $\FM_i=0 $ for all $ i \geq 2$ while $ \FM_1$ has one generator $ \theta$.
Since this generator $\theta $ is odd,  for every decorated tree $t$ with $n \geq 2 $ leaves, $t[ \theta , \dots, \theta ]$ has a class in the quotient space  $ \CT[\FM_\bullet]$ which is equal to $0$.  As a consequence, the whole construction of Section \ref{sec:arborescent} becomes trivial, and the arborescent Koszul-Tate resolution coincides with the initial Koszul resolution.
\end{example}

\vspace{0.1cm}

\noindent
Example \ref{ex:K1_2} is a very particular case. It follows from Proposition \ref{prop:backtoKoszul} that the construction of Section \ref{sec:arborescent}, applied to the  $ \CO$-module resolution given by the Koszul resolution $(K_\bullet(k), \delta) $ of a regular sequence  of length  $k \geq 2$ yields an  arborescent Koszul-Tate resolution that will be much larger than $(K_\bullet(k), \delta) $.
For Koszul complexes, there is no gain in considering arborescent Kozul-Tate resolution, even on the contrary, it makes things more complicated there. However, in the following we provide examples for which our construction is very useful: one is given an $\CO$-resolution that happens to be a DGCA but not a Koszul-Tate resolution and obtain an aborescent Koszul-Tate resolution by the use of Proposition \ref{prop:psi.dga}.

\subsubsection{Taylor resolutions of monomial ideals}
\label{sec:taylor.monomials}

\vspace{0.1cm}

\noindent
Let $\CO$ be the polynomial ring $\CO =\mathbb K[x_1, \dots, x_n]$. An ideal $\CI \subset \CO $ is said to be a {\emph{monomial ideal}} if it admits a set  $\varphi = \lbrace \varphi_1, \dots, \varphi_k \rbrace$ of monomial generators. Then $ \CO/\CI$  admits resolutions equipped with a differential graded commutative algebra structure,
the so-called \emph{Taylor resolutions}, that we now describe  following [31].
% We denote by $\CJ$ the \emph{irrelevant ideal}, i.e the maximal ideal generated by $x_1, \dots, x_n$.  Notice that $ \CI \subset \CJ $.

\vspace{1mm}
\noindent
\begin{construction}[Taylor resolutions]
\normalfont
The \emph{Taylor complex} is the chain complex
\[
\begin{tikzcd}[row sep=2.5em, column sep=2.5em]
\cdots \arrow[r, "d"] & C_i(k) \arrow[r, "d"] & \cdots \arrow[r, "d"] & \mathcal C_0(k)
\end{tikzcd}
\]
of free $\CO$-modules $C_i(\varphi)$  constructed as follows:
\begin{itemize}
\item The graded vector space $C_\bullet(k) $ only depends on the number $k$ of monomials generating $ \CI$.
Let $C_\bullet $ be the free $ \CO$-module  generated by elements $e_\tau $ indexed by subsets  $ \tau \subset  \{1, \dots, k\} $:
$$ C_\bullet(k) = \oplus_{\tau \subset \{1, \dots, k\}}  \CO \, e_{\tau} .$$
We equip $C_\bullet $ with a grading by imposing the degree of $ e_{\tau}$ to be the cardinality of $ \tau$.  For instance, $ C_0(k) = \CO \, e_{{\emptyset}} \simeq \CO$ while $C_1(k)= \oplus_{i=1}^k \CO \, e_{\{i\}} $.
\item For every subset $\tau$ of $ \{1, \dots,k\}$, let $m_\tau $ be
the least common multiple of the monomials $ \varphi_i$ whose index $i$ is in $ \tau$. Explicitly, for $ \tau= \{i_1,\dots, i_m\}  \subset \{1, \dots,k\}$, we have $m_\tau = \mathrm{lcm} \left( \varphi_{i_1}, \ldots,  \varphi_{i_m}  \right)$.
Then the differential $d$ is defined by:
d(e_{\tau}) = \sum_{j \in \tau} \, \mathrm{sign}(j, \tau) \, \frac{m_\tau}{m_{\tau \backslash \{j\}}} \, e_{\tau \backslash \{j\}}
where $\mathrm{sign}(j, \tau)$ is $+1$ if $ \tau$ contains an even number of elements strictly smaller than $j$ and $ -1$ otherwise.
%$(-1)^{p_{j} +1}$, with $p_{j}$ being the position of $j$ in the ordered subset $\tau$.
\end{itemize}
\end{construction}

% \begin{remark}
\noindent
The complex $(C_\bullet(k),d)$ is an $\CO$-module resolution of $\CO/\CI$, see e.g. Theorem 26.7 in [31] (the minimality condition on the generators imposed there can be dropped). This complex is called the \emph{Taylor resolution} of $ \CO/\CI$.

\begin{remark}
\normalfont
Taylor resolutions need not be minimal $\CO $-resolutions in general. For instance, for the ideal of $\mathbb C[x,y]$ generated by $ x^2,xy,y^2$, the Taylor resolution has length three, while the $ \CO$-resolution in Section \ref{sec:basicExamples} case I has length two.
\end{remark}

\vspace{0.1cm}

\noindent
As explained in [8],
the Taylor resolution is equipped with an associative and graded-commutative multiplication $\star$: for every subsets $ \sigma, \tau \subset \{1, \dots, k\} $ we pose
\begin{equation}
\label{eq:mult.taylor}
e_{\sigma}\star e_{\tau} =  \left\{ \begin{array}{ll}
\mathrm{sign}(\sigma, \tau) \; \dfrac{\, m_{\sigma} \, m_{\tau}\, }{m_{\sigma\cup \tau}} \; \:e_{\sigma\cup \tau},  &
\hbox{if $\sigma \cap \tau= \varnothing $} \\[10pt] 0 &
\hbox{if $\sigma\cap \tau \neq \varnothing $.}
\end{array}\right.
\end{equation}
Here $\mathrm{sign}(\sigma , \tau)=(-1)^N$ where $N$ is the cardinality of pairs $ (i,j)$ with $i \in \sigma, j \in \tau$ such that $ i > j $. It is routine to check that this multiplication is compatible with $d$  as defined above, turning
the Taylor resolution of  $ \CO/\CI$ into a DGCA.

\begin{remark}
\normalfont
If $\varphi_1, \dots, \varphi_k $ is a regular sequence of monomials, then the Taylor resolution  coincides with the Koszul complex $(K_\bullet(k), \delta)$ of $\CO/\CI$ as DGCAs.
\end{remark}

\vspace{1mm}
\noindent
In general, Taylor resolutions are \emph{not} Koszul-Tate resolutions, because they may  not be symmetric algebras.
Being free $\CO$-module resolutions, Proposition  \ref{prop:psi.dga} allows us to construct an arborescent Koszul-Tate resolution out of them.
As far as we know,  this is the first explicit construction of a Koszul-Tate resolution which works for any monomial ideal even if the Koszul complex does not provide a resolution.

\section{Complexity of Koszul-Tate resolutions}
\label{sec:complexity}
\subsection{Counting computations needed for constructing  Koszul-Tate resolutions}

Arborescent Koszul-Tate resolutions are explicit in the sense that finitely many computations often suffice to construct it.
This section intends to make this statement precise and rigorous.

\vspace{1mm}
\noindent
Let us consider the following two classes of problems.
\begin{itemize}
\item "\emph{$\CO $-linear problem of the first type.}" Given free or projective $\CO$-modules $\mathcal N_1 $ and $ \mathcal N_2 $ and an $ \CO$-linear map $d_1 \colon \mathcal N_1 \to \mathcal N_2 $, find a free or projective $\CO$-module $ \mathcal N_0$ and an $ \CO$-linear map $ d_0 \colon \mathcal N_0 \to \mathcal N_1 $ such that $ {\mathrm{im}}(d_0)={\mathrm{ker}}(d_1) $.
\item "\emph{$\CO $-linear problem of the second type.}" Given three $ \CO$-modules $\mathcal N_0,\mathcal N_1, \mathcal N_2 $ with $ \mathcal N_0$ free or projective, and given
$ \chi_1\colon \mathcal N_0 \to \mathcal N_2$ and $ \chi_2 \colon \mathcal N_1 \to \mathcal N_2$ $\CO $-linear maps with $ \chi_2$ surjective,
construct an $ \CO$-linear map $\psi $ making the following diagram commutative:
$$ \begin{tikzcd} \mathcal N_0 \arrow[r, dotted,"\psi"]  \ar[dr,"\chi_1"']& \mathcal N_1 \ar[d,->>,"\chi_2"]\\ & \mathcal N_2  \end{tikzcd} .$$
\end{itemize}
For both problems above a solution exists.

\vspace{1mm}
\noindent
Recall that an $ \CO$-module resolution $ (\FM_\bullet, d)$ of $\CO/\CI $ is said to have \emph{finite length} if $\exists N \in \mathbb N$ s.t. $ \forall i > N$, $ \FM_i=0$.

\begin{theorem}
\label{th:finitelyMany}
For every  $ \CO$-resolution $(\FM_\bullet, d)$ of $\CO/\CI $ of finite length,  its arborescent Koszul-Tate resolutions can be computed by solving finitely many $\CO$-linear problems of the second type.
\end{theorem}
\begin{proof}
The recursion describing the construction of the arborescent operations $\psi $ such that $ \delta_\psi=0$ is given in Equation \eqref{eq:psi3}. It consists in solving an $ \CO$-linear problem  of the second type.
Let $N$ be the length of $(\FM_\bullet, d) $. For degree reasons, $ \psi_t(a_1, \dots, a_n)$ is zero if $t[a_1, \dots, a_n]$ is of degree greater than $N+1$. There are therefore finitely many trees for which $\psi_t \neq 0 $.
One therefore needs to solve finitely many $ \CO$-linear problems of the second type only.
\end{proof}

% at least for Syzygy algebras.
\vspace{1mm}
\noindent
We say that a Noetherian  algebra $\CO$ is a \emph{Syzygy algebra} if every finitely generated free $\CO$-module has a  finite length $\CO $-module resolution by finitely generated free   $\CO $-modules.
For example, by Hilbert's Syzygy theorem and its generalizations [13], polynomial rings in $N$ variables, germs at a point  of holomorphic or real analytic functions in $N$ variables, or formal power series in $N$ variables are all Syzygy algebras [38]. Moreover every finitely generated module over the previous Syzygy algebras admit free $ \CO$-module resolutions of length less than or equal to~$N$.

\begin{corollary}
\label{cor:finmanylinear}
Let $ \CO$ be a Syzygy algebra.
For every ideal $\CI \subset \CO$, one can construct an arborescent Koszul-Tate resolution of $ \CO/\CI$  by solving a finite number of $\mathcal O$-linear problems involving free $ \CO$-modules of finite ranks only.
\end{corollary}

\begin{proof}
The $n$ first terms of a free $ \CO$-module resolution of $\CO/\CI $ can be computed for all $n$ by solving finitely many $\mathcal O$-linear problems of the first type. Since $\CO$ is a Syzygy algebra, this free $\CO$-module resolution $(\FM_\bullet, d) $  can be chosen of finite length. The result then follows from Theorem~\ref{th:finitelyMany}.
\end{proof}

\noindent
If one restricts further to the polynomial algebra,  $ \CO=\mathbb K[x_1, \dots,x_n]$, one can obtain even a much stronger statement  than
Corollary \ref{cor:finmanylinear}.
$\CO $-linear problems of the first or the second type, with all $ \CO$-modules involved being of finite ranks, can be solved by an algorithm that uses only finitely many algebraic operations  $ +, - ,\times, / $. This follows from the theory of Gr\"obner basis and Schreyer’s theorem, see e.g. Theorems 2.4.2 and 2.4.4 in [1], or Chapter 15 in [13].

\begin{corollary}
\label{cor:finmanyoperations}
For any ideal $\CI $ of $\CO=\mathbb K[x_1, \dots,x_n]$, an arborescent Koszul-Tate resolution of $ \CO/\CI$ can be constructed through an algorithm that uses finitely many algebraic operations.
\end{corollary}

\vspace{.4cm}

\noindent
In contrast, the Tate algorithm requires solving infinitely many $\CO $-linear problems of the first type, even for a Syzygy or polynomial algebra. The next proposition and the discussion below make this statement precise. The ranks  of free $\CO $-modules $\CEb=\oplus_{i \geq 1} \CE_i$ entering a Koszul-Tate resolution of $\CO/\CI $ can not be arbitrarily small.
In Appendix \ref{app:reduced}, we give a lower bound to these ranks. %In Appendix \ref{app:minimal}, we even show these bounds are reached under some assumptions on $ \CO$ or $ \CI$.
%over $ \CO$ of $\CE_i $, where $ S( \CE)$ %I sent you the zoom link
It goes through the notion of a reduced complex $(\CEb\otimes \CO/\CI, \underline \delta)$ of $\CO/\CI $, which is a complex of free or projective $\CO/\CI $-modules.
We show that it is unique up to homotopy equivalence. Therefore, its
%denoted by $H(\CEb \otimes \CO/\CI, \underline \delta)$,
does not depend on the choice of a Koszul-Tate resolution.
The same holds for its homology of the complex of $\mathbb K $-vector spaces obtained by tensoring by $ \mathbb K \simeq \CO/\CJ$ with $ \CJ$ a maximal ideal containing $ \CI$. This  homology is a graded vector space of dimension $ b_i(\CO/\CJ,\CI) \in \mathbb N_0 \cap \{+\infty\}$ in degree $ i \geq 1$.  By Theorem \ref{thm:lowernboundonrank}, for any free $\CO $-module $\CEb=\oplus_{i \geq 1} \CE_i$ entering a Koszul-Tate resolution of $\CO/\CI $, the rank $\CE_i $ is greater or equal to $b_i(\CO/\CI,\CJ)$. This has consequences for the complexity of the Tate algorithm.

\begin{proposition}
\label{prop:TateIsInfinite}
If infinitely many of the integers $(b_i(\CO/\CI,\CJ))_{i \in \mathbb N}$ are non-zero for a maximal ideal $ \CJ$ containing $ \CI$, then the Tate algorithm requires to solve an infinite number of $ \CO$-linear problems of the first type.
\end{proposition}
\begin{proof}
Consider a Koszul-Tate resolution $ (S(\CE), \delta)$  coming from the Tate algorithm as in Construction \ref{const:tate.algorithm}.
Theorem \ref{thm:lowernboundonrank} implies that $\CE_i \neq 0$ for every $i \in \mathbb N$ such that $b_i(\CO/\CI,\CJ) \neq 0$.
When  $ \CE_i$ is non-zero, the $i$-th step of the Tate algorithm requires to solve an $ \CO$-linear problem of the first type, since the $i$-th step consists in finding generators of the kernel of $ \delta_{i-1}$.
%for every $i$ such that $ d_i \neq 0$.
\end{proof}

\vspace{0.1cm}

\noindent
Proposition \ref{prop:TateIsInfinite} is useful when one can compute the homology of the reduced complex of $\CO/\CI $ tensored with $\CO/\CJ $ for some maximal ideal $ \CJ$.
Arborescent Koszul-Tate resolutions allow precisely to make such computations, since it allows to describe explicitly a reduced complex of $ \CO/\CI$.

\vspace{0.1cm}

\noindent
We conclude this section by applying this strategy to monomial ideals of a polynomial algebra.

\begin{proposition}
\label{prop:nonregular}
Let $\CI \subset \CO = \mathbb K[x_1, \dots, x_n]$ be a monomial ideal minimally generated by a sequence $\phi_1, \dots, \phi_k $ which is not regular.
Then the Tate algorithm requires infinitely many algebraic operations.
\end{proposition}
\vspace{0.2cm}
\noindent
We start with a technical lemma.
\begin{lemma}
\label{lem:strictdivisor}
Let $\varphi_1, \dots, \varphi_k$ be a not a regular sequence of monomials, then there exist distinct $i ,j \in \{1, \dots, k\}$ such that $\mathrm{lcm}\lbrace \varphi_i, \varphi_j \rbrace \neq \varphi_i\cdot \varphi_j$.
\end{lemma}
\begin{proof}
Let $\varphi_1, \dots, \varphi_k$ be a not regular sequence in $\CO$. By definition, there exists $m\geq 2$ (case $m =1$ is excluded) such that $a\cdot \varphi_m = \alpha_1 \cdot \varphi_1 + \dots \alpha_{m-1}\cdot \varphi_{m-1}$, $a, \alpha_{\bullet} \in \CO$ and $a\notin \langle \varphi_1, \dots, \varphi_m \rangle_{\CO}$. Without any loss of generality we choose $a, \alpha_{\bullet}$ to be monomials and we impose that any non-zero summand $\alpha_i\varphi_i$ has the same multipolynomial degree as $a\varphi_m$. Then it easily follows if $\alpha_i\varphi_i \neq 0$ there exists $b\in \mathbb K^*$ such that $b\alpha_i\varphi_i = a\varphi_m$. Thus $\varphi_i$ divides  $a\varphi_m$. Since $\varphi_i$ does not divide $a$ alone according to the assumptions, the greatest common divisor of $\varphi_i$ and $\varphi_m$ is non-trivial and therefore their least common multiplier is not just given by their product. Therefore there exists a monomial $\chi \in \CJ $ such that $ \varphi_i\cdot \varphi_j=\chi \cdot \,\mathrm{lcm}\lbrace \varphi_i, \varphi_j \rbrace$.
\end{proof}

\vspace{0.2cm}
\noindent
Let $ \CJ \subset \CO$ be the maximal ideal generated by $x_1,\dots,x_n$. Notice that $ \CI \subset \CJ$.

\begin{lemma}
\label{lem:nonregular}
For every  $i \geq 0 $, we have $ b_{2i+1}(\CO/\CI,\CJ) \neq 0$.

\end{lemma}
\begin{proof}
By Lemma \ref{lem:strictdivisor}, there exists indices $i,j$ and a monomial $ \chi \in \CJ$ such that \begin{equation}\label{eq:explicitproductmonomials} e_{\{i\}} \star e_{\{j\}}= \chi \, e_{\{i,j\}}.\end{equation}

\vspace{1mm}
\noindent
Consider the arborescent Koszul-Tate resolution $(S(\CT[C(k)]), \delta_\psi)$, where the arborescent operations are chosen as in Proposition \ref{prop:psi.dga}.
In particular
\begin{equation}\label{eq:explicitproductmonomials2} \psi_\vee (e_{\{i\}} , e_{\{j\}})= \chi \, e_{\{i,j\}}\end{equation}
takes values in $\CJ \CT[C(k)]$.
Let us now examine the ordered decorated tree $T_m$ of degree $ 2m+1$
$$\adjustbox{valign=c}{$\begin{array}{c} \\T_m \, := \, \end{array}$}
\adjustbox{valign=c}{\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[,label = {[mystyle] \scalebox{2}{$ A_1 \quad$}}
[, edge = dashed,
[,label = {[mystyle] \scalebox{2}{$ A_m \quad$}}, edge = dashed,
[\scalebox{2}{$e_{\{i\}}$}, tier =1 ]
[\scalebox{2}{$e_{\{j\}}$}, tier =1]
[\scalebox{2}{$e_{\{j\}}$}, tier = 1 ]
\scalebox{2}{$e_{\{j\}}$}, tier =1
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!11.child anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt];
\end{forest}
A direct computation using Equation \eqref{eq:deltaexpression} gives:
$$\adjustbox{valign=c}{$\begin{array}{c} \\
\delta_\psi T_m = (-1)^m\end{array}$}
\adjustbox{valign=c}{\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[,label = {[mystyle] \scalebox{2}{$ A_1 \quad$}}
[, edge = dashed,
[,label = {[mystyle] \scalebox{2}{$ A_{m-1} \quad\quad $}}, edge = dashed,
[\scalebox{2}{$\psi_\vee\left(e_{\{i\}}, e_{\{j\}}\right)$}, tier =1 ]
[\scalebox{2}{$e_{\{j\}}$}, tier =1]
[\scalebox{2}{$e_{\{j\}}$}, tier = 1 ]
\scalebox{2}{$e_{\{j\}}$}, tier =1
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!11.child anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt];
\end{forest}
}} \adjustbox{valign=c}{$\begin{array}{c} \\+\end{array}$}\adjustbox{valign=c}{
\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[,label = {[mystyle] \scalebox{2}{$ A_1 \quad$}}
[, edge = dashed,
[,label = {[mystyle] \scalebox{2}{$ A_{m-1} \quad \quad$}}, edge = dashed,
[\scalebox{2}{$e_{\{i\}}$}, tier =1 ]
[\scalebox{2}{$e_{\{j\}}$}, tier =1]
[\scalebox{2}{$e_{\{j\}}$}, tier = 1 ]
\scalebox{2}{$e_{\{j\}}$}, tier =1
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!11.child anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt];
\end{forest}
}}\adjustbox{valign=c}{$\begin{array}{c} \\
\odot \end{array}$}\adjustbox{valign=c}{
\scalebox{1.4}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{0.714}{$e_{\{j\}}$}, tier =1]
\end{forest}} \\ \vspace{0.5cm}}\adjustbox{valign=c}{$\begin{array}{c} \\,\end{array}$}
all the remaining terms being zero either for symmetry reasons, or because the only non-vanishing arborescent operations $  \psi_t$ are those where $t$ is of the form given in Equation \eqref{eq:onerootnleaves}.
The first term on the r.h.s. is in $ \CJ \, \CT[C(k)]$ in view of Equation \eqref{eq:explicitproductmonomials2}.
Since the second term is in $\Sdeg (\CT[C(k)])$,
we have that $$\delta_\psi  (T_m ) \in \CJ \, \CT[C(k)]\oplus \Sdeg (\CT[C(k)]).$$
Therefore, $T_m$ induces a closed element $[T_m]$ in the reduced complex $(\CT[C(k)] \otimes \CO/\CJ, \underline{\delta_\psi})$.
Also, $[T_m]$ can not be exact: this follows from Equation \eqref{eq:deltaexpression}  since the image of $\underline{\delta_\psi}$ is made of linear combinations of trees which are either  non-binary trees, or trees with $d$-exact decorations, or trees with decorations in the image of $\psi_t$ for some tree $t$ as in Equation \eqref{eq:onerootnleaves} and therefore valued in  $C(k)_{\geq 2}$.
Since $ T_m$ has degree $ 2m+1$, we have $b_{2m+1}(\CO/\CI, \CJ) \neq 0$.
\end{proof}

\vspace{0.1cm}
\noindent Similarly to Lemma \ref{lem:nonregular}, one can also prove $b_{2i}(\CO/\CI,\CJ) \neq 0$, but since we do not need it for proving Proposition \ref{prop:nonregular}, we omit it here.

\vspace{0.1cm}
\noindent
\begin{proof}[Proof of Proposition \ref{prop:nonregular}.]
The statement is an immediate consequence of Proposition \ref{prop:TateIsInfinite} and Lemma \ref{lem:nonregular}.
\end{proof}

\begin{remark}
\normalfont
A statement similar
to Proposition \ref{prop:nonregular}
appears under the name "Rigidity Theorem" in Avramov's [4], where Koszul-Tate resolutions are denoted by $ \hat{S}$ but are not given a name.
\end{remark}

\subsection{A  word of conclusion about simplicity}

\vspace{0.1cm}

\noindent
We now have two main types of Koszul-Tate resolutions:  the arborescent Kozul-Tate resolutions obtained of a well-chosen free or projective $\CO$-module resolution, and those obtained out of the Tate algorithm. Hence a natural question: Which of the two is simpler?

\vspace{0.1cm}

\noindent
From the point of view of complexity, i.e. the quantity of computations, arborescent Koszul-Tate resolutions are significantly simpler, as can be seen by comparing, e.g., Theorem \ref{th:finitelyMany}  with Proposition \ref{prop:nonregular}.
In particular, they can often be computed using finitely many elementary operations, see, e.g. Corollary \ref{cor:finmanyoperations}.

\vspace{0.1cm}

\noindent
as we see in Appendix \ref{app:minimal}, if properly constructed, the Tate algorithm is minimal. It suffices to choose a minimal set of generators of the homology of the truncated complex $$H_k\left( S({\mathcal E}_{1}\oplus \dots \oplus  {\mathcal E}_{k}), \delta_k \right)$$ at each step of Construction \ref{const:tate.algorithm}, thus forming a "smallest possible" free $\CO$-module $\CE_{k+1}$.
In contrast, the arborescent Koszul-Tate resolution is almost never minimal.

\vspace{0.1cm}
\noindent
In conclusion, arborescent Koszul-Tate resolutions are simpler from the point of view of complexity, but not from the point of view of  size.

\vspace{0.1cm}
\noindent
Also, arborescent Koszul-Tate resolutions have easier homotopies.
Assume a module resolution $(\FM_\bullet,d) $  of $ \CO/\CI$ of length $N$ is given.

\vspace{0.1cm}
\noindent
both its arborescent Koszul-Tate resolutions and any Koszul-Tate resolution obtained by the Tate algorithm are homotopy equivalent to $(FM_\bullet,d) $. While in our construction this homotopy is completely explicit, see Proposition \ref{prop:Homotopy}, while in the second case,  the construction of such a homotopy equivalence would again need  infinitely many computations.

\vspace{0.1cm}
\noindent
Also, Proposition \ref{prop:Homotopy} gives explicit expression of a homotopy $h$ for $\delta_\psi $ upon restricting to a subspace of the kernel of ${\mathrm{Proj}}_\psi $, i.e.\ in particular to $\CT[\FM_\bullet]_{\geq N+1}$. This is very useful to have in applications like the BV or BFV constructions, see [18].

\section*{Acknowledgements}
\noindent A.H. and T.S. are thankful to Erwin Schrodinger Institute for a stay during a thematic programme "Higher structures and Field Theory",  where this work was partially conducted.
C.L.-G.  would like to thank the Tsing Hua University \begin{CJK*}{UTF8}{bsmi}國立清華大學\end{CJK*}  and the NCTS \begin{CJK*}{UTF8}{bsmi}國家理論科學研究中心\end{CJK*}  for their hospitality.
Last, all three authors thank the Institut Henri Poincaré for its hospitality in 2023.\\

\vspace{0.1cm}

\noindent
We thank Ruben Louis, Vladimir Salnikov, Rong Tang and, in particular, Barbara Fantechi, Bernhard Keller, Oleksii Kotov and Pavel Mnev for useful  discussions.

\appendix

\section{An ab initio proof of Theorem \ref{thm:ainfty}}

\label{app:alternative}
\noindent The purpose of this Appendix is to give a detailed proof of Theorem \ref{thm:ainfty} without using the homotopy transfer theorem. To be more in line with the rest of the paper, we keep unshifted  degrees.
%and use the following definition of an $ A_\infty$-algebra.
This requires to reformulate the defining properties of  $ A_\infty$ and $ C_\infty$ algebras correspondingly. For the case of the $ A_\infty$ algebra this then actually agrees with its original definition in the literature.
%new definition are well-known to be equivalent to the previous ones, using the so-called décalage isomorphism.

\begin{definition} [34, 35, 22].
Let $ (\mathfrak A,d)$ be a complex of $ \CO$-modules. A sequence of $\CO $-linear maps $\mu_i \colon \otimes^i C_\bullet \to C_{\bullet} $,
with $\mu_i $  of degree $i-2 $
is said to be a $ A_\infty$-algebra if it satisfies the "higher associativity" conditions for all $n \geq 1$:
\begin{equation}
\label{eq:ainfty2}
\sum_{\scalebox{0.5}{$\begin{array}{c} i,k \geq 0, j \geq 1 \\ i+j+k = n \end{array}$}}
\mu_{n-j+1}(\id^{\otimes i}\otimes \mu_{j} \otimes \id^{\otimes k}) = 0
\end{equation}
with the convention that $\mu_1=
d $.
An $A_\infty$-algebra structure $(\mu_n)_{n \geq 1} $  on a graded $ \CO$-module ${\mathfrak{A}}_\bullet $  is said to be a \emph{$ C_\infty$-algebra} if the following condition is satisfied: for every $ n \geq 1$ and all homogeneous $ a_1, \dots,a_n \in {\mathfrak A}_\bullet$, one has
\begin{equation}\label{eq:Cinfty2} \sum_{\sigma \in {\mathrm{Sh}}(i,n)}
\theta(\sigma,a)  \mu_{n}(a_{\sigma(1)}, \dots, a_{\sigma(n)})  =0 .\end{equation}
Here ${\mathrm{Sh}}(i,n)$ stands for the set of all $(i,n)$-shuffles of $\{1, \dots, n\} $, i.e. all permutations $ \sigma$ of $ \{1, \dots,n\}$ such that  $ \sigma(a) \leq \sigma (b) $ if   $1 \leq a \leq b \leq i$ or $i+1 \leq a \leq b \leq n$.
$e(\sigma) \in \{-1,1\}$ is  the ordinary signature of $ \sigma$ and
$\theta(\sigma,a) $ is the Koszul signature of $\sigma$  defined to be the unique sign that satisfies
$$ a_{\sigma(1)} \odot \dots \odot a_{\sigma(n)} = \theta(\sigma,a) \, a_1 \odot \dots \odot a_n  $$
within the graded symmetric algebra $ S(\mathfrak A)$  generated by $\mathfrak A $.
\end{definition}

\noindent
For the non-shifted degree,  theorem  \ref{thm:ainfty} becomes:

\begin{theorem}
\label{thm:ainfty2}
Consider an arborescent Koszul-Tate resolution of $\CO/\CI $ associated to a free or projective $\CO$-module resolution $(\FM_\bullet,d) $ with arborescent operations $ \psi$  and differential  $\delta_\psi $.
Then the following structure maps $(\mu_n)_{n\geq 1} $  define an $A_\infty $-algebra on $ \FM_\bullet \oplus \CO$:

\begin{enumerate}
\item For $n=1 $, we set $ \mu_1=
\item For $ n=2$, we set for all $ a_1,a_2 \in \FM_\bullet$ and $ F_1,F_2 \in \CO$  $$\begin{array}{rrcl}
\mu_2\colon & \otimes^2  \left( {\FM}_\bullet \oplus \CO \right) & \rightarrow &  {\FM}_\bullet  \oplus \CO \\
& (a_1 + F_1 \, ,  \, a_2+F_2) & \mapsto & F_1a_2+ F_2 a_1  +\psi_\vee (a_1,a_2) + F_1 F_2
\end{array} .$$
\item For all $n \geq 3$, we set for all homogeneous $ a_1, \dots, a_n \in \FM_\bullet$ and $F_1, \dots, F_n \in\CO$
\begin{equation}
\label{eq:homotopytransfer2} \begin{array}{rrcl}
\mu_n\colon & \otimes^n  \left( {\FM}_\bullet \oplus \CO\right) & \rightarrow & {\FM}_\bullet  \oplus  \CO \\
& (a_1+F_1, \dots, a_n+F_n) & \mapsto & \sum_{ t \in Y_n} (-1)^{P(t[a_1, \dots,a_n])
\, \,  \psi_t(a_1, \dots, a_n),
\end{array}
\end{equation}
where $Y_n$ is the set of ordered binary trees with $n$ leaves, and
$P(t[a_1, \dots,a_n])$ is the sum of the degrees of all left subtrees
of vertices in $t[a_1, \dots,a_n]$.
\end{enumerate}
{ Moreover, the structure maps $(\mu_n)_{n \geq 1} $ satisfy \eqref{eq:Cinfty2}, proving that one even has a $C_\infty$-structure.
\end{theorem}

\vspace{0.1cm}
\noindent
Let us explain  the third item.
Let $t[a]$ be an ordered decorated tree with $k$ leaves where $a$ denotes a list of $k$ elements in  $\FM_\bullet$.  A strict subtree $t_1[b]$, with $ b$ a sublist of $a$,  can be located to  the left or to the right
of a vertex $A$ of $t$:
For instance, in the following ordered binary tree, both $ t_2[c] $ and $t_3[e] $ have  the vertex $ A$ as parent of their roots, but only $ t_2[c]$ is located to the left of $A$. %Also, $ t_1[b]$ is attached to the left of the vertex of $t$ to which is it attached, but the subtree of all descendants of $ A$ is not.
\begin{equation}
\label{eq:exampleRightLeft}
\adjustbox{valign =c}{ $\begin{array}{c} t[a] =\\ \\ \\ \\ \\  \end{array}$\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[ , edge = dashed
[\scalebox{2}{$t_1[b]$}, tier = 1 ]
[,label = {[mystyle] \scalebox{2}{$A\enspace$}},
[\scalebox{2}{$t_{2}[c]$}, tier =1]
[\scalebox{2}{$t_3[e]$}, tier =1]
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt]
(!12.child anchor) circle[radius=3pt];
\end{forest}
\end{equation}
\vspace{-10mm}

\noindent As an example, let us describe  $\mu_3 $ with its signs.
We have $ Y_3=\{\ltree,\rtree\}$.
For the decorated tree $\ltree[a_1,a_2,a_3] $, the left subtrees are the subtrees $| \otimes a_1 $  and $ \vtree[a_1,a_2]$. Hence $ P(\ltree[a_1,a_2,a_3])=|a_2|+1$.
For the tree $\rtree[a_1,a_2,a_3] $, the left subtrees are $|\otimes a_1$ and $|\otimes a_2$, so that
$ P(\rtree[a_1,a_2,a_3])=|a_1|+|a_2|$.
One then computes
\begin{equation}
\label{eq:mu3}
\mu_3(a_1,a_2,a_3)= - \psi_{\ltree}(a_1,a_2,a_3) + (-1)^{|a_1|}\psi_{\rtree}(a_1,a_2,a_3)  .
\end{equation}
(Notice that this is consistent with Equation \eqref{eq:m3}).

\vspace{0.1cm}
\noindent
%(we consider the trivial tree as a binary tree).
Also  the set $Y_n$ of ordered binary trees with  $n$ leaves:
$$Y_1 = \lbrace \vert \rbrace, \enspace Y_2 = \left\{ \begin{array}{c}\scalebox{0.55}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes, for tree={ inner sep=0 pt, s sep= 0 pt, fit=band,
[[, tier =1] [, tier= 1]]
\path[fill=black] (.parent anchor) circle[radius=2pt];
\end{forest}} \end{array} \right\}, \quad
Y_3 = \left\{ \begin{array}{c}\scalebox{0.3}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[, tier =1]
[, tier = 1]
[, tier =1]
\path[fill=black] (.parent anchor) circle[radius=2pt]
(!1.child anchor) circle[radius=2pt];
\end{forest} }
\scalebox{0.3}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[, tier =1]
[, tier =1]
[, tier = 1]
\path[fill=black] (.parent anchor) circle[radius=2pt]
(!2.child anchor) circle[radius=2pt];
\end{forest}}
\end{array}\right\}, \cdots
is described recursively [26] by:
\begin{equation}\label{eq:recYn}\quad Y_n =  \left\{ \begin{array}{c}  \scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{2}{$Y_i$}, tier =1]
[\scalebox{2}{$Y_k$}, tier =1]
\path[fill=black] (.parent anchor) circle[radius=4pt];
\end{forest}}\\ \end{array}, \, i+k =n \right\} \simeq \bigsqcup_{i+k=n}Y_i\times Y_k.\end{equation}

\vspace{1mm}
\noindent
Before proving Theorem \ref{thm:ainfty2}, some considerations about ordered binary trees are required.
For any homogeneous $ a_1, \dots, a_n \in \FM_\bullet \oplus \CO$,
we define an element $ k_n[a_1, \dots, a_n] \in \
Tree[\FMb]$ by the recursive formula
\begin{equation}
\label{eq:defkn}
k_n[a_1, \dots,a_n] = \sum_{j=1}^{n-1} (-1)^{|k_j[a_1, \dots, a_j]|} \r \left( k_j[a_1, \dots, a_j] \otimes k_{n-j}[a_{j+1}, \dots, a_n]  \right)
\end{equation}
with initial condition $ k_1(a_1)= |[a_1]$ for all $ a_1\in \FMb$. Here $\r $ is the root map defined on ordered decorated trees, and $|\cdot|$ is the degree of a decorated tree as defined in Section \ref{sec:space}.
Equation \eqref{eq:defkn} may be also depicted graphically as follows:
\adjustbox{valign =c}{
k_n[a] =\sum_{j = 1}^{n-1} (-1)^{|k_j[a_{\leq j}]|}$}
\adjustbox{valign =c}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[$k_j{[a_{\leq j}]}$, tier =1]
[$k_{m-j} {[a_{>j}]}$, tier =1]
\path[fill=black] (.parent anchor) circle[radius=2pt];
\end{forest}}
%\adjustbox{valign =c}{ }
where now $ a= a_1, \dots, a_n$,  $ a_{\leq j} =a_1, \dots, a_j$, and  $a_{>j} = a_{j+1}, \dots, a_n$.
\vspace{1mm}
\noindent

%so that $|k_j[a_1, \dots, a_j]|$ is well-defined.

\begin{example}
\normalfont
Let us compute the first terms. For $n=2,3$, we obtain:
\adjustbox{valign =c}{$k_2[a_1, a_2] = (-1)^{|a_1|}$}
\adjustbox{valign =c}{\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[$a_1$, tier =1]
[$a_2$, tier =1]
\path[fill=black] (.parent anchor) circle[radius=2pt];
\end{forest}}

\begin{equation}
\label{eq:k3}
\adjustbox{valign =c}{$\quad k_3[a_1, a_2, a_3] = (-1)^{|a_1| + |a_2|}$}
\adjustbox{valign =c}{
\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{2}{$a_1$}, tier =1]
[\scalebox{2}{$a_2$}, tier =1]
[\scalebox{2}{$a_3$}, tier = 1]
\path[fill=black] (.parent anchor) circle[radius=4pt]
(!2.child anchor) circle[radius=4pt];
\end{forest}}
\adjustbox{valign =c}{$
(-1)^{2|a_1| +|a_2| +1}$}
\adjustbox{valign =c}{
\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[\scalebox{2}{$a_1$}, tier =1]
[\scalebox{2}{$a_2$}, tier =1]
[\scalebox{2}{$a_3$}, tier = 1]
\path[fill=black] (.parent anchor) circle[radius=4pt]
(!1.child anchor) circle[radius=4pt];
\end{forest}}}.
\end{equation}
\end{example}

\vspace{1mm}\noindent We intend to construct $ \mu_n(a_1,\dots,a_n)$ out of  $ \psi$ and the fixed element $k_n[a_1, \dots,a_n]$, see Equation \eqref{eq:defmuoutofk} below. For this purpose it is  useful to first study some properties of $k_n[a_1, \dots, a_n]$:
\begin{lemma}
\label{lem:knexplicit}
For every $n \geq 1$, and every $a=(a_1, \dots,a_n) \in (\FM_\bullet)^n$,
$k_n[a]$ has degree $\sum_{i=1}^n |a_i| + n-1 $. It is
a linear combination  of all  ordered binary trees with $n$ leaves decorated with $ a_1, \dots, a_n$. The prefactor of all terms of this linear combination takes values in $\{ +1,-1\}$ and   explicitly one has:
\begin{equation}
\label{eq:kmExpl} k_n [a] = \sum_{t \in Y_n} (-1)^{P(t[a])} \,t[a]
\end{equation}
where $P(t[a])$ is the same integer that was defined in Theorem \ref{thm:ainfty2} and the text following it.
\end{lemma}
%Let us clarify the sign expression $P(t[a])$.

\begin{proof} Equation \eqref{eq:recYn}
implies that all $ t[a]$, with $t \in Y_n$, appears once and only once in the expression of $ k_n[a]$ given by the recurrence relation \eqref{eq:defkn}.
%[Proof of Lemma \ref{lem:knexplicit}]
%The formula giving the degree can be proven by recursion.
This proves that $k_n[a] $ is a linear combination of all ordered  binary trees
decorated by $ a_1, \dots,a_n$. Being binary, such trees always have $n-1 $ vertices, which thus leads to the claimed degree of $ k_n[a]$.

\vspace{1mm}\noindent
Let us compute the exact value of the prefactors.
We denote by
$\mathcal C_l(t[a])$
list of
all left subtrees of  an ordered tree $t[a] $.  For example,  in the tree $t[a]$ in Equation \eqref{eq:exampleRightLeft}, the subtrees $ t_1[b]$ and $t_2[c] $ would belong to that list (completed by the contributions of all strict left subtrees of $ t_1[b]$, $t_2[c] $,  and $t_3[e] $).  For all ordered binary trees $ t_1$ and  $t_2$ decorated by $b$ and $c$, respectively, one has
$$ \mathcal C_l( \r (t_1 [b] \otimes t_2[c] )) = \mathcal C_l(t_1[b]) \sqcup \mathcal C_l(t_2[c]) \sqcup \{t_1[b]\}.  $$
Since $P(t[a]) $ is the sum of the degrees of the trees in $ \mathcal C_l(t[a])$, we find \begin{equation}\label{eq:Pn} P(\r (t_1 [b] \otimes t_{2}[ c] )) =
P(t_1[b] )+
P (t_2 [c] ) + |t_1[b]|.
\end{equation}
In turn, this relation allows to check that the linear combination given at
the right hand side of \eqref{eq:Pn}
satisfies the  recursion relation  Equation \eqref{eq:defkn}. By uniqueness, both expressions coincide. This completes the proof.
\end{proof}

\vspace{.3cm}

%Another property of $k_n$ is that it lies in the kernel of $\partial$ (see Section \ref{} ), i.e.
\begin{lemma}
\label{lem:a_infty1}
Each element of $\lbrace k_n[a] \rbrace$ lies in  $\ker(\partial)$, i.e. for all $ n \geq 1$:
$$   \partial k_n[a] =0,
where $\partial $
is the differential on $Tree[\FMb]$ defined in Section \ref{sec:constructingdeltapsi}. \end{lemma}
\begin{proof}
Let $t[a] $ be a binary decorated tree with $n$ leaves, $a$ a list of $n$ homogeneous elements in $ \FM_\bullet$, and $A$ an inner vertex of $ t$.
For every pair $(t[a],A)$ there is a dual one, $ \overline{(t[a],A)} =(\bar{t}_A[a],\bar{A})$, constructed as follows. Assume that $A$ is the root of the right subtree of its parent $P_A $ as in the following picture,
\begin{equation}
\label{tree:Aright}
\adjustbox{valign =c}{$t{[a]} =$}
\adjustbox{valign =c}{ \scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[ ,label = {[mystyle] \scalebox{2}{$P_A \enspace$}}, edge = dashed
[\scalebox{2}{$t_1{[b]}$}, tier = 1 ]
[,label = {[mystyle] \scalebox{2}{$A \enspace$}},
[\scalebox{2}{$t_2{[c]}$}, tier =1]
[\scalebox{2}{$t_3{[e]}$}, tier =1]
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt]
(!12.child anchor) circle[radius=3pt];
\end{forest}}
\end{equation}
where $t_1[b], t_2[c], t_3[d]$ are  subtrees of $ t[a]$. Then we define $\overline{ (t[a],A)}=(\bar{t}_A[a],\bar{A})$  as follows:
%that differs from $T$ by the location of $A$:

\begin{equation}
\label{tree:Aleft}
\adjustbox{valign =c}{$\bar{t}_A[a] =$}
\adjustbox{valign =c}{\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[ ,label = {[mystyle] \scalebox{2}{$P_{\bar A} \enspace \enspace $}}, edge = dashed
[,label = {[mystyle] \scalebox{2}{$\bar A \enspace \enspace$}},
[\scalebox{2}{$t_1{[b]}$}, tier =1]
[\scalebox{2}{$t_2{[c]}$}, tier =1]
[\scalebox{2}{$t_3{[e]}$}, tier = 1 ]
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt]
(!11.child anchor) circle[radius=3pt];
\end{forest}
\end{equation}
with the rest of the tree $\overline{t}_A $  being identical to the corresponding part in $t$. We then complete the description of the duality by imposing that it has to be an involution.
Notice that no pair $(t[a],A) $ is dual to itself.

\vspace{0.1cm}
\noindent
%Fix $ a= (a_1, \dots,a_n) \in \FM_\bullet^n$.
In view of the explicit description of $ \partial $ in
Equation \eqref{eq:differential_on_trees}
and the explicit description of $k_n[a] $ in
Equation \eqref{eq:defkn}, we see that
%Lemma \ref{lem:knexplicit}
$\partial k_n[a]$ is given by
%for all $m \geq 2$, $\partial k_m[a] $ is a sum over
%pairs $(T,A) $, with $T \in Y_n$ and $ A$ an inner vertex of~$T$:
\begin{equation}
\label{eq:deltaasdoublesum}
\partial k_m[a] = \sum_{(t[a],A) } (-1)^{W_A+P(t[a])}  \partial_A (t[a]) \: .
\end{equation}
Here the sum runs over all pairs $ (t[a],A)$, where $t \in Y_n$ is an ordered decorated tree with $n$ leaves, and $A$ is an inner vertex of $t$.

\vspace{0.2cm}
\noindent
Note that in the above sum, for every pair $(t[a],A)$ there is also a contribution from its dual pair  $\overline{ (t[a],A)}=(\bar{t}_A,\bar A))$.
We will now show that they cancel against each other. Note first  that
%these two contributions give the same tree, i.e. $\partial_A(t[a])=\partial_{\bar{A}}(\bar t_A[a]) $. The later tree is a binary tree except for one inner vertex from which three subtrees descend, and is obtained by erasing the edges going from $A$ and $ \bar{A}$ to their parents:
\partial_A(t[a])=\adjustbox{valign =c}{\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes,
[ ,label = {[mystyle] \scalebox{2}{$A \mapsto P_{A} \enspace \enspace \quad \quad $}}, edge = dashed,
[\scalebox{2}{$t_1{[b]}$}, tier =1]
[\scalebox{2}{$t_2{[c]}$}, tier =1]
[\scalebox{2}{$t_3{[e]}$}, tier = 1 ]
\path[fill=black] (.parent anchor) circle[radius=3pt]
(!1.child anchor) circle[radius=3pt];
\end{forest}
}} =\partial_{\bar{A}}(\bar t_A [a]  )
where we used the pictorial description in Equations \eqref{tree:Aleft} and \eqref{tree:Aright}.
%which becomes a three with three descendents.
We are therefore just left with showing that
\begin{equation}\label{eq:oppositesigns} (-1)^{P(t[a])+W_A} = -(-1)^{ P(\bar t_A[a])+W_{\bar A}}.\end{equation}
%the two prefactors $(-1)^{P(t[a])+W_A}$ and are opposite.
%Since these prefactors are valued , we just have to check that they are opposite.
Recall that the integer $P(t[a])$
%and $P(\overline t_A[a])$
is obtained by summing up all the degrees of the left subtrees of every vertex in $t$, its root included.
%over vertices in $t$, root included, all the degrees of all subtrees located to the left of their parents.
But the degrees of these subtrees coincide for both trees $t[a] $ and $\bar{t}_A[a] $, except for the ones that we now describe. The subtree $ t_2[c]$  only appears in the tree \eqref{tree:Aright} and not in the tree \eqref{tree:Aleft}, while the tree $ \r(t_1[b],t_2[c])$  appears as a subtree of the tree \eqref{tree:Aleft} but not of the tree $ \eqref{tree:Aright}$. The degree of these trees are $|t_2[c]| $ and $|t_1[b]|+|t_2[c]| +1 $, respectively, which yields

$$(-1)^{P(t[a])}= -(-1)^{P(\bar{t}_A[a])} (-1)^{|t_2[c]|}. $$
Likewise, the weight $ W_A$ is the sum of the degrees of all trees on the left of the path from the root to $A$, to which the length of that path is added.
Comparing $W_A$ and $ W_{\bar A}$ as in the trees \eqref{tree:Aleft} and \eqref{tree:Aright}, one easily finds
$$  (-1)^{W_A}=(-1)^{W_{\bar{A}}} (-1)^{|t_2[c]|}.$$
%respectively given by
%${|T_1[b]|}$ and $ 0$.
This then shows that the equality in Equation
\eqref{eq:oppositesigns}  holds true.

\end{proof}

\begin{lemma}
\label{lem:aboutmuexplicit}
The following relation holds:
\begin{equation}\label{eq:meaningmu}
\begin{array}{ll}&(-1)^{i+jk}\mu_{n-j+1}(\id^{\otimes i}\otimes \mu_{j} \otimes \id^{\otimes k})  (a_1, \dots, a_n)\\[2pt]=&
\sum\limits_{(t_1,t_2) \in Y_{n-j+1} \times Y_{j}}  \, \epsilon_{t_1,t_2}  \,   \psi_{t_1}(a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n)
\end{array}\end{equation}
\begin{equation}\label{eq:defepsilon} \begin{array}{rcl}\epsilon_{t_1,t_2} &=&  \eta \,  (-1)^{i +\sum_{r=1}^{i} |a_{r}| +P(t_2[a_{i+1},
\dots,a_{i+j}])+ P(t_1[a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n])}

\end{array} \end{equation}
with $ \eta =(-1)^{\sum_{r=1}^n |a_r|(n-r)}$.
\end{lemma}
\begin{proof}
A direct computation gives
$$\begin{array}{ll}&(-1)^{i+jk}\mu_{n-j+1}(\id^{\otimes i}\otimes \mu_{j} \otimes \id^{\otimes k})  (a_1, \dots, a_n)\\[2pt]=&(-1)^{i+jk +(j-2)(|a_1|+\dots + |a_i|)}\mu_{n-j+1}(a_{1}, \dots, a_i, \mu_{j}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1}, \dots, a_{i+j+k}) \\[2pt]=&
\sum\limits_{(t_1,t_2) \in Y_{n-j+1} \times Y_{j}}  \, \epsilon_{t_1,t_2}  \,   \psi_{t_1}(a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n)
\end{array}
Here we used $ |\mu_j| = j-2$ and the Koszul sign convention when exchanging the order of elements in $\FMb$  to go from the first to the second line. To go from the second to the third line one makes use of the  definition of $ \mu_\bullet$ given in Equation \eqref{eq:homotopytransfer2}.
Furthermore, %we used the abbreviation
$\epsilon_{t_1,t_2} $ is given  by the following complicated but explicit expression:
$$ \begin{array}{rcll}\epsilon_{t_1,t_2} &=& & (-1)^{i+jk +(\sum_{r=1}^i|a_i|)(j-2)}\\
& & \times &(-1)^{P(t_2[a_{i+1},
\dots,a_{i+j}])+ \sum_{r=i+1}^{i+j} |a_{r}|(i+j-r)} \\
& & \times & (-1)^{P(t_1[a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n])} \\
& & \times& (-1)^{  \sum_{r=1}^{i} |a_{r}| (n+1-j-r)+ |\psi_{t_2}[a_{i+1}, \dots, a_{i+j}]|(n-i-j)+\sum_{r=i+j+1}^{n} |a_{r}| (n-r)   } .\end{array}$$
Using $ |\psi_{t_2}(a_{i+1, \dots, a_{i+j}})| =  \sum_{r=i+1}^{i+j} |a_r| + j-2 $  and $ n-i-j=k$, this gives, by regrouping  terms in an appropriate manner:
\begin{array}{rcll} \epsilon_{t_1,t_2} &=& & (-1)^{i+jk }\\
& & \times &(-1)^{P(t_2[a_{i+1},
\dots,a_{i+j}])+
P(t_1[a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n])} \\ & & \times & (-1)^{\sum_{r=1}^{i} |a_{r}| }
\\ & & \times&  (-1)^{\sum_{r=1}^{i} |a_{r}| (n-r)+ \sum_{r=i+1}^{i+j}|a_r|(n-r)+(2-j)k+\sum_{r=i+j+1}^{n} |a_{r}| (n-r)   } \\
&=& & (-1)^{i}\\
& & \times &(-1)^{P(t_2[a_{i+1},
\dots,a_{i+j}])+ P(t_1[a_1, \dots, a_i  ,
\psi_{t_2}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n])} \\ & & \times & (-1)^{\sum_{r=1}^{i} |a_{r}| }
\\ & & \times&  (-1)^{\sum_{r=1}^{n} |a_{r}| (n-r)}
.\end{array} $$
This is the desired expression.
\end{proof}

%Let us prove
%Theorem \ref{thm:ainfty}.

\begin{lemma}
\label{lem:explicitrelations}
Let $ t[a_1, \dots,a_n]$ be an ordered decorated tree, $A$ be a vertex of $t$, and $ a_{i+1}, \dots,a_{i+j}$ the decorations that descend from $A$. We have
\begin{equation*}  \begin{array}{rcl}
W_{A} &=& i + \sum_{r=1}^i|a_r| + b_A \\
P(t[a]) &=& P(t_{\uparrow A}[ a_{i+1}, \dots, a_{i+j}]) + P(t_{\downarrow A}[a_{1}, \dots, a_{i}, \psi_{t_2}(a_{i+1}, \dots, a_{i+j}),a_{i+j+1}, \dots, a_{n} ])+ b_A.
\end{array}
\end{equation*}
Here $b_A$ is the number of vertices $B$ of $t[a]$ such that $A$ belongs to the left subtree of $B$.
\end{lemma}
\noindent For the tree in Equation \eqref{eq:AB}, $b_A=1$ since the vertex $B$ is the only one that satisfies the condition above. Furthermore, $ b_B=b_R=0$.
\begin{proof}[Proof of Lemma \ref{lem:explicitrelations}]
We leave the first relation to the reader and only prove the second one.
Let us use the shorthand $a$ for the list $ a_1,\dots,a_n$ and $a_A$ for the list $a_{i+1}, \dots, a_{i+j} $.
Moreover, we denote the vertices of $t$ which are not a leaf by
${\mathrm{RVer}}(t)$, i.e.
{\mathrm{RVer}}(t) =
\{ {\mathrm{Root}}(t)\} \, \cup  \,   {\mathrm{InnVer}}(t), $$ and the left subtree of $B \in {\mathrm{RVer}}(t) $ by $\ell(B,t[a]) $. For the trees $t_{\uparrow A} $ and $t_{\downarrow A} $, we use the notations correspondingly.
A vertex $B \in {\mathrm{RVer}}(t)$ can be either in $ {\mathrm{RVer}}(t_{\uparrow A})$, or in ${\mathrm{RVer}}(t_{\downarrow A}) $. In the later case, $A$ either belongs to the left tree of $B$ or  not. Since by definition
$P(t[a])= \sum_{B \in {\mathrm{RVer}}(t) } |\ell(B,t[a])|$, we have
\begin{equation}\label{eq:firstone0}  P(t[a])= \left( \sum_{B \in {\mathrm{RVer}}(t_{\uparrow A}[a])}  + \sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \notin \ell(B,t[a]) \end{array}$ }} +\sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \in \ell(B,t[a]) \end{array}$ }}   \right) |\ell(B,t[a])| .\end{equation}
For  $B\in {\mathrm{RVer}}(t_{\uparrow A}) $, the left subtrees $ \ell(B,t[a])$ and $\ell(B,t_{\uparrow A}[a_A] ) $ coincide and
\begin{equation}\label{eq:firstone1}     \sum_{B \in {\mathrm{RVer}}(t_{\uparrow A})}|\ell(B,t[a])|=\sum_{B \in {\mathrm{RVer}}(t_{\uparrow A})}  |\ell(B,t_{\uparrow A}[a_A])|=P(t_{\uparrow}[a]).\end{equation}
For a vertex $B \in {\mathrm{RVer}}(t_{\downarrow A}) $ such that $A$ does not belong to  $\ell(B,t[a])$, the left subtrees
$ \ell(B,t[a]) $ and $\ell\left(B,t_{\downarrow A}[a_1, \dots,a_i,\psi_{t_{\uparrow}}(a_A), a_{i+j+1}, \dots, a_n] \right) $ coincide, and
\begin{equation}\label{eq:firstone2}  \sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \notin \ell(B,t[a]) \end{array}$ }}   |\ell(B,t[a])|   = \sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \notin \ell(B,t[a]) \end{array}$ }}  |\ell(B,t_{\downarrow A}[a_1, \dots,a_i,\psi_{t_{\uparrow}}(a_A), a_{i+j+1}, \dots, a_n] )|  . \end{equation}
If $A$ belongs to the left subtree of $B$ in $ t[a]$, then  $\ell(B,t_{\downarrow A}[a_1, \dots,a_i,\psi_{t_{\uparrow}}(a_A), a_{i+j+1}, \dots, a_n] ) $ is obtained out of $ \ell(B,t[a])$ by replacing the subtree $t_{\uparrow A}(a_A)$  by a new leaf with decoration $ \psi_t[a_{i+1}, \dots, a_{i+j}]$. This substitution reduces the degree by
one.  Since there are $b_A$ such vertices,  one obtains
\begin{equation}\label{eq:firstone3} \sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \in \ell(B,t[a]) \end{array} $}}   |\ell(B,t[a])|   = \sum_{\scalebox{0.5}{ $\begin{array}{c}B \in {\mathrm{RVer}}(t_{\downarrow A}[a]) \\ A \in \ell(B,t[a]) \end{array} $}}  |\ell(B,t[a_1, \dots,a_i,\psi_{t_{\uparrow}}(a_A), a_{i+j+1}, \dots, a_n] )|  +b_A. \end{equation}
\eqref{eq:firstone1}, \eqref{eq:firstone2}, and \eqref{eq:firstone3}
give an expression for each one of the three terms that appear in
\eqref{eq:firstone0}. When combined, this gives the desired result.
% \begin{equation}\label{eq:firstone3} }.
\end{proof}
%where $\eta$ is as in Lemma\ref{lem:aboutmuexplicit}.

\begin{proof}[Proof of Theorem \ref{thm:ainfty2}]
We have to check that the collection of the $\CO$-linear maps
(\mu_n)_{n \geq 1}
%\FM^{\otimes n}\longrightarrow \FM, \, a_1\otimes \dots \otimes a_n \mapsto (-1)^{\sum_{i=1}^n|a_i|(n-i)}\psi(k_n[a_1, \dots a_n]), \, n>1
$ as in Theorem \ref{thm:ainfty2}
satisfy the axioms of an $A_\infty$-algebra given in Equation \eqref{eq:ainfty2}.
For $n=1$, this condition simply means that $ d^2=0$, which is true. For $n=2$, it means that $\mu_2 $ is a chain map, which one checks by a direct calculation.

\vspace{2mm} \noindent
In order to check it for all $n \geq 3$, we first notice that
\begin{equation}
\label{eq:defmuoutofk}\mu_n (a_1, \dots , a_n )  (-1)^{\sum_{i=1}^n|a_i|(n-i)}(\psi\circ \mathrm{pr})(k_n[a_1, \dots, a_n]),\end{equation}
where $\psi$ is the chosen arborescent operation and $\mathrm{pr} $ is the projection from ordered to symmetrized trees as in Equation \eqref{eq:psitDef}.
To show this, we  use the properties of $k_n[a_1, \dots, a_n] $, i.e Lemmas \ref{lem:knexplicit} and \ref{lem:a_infty1}, and of $\psi$, i.e. Equation \eqref{eq:deltaPsiSquare}.

\vspace{0.1cm}

\noindent
%Let us assume $ n \geq 3$ and
Assume that the axioms Equation \eqref{eq:ainfty2} are satisfied up to order $n-1$.
%that $\mu_k(a_1, \dots,a_k)=0 $ for all $k \geq 3 $:
%Let us check it at order $n$.
First let us verify the relation \eqref{eq:ainfty2}  when applied to an $n$-tuple $ a_1 , \dots , a_n \in \FM_\bullet \oplus \CO$ where for at least one index $l$ we have $a_l=F \in \CO$.
%Convention \ref{conv:extensionO} implies that
By definition, we have $\mu_k(a_{i_1}, \dots, a_{i_k}) =0$ for all $ k \geq 3$ provided one of the indices $ i_1, \dots, i_k$ is equal to $ l$.
There are only two non-zero contributions in \eqref{eq:ainfty2}.
When $ l $ is neither $1$ nor $n$, the two non-zero contributions  are
\begin{align*}
(-1)^{l-1}\mu_{n-1}(a_1, \dots,  a_{l-1}, \mu_2(F, a_{l+1}), \dots, a_n )+ (-1)^{l-2}\mu_{n-1}(a_1, \dots,  a_{l-2}, \mu_2(a_{l-1}, F), \dots, a_n )   .
\end{align*}
They add up to zero since
$ \mu_2(F,b+G)=F (b+G)=\mu_2(b+G,F)$ for every $b \in \FM_\bullet, G \in \CO $ and because $ \mu_{n-1}$ is $ \CO$-linear.
For $l = 1$ and $l=n$, the two non-zero contributions are
\mu_{n-1}(\mu_2(F, a_2), a_3, \dots, a_n )- \mu_2(F, \mu_{n-1}(a_2, \dots, a_n)),
$$  (-1)^{n-2}\mu_{n-1}(a_1, \dots, \mu_2(a_{n-1}, F) )+(-1)^{n-1} \mu_2( \mu_{n-1}(a_1, \dots, a_{n-1}), F),
respectively. In both cases,  the two terms add up to zero for the same reasons as  before.

%Since $\mu_2(b, F) = Fb$ for all $b \in \FMb $ and $\mu_{n-1}$ is $\CO$-linear,  the relation holds true also in this case.

\vspace{0.1cm}
\noindent
It remains to check that Equation \eqref{eq:ainfty2}  holds when each one of the elements $a_1 , \dots , a_n \in \FMb$ is homogeneous.
Since $ \psi$ is an arborescent operation such that  $ \delta_\psi^2=0$, Equation \eqref{eq:psi4} holds for every ordered tree $t$ with $n$ leaves.
We sum  Equation \eqref{eq:psi4}  over all binary ordered trees $t \in Y_n$, with the prefactor $ (-1)^{P(t[a])}$ defined in Lemma \ref{lem:knexplicit}:
\begin{equation}
\label{eq:sumup}\sum_{t \in Y_n}(-1)^{P(t[a])}  \sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{Ver}(t)\end{array}$}} \psi_{t_{\downarrow A}} \circ_{A} \psi_ {t_{\uparrow A}} (a_1, \dots, a_n)=
\sum_{t \in Y_n}(-1)^{P(t[a])} \sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{InnVer}(t)\\  \end{array}$}}  (-1)^{W_A} \psi_{\partial_A t} (a_1, \dots, a_n).
\     \end{equation}
The right-hand side of Equation \eqref{eq:sumup} vanishes for the following reason:
\begin{equation*}
\begin{array}{ll}
&\sum_{t \in Y_n}(-1)^{P(t[a])} \sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{InnVer}(t)   \end{array}$}}  (-1)^{W_A}\psi_{\partial_A t} (a_1, \dots, a_n)    \\ =& \psi \circ {\mathrm{pr}} \left( \sum_{t \in Y_n}\sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{InnVer}(t)\\ \end{array}$}} (-1)^{P(t[a]) + W_A} \partial_A t[a_1, \dots, a_n] \right)  \\
=&  \psi \circ {\mathrm{pr}} \, \, (\partial (k_n[a_1, \dots, a_n])) = 0 .
\end{array}
\end{equation*}
The first equality follows from Equation \eqref{eq:psitDef}, the second one from Lemma \ref{lem:knexplicit} and the
definition of $ \partial$ given in Equation \eqref{eq:differential_on_trees}. The expression vanishes due to Lemma \ref{lem:a_infty1}. By the convention given in first line of Equation \eqref{eq:conventions}, Equation \eqref{eq:sumup} now becomes:
\begin{equation}
\label{eq:TYn}
\sum_{t\in Y_n}
\sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{Ver}(t)\end{array}$}}
(-1)^{P(t[a])+W_A} \psi_{t_{\downarrow A}}(a_1, \dots, a_i, \psi_{t_{\uparrow A}}(a_{i+1}, \dots, a_{i+j}),a_{i+j+1} \dots, a_n)=0.
\end{equation}
\vspace{-2mm}

\noindent
We have to prove that the latter expression is equivalent to Equation \eqref{eq:ainfty2}.
Consider a triple $ (i,j,k) \in \mathbb N_0 \times \mathbb N \times \mathbb N_0$  such that $i+j+k =n $. We say that a pair $ (t[a],A)$, with $ t \in Y_n$  and $A $ a vertex (root included) or a leaf of $t$,  is of type $ (i,j,k)$ if the decorations of  $t_{\uparrow A} $ are $a_{i+1} , \dots , a_{i+j} $.
For instance, the following pair $(t,A) $ is of type $ (1,3,1)$:
\begin{equation}
\label{eq:tree131}
\adjustbox{valign=c}{\scalebox{0.5}{
\begin{forest}
for tree = {grow' = 90}, nice empty nodes, for tree={ inner sep= 0 pt, s sep= 0 pt,
[\scalebox{2}{$a_1$}, tier =1]
[, ,edge label={node[left]{\scalebox{2}{$A\,\,$}}},
[\scalebox{2}{$a_2$}, tier = 1]
[\scalebox{2}{$a_3$}, tier = 1]
[\scalebox{2}{$a_4$}, tier = 1]
[\scalebox{2}{$a_5$}, tier =1]
\path[fill=black] (.parent anchor) circle[radius=4pt]
(!2.child anchor) circle[radius=4pt]
(!21.child anchor) circle[radius=4pt]
(!212.child anchor) circle[radius=4pt];
\end{forest}}}
\end{equation}
since the leaves that descend from $A$ are $(a_2,a_3,a_4) $.
The summation that appears in Equation \eqref{eq:TYn} can be rearranged as:
\sum_{t\in Y_n}
\sum_{\scalebox{0.5}{$\begin{array}{r}A \in \mathrm{Ver}(t)\end{array}$}} =
\sum_{\scalebox{0.5}{$\begin{array}{c}i,k \geq 0 , j \geq 1 \\
\end{array}$}} \sum_{\scalebox{0.5}{$\begin{array}{c}(t,A)   \hbox{ of } \\ \hbox{ type }  (i,j,k)\end{array}$}} .
Equation \eqref{eq:TYn} then becomes:
\begin{equation}
\label{eq:TYn2}
\sum_{\scalebox{0.5}{$\begin{array}{c}i,k \geq 0 , j \geq 1 \\
\end{array}$}} \sum_{\scalebox{0.5}{$\begin{array}{c}(t,A)   \hbox{ of } \\ \hbox{ type }  (i,j,k)\end{array}$}}
(-1)^{P(t[a])+W_A} \psi_{t_{\downarrow A}}(a_1, \dots, a_i, \psi_{t_{\uparrow A}}(a_{i+1}, \dots,a_{i+j}),a_{i+j+1}, \dots, a_n)=0.
\end{equation}
\vspace{-2mm}

%can be any constant non-zero number: we will show that it is equal to $ (-1)^{\sum_{r=1}^n |a_r|(n-r)}$.

\noindent
There is a natural bijection  \begin{equation}\label{eq:treestrees}  \left\{ \hbox{ Pairs $ (t,A)$ of type $ (i,j,k)$}\right\}
\, \simeq  \,  Y_{n-j+1} \times Y_{j}
This bijection maps $ (t,A)$ to the pair $(t_{\downarrow A} , t_{\uparrow A}) $. We recall that when $A$ is a leaf,  $t_{\uparrow A} $ is the trivial tree, and when it is the root $t_{\downarrow A} $ is the trivial tree. Its inverse maps $ (t_1,t_2) \in Y_{n-j+1 \times Y_i}  $ to the pair $ (t,A)$ with $ t \in Y_n$ and  $A $ an inner vertex, a root, or a leaf of $t$ such that
$t_1=t_{\downarrow A}  $ and $ t_2 = t_{\uparrow A}$. For instance, the pair $(t,A) $ of type $(1,3,1) $ described in Equation \eqref{eq:tree131} corresponds to the pair $ (\rtree,\rtree)\in Y_3 \times Y_3$.
In view of this bijection, the identity \eqref{eq:meaningmu} given in
Lemma \ref{lem:aboutmuexplicit} becomes
%the desired equality \eqref{eq:step1} becomes
\begin{equation}
\label{eq:step2}
\begin{array}{ll}&  (-1)^{i+jk}\mu_{n-j+1}(\id^{\otimes i}\otimes \mu_{j} \otimes \id^{\otimes k})  (a_1, \dots, a_n)
%\\[4pt] =& (-1)^{i+jk +(j-2)(|a_1|+\dots + |a_i|)}\mu_{n-j+1}(a_{1}, \dots, a_i, \mu_{j}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1}, \dots, a_{i+j+k})..
\\=&
\sum\limits_{\scalebox{0.5}{$\begin{array}{c}(t,A)   \hbox{ of } \\ \hbox{ type }   (i,j,k)\end{array}$}}\epsilon_{t_{\downarrow A},t_{ \uparrow A}}  \, \psi_{t_{\downarrow A}}(a_1, \dots, a_i, \psi_{t_{\uparrow A}}(a_{i+1}, \dots, a_{i+j}),  a_{i+j+1},\dots, a_n).
\end{array}
\end{equation}
%In the above expression, $ (-1)^{P(t[a])+W_A}$ refers to the pair $ (t,A)$ associated to $ (t_1,t_2) $ through the bijection \eqref{eq:treestrees}.
Now one checks  that $ \epsilon_{t_{\downarrow A},t_{\uparrow A}}= \eta \, (-1)^{P(t[a])+W_A}$.
in view of the definition of $ \epsilon_{t_{\downarrow A},t_{\uparrow A}} $ given in Equation \eqref{eq:defepsilon} and Lemma \ref{lem:explicitrelations}.
As a consequence, Equation \eqref{eq:step2} becomes
\begin{equation}
\label{eq:step1}
\begin{array}{ll}  & (-1)^{i+jk}\mu_{n-j+1}(\id^{\otimes i}\otimes \mu_{j} \otimes \id^{\otimes k})  (a_1, \dots, a_n)
%\\[4pt] =& (-1)^{i+jk +(j-2)(|a_1|+\dots + |a_i|)}\mu_{n-j+1}(a_{1}, \dots, a_i, \mu_{j}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1}, \dots, a_{i+j+k})..
\\=&
\eta \sum\limits_{\scalebox{0.5}{$\begin{array}{c}(t,A)   \hbox{ of } \\ \hbox{ type }   (i,j,k)\end{array}$}} (-1)^{P(t[a])+W_A}  \psi_{t_{\downarrow A}}(a_1, \dots, a_i, \psi_{t_{\uparrow A}}(a_{i+1}, \dots, a_{i+j}), a_{i+j+1},\dots, a_n)
\end{array}\end{equation}
In view of Equation \eqref{eq:TYn2}, this indeed yields
the $n$-th $A_\infty $-axiom Equation \eqref{eq:ainfty2}.

\vspace{1mm}
\noindent
We are left with the task of checking the axiom defining a $C_\infty$-algebra, i.e.  Equation \eqref{eq:Cinfty2}.
In view of the definition of $\mu_n $ given in Theorem \ref{thm:ainfty2},  it suffices to show that for all $ i \leq n$ and all homogeneous $ a=(a_1, \dots,a_n) \in \FMb^n$:
$$ \sum_{\sigma \in {\mathrm{Sh}}(i,n)} \epsilon(\sigma,a)  k_n[a_{\sigma}]= 0$$
lies in the kernel of the projection $ {\mathrm{pr}}$ from ordered decorated trees to symetrized decorated trees.
Above, $ a_{\sigma} $ is an abbreviation for $a_{\sigma(1)}, \dots, a_{\sigma(n)}$ and $\epsilon(\sigma,a)$ is the Koszul sign of the permutation $ \sigma$ with the convention that the degree of $a_i $ is  $ |a_i|+1$.
We therefore have to prove that
\begin{equation}\label{label:forCinfty}
\sum\limits_{(\sigma,t) \in {\mathrm{Sh}}(i,n) \times Y_n} \epsilon(\sigma,a)(-1)^{P(t[a_\sigma])}  {\mathrm{pr}}( t[a_{\sigma}])= 0
\end{equation}

\vspace{1mm}
\noindent
We proceed as follows. We say that a subtree of $t[ a_{\sigma(1)}, \dots, a_{\sigma(n)}] $ is of \emph{$(\leq i)$-type} if all its leaves are in $\{a_1, \dots,a_i \} $ and of \emph{($> i)$-type} if all its leaves are in $\{a_{i+1}, \dots,a_{n} \} $.
We say that a vertex $A$ in an ordered decorated tree $t[a_{\sigma(1)}, \dots, a_{\sigma(n)} ] $ is of \emph{mixed type} if either its left subtree is of $ (\leq i) $-type and its right subtree is of $ (>i) $-type, or its left subtree is of $(>i)$- type and its right subtree is of $(\leq i)$-type. For every $(\sigma,t) \in {\mathrm{Sh}}(i,n)\times Y_n $,  there exists at least one vertex of mixed type.  We say that two pairs $(t,\sigma), (t',\sigma') \in {\mathrm{Sh}}(i,n) \times Y_n$ with leftmost mixed vertices $A$ and $A'$
are \emph{dual} if the right subtree of $A$  is the left subtree $\ell(A',t'[a_{\sigma'}])$ of $A'$, the left subtree $\ell(A,t[a_\sigma])$ of $A$  is the right subtree of $A'$, while the rest of the trees $ t[a_\sigma]$ and $t'[{a_{\sigma'}}]$ coincide.

\vspace{1mm}
\noindent
Let us study several properties of dual pairs.
In view of the equivalence relation \eqref{eq:quotientEquiv} whose equivalence classes define the projection $ {\mathrm{pr}}$, we have
$$ {\mathrm{pr}}(t[a_\sigma]) =  (-1)^{|\ell(A,t[a_\sigma])||\ell(A',t'[a_{\sigma'}])|} {\mathrm{pr}}(t'[a_{\sigma'}]). $$
The decorated trees $t[a] $ and $ t'[a_{\sigma'}]$ have left subtrees with exactly the same degrees, except for the left subtrees of the vertices $A$ and $A'$, which are $\ell(A,t[a_\sigma]$ and $\ell(A',t'[a_{\sigma'}]$, respectively. Hence:
\begin{array}{rcl} P(t[a_\sigma])+|\ell(A,t[a_\sigma])|&=&P(t'[a_{\sigma'}]) +| \ell(A',t'[a_{\sigma'}])|.
\end{array}
% with $I,I' \subset \{1, \dots,n\}$ being the indexes of the leaves of  $\ell(A,t[a_\sigma])$
%and $\ell(A',t'[a_{\sigma'}])$.
By definition of the Koszul sign
$$ \epsilon(\sigma',a)=\epsilon(\sigma,a)(-1)^{ (\sum_{r \in I}  |a_r|-|I_l|)(\sum_{r \in I'}  |a_r|-|I'|)}  = \epsilon(\sigma) (-1)^{(|\ell(A,t[a_{\sigma}])|+1)(|\ell(A',t'[a_{\sigma'}])|) +1) }.$$
Here $I,I'\subset \{1, \dots,n\}$ are the indices of the leaves of $ \ell(A,t[a_{\sigma}])$ and $\ell(A',t'[a_{\sigma'}]) $, respectively.
The three relations above imply that the contribution of a pair and its dual pair  come with opposite signs in the left hand side Equation \eqref{label:forCinfty}.
Since any pair in $({\mathrm{Sh}}(i,n) \times Y_n) $ admits one and only one dual pair, Equation \eqref{label:forCinfty} holds. This completes the proof of Theorem \ref{thm:ainfty2}.

\end{proof}

\section{Ideals invariant under a finite group action}
\label{sec:finitegroups}

\noindent Consider an algebra  $ \CO$ on which a finite group $G$ acts by algebra automorphisms, and let $ \CI \subset \CO$ be a $G$-invariant ideal.
It is natural to look for \emph{$G$-equivariant Koszul-Tate resolutions} $(S(\CE),\delta) $. This means that we require an action  of $ G$ by DGCA automorphisms of $(S(\CE),\delta) $ whose restriction to elements of degree $0$, which are isomorphic to $ \CO$, coincides with the initial $G$-action.
%induced action on $ H_0(S(\CE),\delta) \simeq \CO/\CI$ coincides with the natural action of $G$ on $ \CO/\CI$.

\vspace{1mm}\noindent
A pleasant feature of the Tate algorithm presented in Construction \ref{const:tate.algorithm} is that it can be adapted to yield such a  $G$-equivariant Koszul-Tate resolution: it suffices to average at every step (despite the fact that there are infinitely many such steps in general). We prove that the arborescent Koszul-Tate resolution offers the same advantage upon averaging at every step (and there are finitely many such steps in many situations, see Section \ref{sec:complexity}).

\begin{proposition}
Let $G$ be a finite group acting on an algebra $\mathcal O $ over $\mathbb R$ or $\mathbb C $ and $\mathcal I $ a $G $-invariant ideal. There exists a $G$-equivariant arborescent Koszul-Tate resolution.
\end{proposition}
\begin{proof}
The proof goes in two steps.

\vspace{1mm}
\noindent
{\textbf{Step 1.}} Let us show that a $G$-invariant ideal $\CI $ admits a $G$-equivariant free $\CO $-module resolution.

\vspace{0.1cm}

\noindent
We denote by $(g,F) \mapsto g\triangleright F$ the action of $g \in G$ on $F \in  \CO $.
Since $G$ acts on $\CO $, it acts on any free $\CO $-module.
Let $\FM_1 $ be a free $\CO $-module and $d \colon \FM_1 \to \CO $ an $\CO$-module morphism whose image is $ \CI$. In general, $d$ is not $G$-equivariant. Consider the tensor product
$$ \FM_{1}' := \FM_{1} \otimes_{\mathbb K} \mathbb K [G]  $$
with $\mathbb K [G]$ being the free vector space generated by the group $G  $.  There is natural $G$-action on this space by
$$   h \triangleright (a \otimes g) := (h \triangleright a) \otimes hg \hbox{ for all $a \in \FM_1, g,h \in G$ } . $$
With respect to this $G$-action, the $\CO$-linear morphism \begin{equation} \label{eq:equivD} d' ( a \otimes g) = g \triangleright d  \left(  g^{-1}  \triangleright a \right) \hbox{ for all $a \in \FM_{1}, g \in G$} \end{equation}
is $G$-equivariant. Its image is still $\CI $. Since $d$ is $G$-equivariant, its kernel is preserved under the $G$-action, and for all $g \in G$
$$  a \mapsto g \triangleright a $$
is an invertible endomorphism of ${\mathrm{Ker}}(d) $.
Let $\FM_2 $ be a free $\CO$-module equipped with a surjective $\CO$-linear morphism $\FM_2 \longrightarrow {\mathrm{Ker}}(d) $, then consider
$$ \FM_{2}' := \FM_{2} \otimes_{\K} \mathbb K [G] . $$ The map $d'$ constructed as in \eqref{eq:equivD} is $G$-equivariant and its image is ${\mathrm{Ker}}(d)$, so that
$$ \xymatrix{\FM_{2}' \ar[r]^{d'} & \FM_{1}' \ar[r]^{d'} & \CO \ar[r] & \CO/\CI}  $$
are the first terms of a $G$-equivariant free $\CO$-module resolution of $\CI$.
The procedure continues by recursion.

\vspace{.5cm}

\noindent
{\textbf{Step 2.}}
Let $(\FM_\bullet,d') $ be an $K$-equivariant free $\CO$-module resolution as obtained out of Step 1. In the construction of its arborescent Koszul-Tate resolution,
all operations described in Section \ref{sec:arborescent} are $G$-equivariant except maybe those that involve the arborescent operations
$t \mapsto \psi_t  $, so that $\delta_\psi $ will be $G$-equivariant if the maps $ \psi_t $ can be chosen to be $G $-equivariant.

\vspace{0.1cm}

\noindent
We prove by recursion on the degree $k$ that such choices exist. Equation \eqref{eq:deltaPsiSquare} gives the recursion relation that an  arborescent operation $ \psi$ needs to satisfy in order to have $ \delta_\psi^2=0$:

\begin{equation}
\label{eq:psi30}
\begin{gathered}
d' \circ \psi_t (a_1, \dots, a_n)
\, = \\ \,\left(  \sum_{A \in InnVer(t)} \left(\psi_{t_{\downarrow A}} \circ_{A} \psi_ {t_{\uparrow A}} - (-1)^{W_A}\psi_{\partial_A t}\right) -\sum_{i=1}^n(-1)^{W_i}\psi_t(a_1, \dots, d'a_i, \dots,a_n) + \pi^1\circ d'\right) (a_1\otimes \dots \otimes a_n).
\end{gathered}
\end{equation}
More precisely, while proving Theorem \ref{thm:isKT} (and more precisely Corollary \ref{cor:differentialexists}), we showed that $ \delta_\psi^2=0$ if and only if the $ \psi_t$ are recursively constructed so that they satisfy \eqref{eq:psi30}.
If one assumes that the assignment $(a_1, \dots, a_n) \mapsto \psi_t[a_1, \dots, a_n] $ is $G$-equivariant for all tree $t$
and all $ a_1, \dots,a_n \FM$ such that the sum $ |t|+|a_1|+ \dots + |a_n|$ of all degrees is $\leq k $, then the right hand side of the previous equation is $G$-equivariant. Since $d'$ is $G$-equivariant, one may redefine  $ \psi_t(a_1, \dots,a_n)$
and all trees $t$ and homogeneous elements $ a_1, \dots,a_n \in \FM_\bullet$ whose degrees add up to $ k+1$ by
$$ \tilde{\psi}_t[a_1, \dots,a_n] :=  \frac{1}{|K|} \sum_{g \in G} g^{-1} \triangleright \psi_{t} ( g \triangleright a_1, \dots,  g \triangleright a_n ) $$
and Equation \eqref{eq:psi3} still holds with this new definition.

\vspace{1mm}
\noindent The construction of $G$-equivariant arborescent operations $t \mapsto \psi_t $ then continues by recursion. %completes the proof.
\end{proof}

\section{The reduced complex of the arborescent Koszul-Tate resolution}
\label{app:reduced}

\vspace{0.1cm}

\noindent
We prove in this Appendix the following fact : any two Koszul-Tate resolutions are homotopy equivalent in a unique up to homotopy manner. Here, "homotopy" is to be taken  in the sense of  $Q$-manifolds, see e.g. Section 3.4.3 in [25].
This result already appeared in [16], but our description is more precise about the properties of this homotopy equivalence.
We deduce from that result that the reduced complex we introduce below is a well-defined notion.

\vspace{0.1cm}

\noindent
Let us adapt to the present context the definitions in [25] about homotopy equivalence of morphisms.
Denote by $(\mathbb K[t,dt],d) $ the complex of polynomial exterior forms on $\mathbb K $. Here, it must be understood that the degree of $t$ is zero, and that the degree of $ dt$ is $-1$, so that the degree of the de Rham differential $d$ is $ -1$.
For $\CEb, \CEb' $ graded $\CO $-modules, consider a graded commutative algebra morphism
$$  \mathcal H \colon S(\CEb') \to S(\CEb) \otimes_\mathbb K  \mathbb K[t,dt]  . $$
Recall that for any vector space  $V$ a linear map   valued in $ V \otimes_\mathbb K \mathbb K[t]  $
can be equivalently seen as a family $ \psi_t$ of maps valued in $V$ and depending on a parameter $ t \in \mathbb K$.
Therefore, in view of the decomposition
$$ S(\CEb) \otimes_\mathbb K \mathbb K[t,dt] = S(\CEb) \otimes_\mathbb K \mathbb K[t] \oplus S(\CEb) \otimes_\mathbb K \mathbb K[t] \, dt ,$$
$ \mathcal H$ may be seen at a family $ \mathcal H_t$ depending on $t \in \mathbb K$ of graded algebra morphisms from
$ S(\CEb)   $ to $ S(\CEb) \oplus S(\CEb)
{dt}$, where $dt$ is seen as a degree $ -1$ element.
Moreover, $\mathcal H_t $ decomposes as a sum
\begin{equation}\label{eq:Ht} \mathcal H_t = \Phi_t + \Delta_t  \, dt\end{equation}
\begin{enumerate}
\item  $\Phi_t $ is a family depending on $ t\in \mathbb K$ of graded commutative algebra morphisms $S(\CEb')\to S(\CEb) $ of degree $0$.
\item and
$\Delta_t \colon  S( \CEb')\to S(\CEb) $ is a family  depending on $ t\in \mathbb K$
of $\Phi_t $-derivations of degree $+1 $.
%for every $ t\in \mathbb K$.
\end{enumerate}
Assume now that $ S(\CEb')$ and $S(\CEb) $ are equipped with degree $-1 $ differentials $\delta$ and $\delta' $ respectively. Then $S(\CEb) \otimes   \mathbb K[t,dt] $ comes equipped with the differential
$$ \delta \otimes_\mathbb K {\mathrm{id}} +{\mathrm{id}} \otimes_\mathbb K  dt \frac{\partial}{\partial t}   .$$
Here, $  dt \frac{\partial}{\partial t} $ stands for the de Rham differential in one variable
$$    dt \frac{\partial}{\partial t} P(t) = P'(t)  \,dt \hbox{ and } dt \frac{\partial}{\partial t} (P(t) \, dt )=0  $$
for any $ P(t) \in \mathbb K[t]$.
If the algebra morphism $\mathcal H $ is a also a chain map, then $\Phi_t $ is a morphism of differential graded commutative algebras and:
\begin{equation} \label{eq:Phitderivations}
-\frac{\partial \Phi_t}{ \partial t} = \delta \circ \Delta_t  + \Delta_t \circ \delta'.
\end{equation}
$\mathcal H $ is called a \emph{homotopy} between the differential graded algebra morphism $\Phi_0 $ and $\Phi_1 $.

\begin{remark}
\normalfont
\label{rmk:aboutgreaterthan2}
Integrating Equation \eqref{eq:Phitderivations} between $0$ and $1$ shows that $\Phi_0 $ and $ \Phi_1$ are homotopic as chain maps with homotopy $\int_0^1 \Delta_t dt $. Notice that for degree reasons, $ \Delta_t$ maps $\CEb' $ to $S^{\geq 1} (\CEb) $ and therefore maps $ S^{\geq 2} (\CEb') $ to $S^{\geq 2} (\CEb) $.
Hence, the same holds for the homotopy $\int_0^1 \Delta_t dt $.
\end{remark}

\begin{definition}
\label{def:hom.equiv}
Two differential graded commutative algebras $(\CA, \delta)$ and $(\CA', \delta')$ are said  to be homotopy equivalent, if there exist two DGA morphisms $\Phi: (\CA, \delta) \longrightarrow (\CA', \delta')$ and $\Phi': (\CA', \delta') \longrightarrow (\CA, \delta)$ such that the compositions $\Phi'\circ \Phi$ and $\Phi'\circ \Phi$ are homotopic to the identity map.
\end{definition}

\vspace{0.1cm}

\noindent
We start with a lemma.
\begin{lemma}
\label{lem:tensorStillKT}
Let $(S(\CEb), \delta)$ be a Koszul-Tate resolution of $\CO/\CI$.
The homology of $$\left(S(\CEb)    \otimes \mathbb K[t,dt], \delta \otimes id + id \otimesk dt \frac{\partial}{\partial t} \right) $$
is zero in every degree except in degree $ 0$ where it is $ \CO/\CI$.
\end{lemma}
\begin{proof}
The complex computing this homology has two lines:
$$ \xymatrix{ \cdots\ar[rr]^{\delta \otimesk {\mathrm{id}}} && S(\CEb)_{k+1} \otimesk \mathbb K[t] dt \ar[r]^{\delta \otimesk {\mathrm{id}}}& S(\CEb)_{k} \otimesk \mathbb K[t] dt \ar[r]^{\delta \otimesk {\mathrm{id}}}& S(\CEb)_{k-1} \otimesk \mathbb K[t] dt  \ar[rr]^{\delta \otimesk {\mathrm{id}}}&&\cdots \\    \cdots\ar[rr]^{\delta \otimesk {\mathrm{id}}}&& \ar[r]^{\delta \otimesk {\mathrm{id}}}S(\CEb)_{k+1} \otimesk \mathbb K[t]   \ar[r]^{\delta \otimesk {\mathrm{id}}}\ar[u]^{{\mathrm{id}} \otimesk dt \frac{\partial }{ \partial t}}&S(\CEb)_k \otimesk \mathbb K[t] \ar[r]^{\delta \otimesk {\mathrm{id}}}\ar[u]^{{\mathrm{id}} \otimesk dt \frac{\partial }{ \partial t}}& S(\CEb)_{k-1} \otimesk \mathbb K[t] \ar[u]^{{\mathrm{id}} \otimesk dt \frac{\partial }{ \partial t}}  \ar[rr]^{\delta \otimesk {\mathrm{id}}}&&\cdots\\  }  $$
By assumption, horizontal lines have no homology except in degree $0$ where it is $\CO/\CI \otimesk \mathbb K[t] dt $ and  $\CO/\CI \otimesk \mathbb K[t] $ for the upper and lower lines respectively. The $k$-th column has homology equal to $S(\CEb)\otimesk \mathbb K $ concentrated in the lower line. The result follows by a simple diagram chasing argument.
\end{proof}

\vspace{0.1cm}

\noindent
Lemma \ref{lem:tensorStillKT} allows to prove that Koszul-Tate resolutions of $\CO/\CI $ are terminal objects in the category where objects are symmetric algebras over a free or projective $\mathbb N $-graded $\CO$-module $ \CEb$ equipped with a differential such that $\delta (\mathcal E_{1}) \subset \CI$, and arrows are homotopy classes of differential graded commutative algebras. Let us state it in a less abstract manner:

\begin{lemma}
\label{lem:dga.morphisms}
Let $(S(\CEb), \delta)$ be a Koszul-Tate resolution of $\CO/\CI$.
Let $ (S(\CEb'), \delta')$ be  differential graded algebra,
with $  ({\CE_{i}'})_{i \geq 1} $ a sequence of projective $\CO $-modules.
If $\delta'(\CE_1') \subset \CI$, then there exists a differential graded commutative algebra morphism $\Phi^{\bullet}: (S(\CEb'), \delta') \longrightarrow (S(\CEb), \delta)$ and any two such morphisms are homotopic.
\end{lemma}
\begin{proof}
By assumption, $ \delta' (\CE_1') = \CI$ and $ \delta(\CE_1) \subset \CI$. Since $ \CE_1'$ is a projective $ \CO$-module, there exists a $\CO $-linear map $f^1$ making the following diagramm commutative
\begin{tikzcd}[row sep=2.5em, column sep=2.5em]
\CE_1' \arrow[r, "\delta"] \arrow[d, "f_1"] & \CO  \arrow[d, "\id"] \\ \CE_1 \arrow[r, "\delta'"] & \CO
\end{tikzcd}
We then construct $ \Phi$ by recursion out of a family of degree $0$ $ \CO$-linear maps:
$$ f^n \colon \CE_n' \longrightarrow S(\CEb)_n  .$$
Assume $f^1, \dots, f^n $  are constructed such that the graded algebra morphism $ \Phi^n$, whose restriction to $ \CE_i'$ is $f^i$ for $ i \geq n$ and $0$ otherwise, is a chain map when applied to elements of degree $0$ to $n$.
Then $ \Phi^n \circ \delta' -  \delta \circ \Phi^n $ is a $\Phi^n $-derivation whose restriction to $\CE_{n+1} $ is valued in the kernel of $\delta $. Since the homology of $\delta $ is zero by assumption, there exists  $ f^{n+1} \colon \CE_{n+1} \to S(\CEb)_{n+1}$ such that
$$   \Phi^n \circ \delta' -  \delta \circ \Phi^n  = \delta \circ f^{n+1}  . $$
It is routine to check that
the graded algebra morphism $ \Phi^{n+1}$ whose restriction to $ \CE_i'$ is $f^i$ for $ i \leq n+1$ and $0$ otherwise, is a chain map when applied to elements of degree $0$ to $n+1$. The graded algebra morphism $ \Phi$ whose restriction to $ \CE_i'$ is $f^i$ for $ i \in \mathbb N$ is a chain map. This proves the first part of the Lemma.

\vspace{0.1cm}
\noindent
%This is a first step of a recursive construction.
Let $ \Phi_0,\Phi_1 \colon (S(\CEb'),\delta') \to (S(\CEb),\delta)$
be two such morphisms, coming from sequences $ (f_0^i)_{i \geq 1}  $
and $ (f_ 1^i)_{i \geq 1}$ respectively.
%More generally,
Let us assume that $ \Phi_0,\Phi_1$ are homotopic up to a certain order $n$, i.e. that we are given a graded algebra
%We associate to this sequence a sequence of graded algebra
$ \mathcal H^{n} \colon S(\CEb') \to S(\CEb) \otimesk \mathbb K[t,dt]$ which is a chain map when applied to elements of degrees ranging from $0$ to $n$, and such that in the decomposition as in Equation \eqref{eq:Ht}
\begin{equation}
\label{eq:HtFor_n}
\mathcal H^{n}_t = \Phi_t^n + \Delta_t^n dt
\end{equation}
we have  $ \Phi_0^n=\Phi_0$ and $\Phi_1^n = \Phi_1 $. For $n$ = 0 we can choose $\mathcal H^0_t = (1-t)\Phi_0 + t\Phi_1$.
Let us construct $\mathcal H^{n+1} $, making use of the decomposition:
$$   \mathcal H^{n+1}_t = \Phi_t^{n+1} + \Delta_t^{n+1} dt  .$$
The values of $ \Phi_t^{n+1}$ at $t=0 $ and $ t=1$ are imposed to be $ \Phi_0^{n+1}$ and $ \Phi_1^{n+1}$ respectively.
Consider the failure of $\mathcal H^n $ to be a chain map, when applied to elements of degree $n+1 $. Let  $g^{n+1}$ be the restriction to $ S(\CEb')_{n+1}$ of
$$ \left( \delta \otimes \id + \id \otimes \mathrm{dt} \frac{\partial}{\partial t} \right) \circ {\mathcal H}^n  -{\mathcal H}^n \circ \delta'  .$$
By assumption, $g^{n+1}$ is zero on $(S^{\geq 2}\CEb')_{n+1}$. Moreover, its component in  $S(\CEb) \otimes \mathbb K[t]$ is $0$ for $t=0,1$.  The image of $ g^{n+1}$ lies inside the kernel of $\delta \otimes \id + \id \otimes dt \frac{\partial}{\partial t} $.
By lemma \ref{lem:tensorStillKT}, $ g^{n+1}$ is therefore valued in the image of
$$ \delta \otimes \id + \id \otimes dt \frac{\partial}{\partial t}  \colon (S(\CEb) \otimes \mathbb K[t,dt])_{n+1}  \to (S(\CEb) \otimes \mathbb K[t,dt])_{n}  .$$
Let us view $g^{n+1}$ as a map $\CE'_{n+1} \longrightarrow (S(\CEb) \otimes \mathbb K[t,dt])_{n}$.
Since $ \CE_{n+1}'$ is a projective module, there exists an $ \CO$-linear map $$\hat h^{n+1} \colon
\CE_{n+1}' \to (S(\CEb) \otimes \mathbb K[t,dt])_{n+1}   $$
such that
g^{n+1} = -\left(\delta \otimes \id + \id \otimes dt \frac{\partial}{\partial t} \right) \circ \hat h^{n+1} .
Let us write $\hat  h^{n+1}$, which is valued in $S(\CEb)\otimesk \mathbb K[t,dt] $, as a decomposition
$$ \hat h^{n+1} = \alpha_t^{n+1} + \beta^{n+1}_t dt .$$
Since $\delta \alpha_t^{n+1} = 0$ for $t = 0, 1$, by adding an appropriate boundary term, we modify $\hat h^{n+1}$ in such a way that $\alpha_t^{n+1} = 0$ for $t =0, 1$.

\vspace{1mm}
\noindent
Define a map $h^{n+1}\colon \CEb' \longrightarrow (S(\CEb) \otimes \mathbb K[t,dt])$
h^{n+1}(a) = \begin{cases}
\mathcal{H}^n(a),\hbox{ if } a\notin \CE'_{n+1} \\
\mathcal{H}^n(a) + \hat h^{n+1}(a), \hbox{ if } a\in \CE'_{n+1}.
\end{cases}

\vspace{1mm}
\noindent
Then $\mathcal H^{n+1}$ is defined as the graded commutative algebra morphism that coincides with $h^{n+1} $ on $ \CEb'$. By construction $\mathcal H^{n+1}$ is a chain map when applied to elements of degrees ranging from $0$ to $n+1$.
\end{proof}

\vspace{0.1cm}
\noindent
As an immediate consequence of Lemma \ref{lem:dga.morphisms},
we recover the following result from [16] - with additional clarification about the ideals  $S^{\geq2 } \CEb \oplus \CI   \CEb  $ and $S^{\geq 2} \CEb' \oplus \CI  \CEb' $.

\begin{proposition}
\label{prop:homotopicAlways}
Two Koszul-Tate resolutions $(S(\CEb), \delta)$ and $(S(\CEb'), \delta')$ of $ \CO/\CI$   are homotopy equivalent in a unique up to homotopy manner:
\vspace{-9mm}
\begin{equation}\label{eq:existHomotopy}
\begin{tikzcd}[column sep = 4em]
(S(\CEb), \delta) \arrow[r, bend left=20, "\Psi", shift ={(0 ,1mm)}]  \arrow[out=225, in=135, looseness=8, loop, "\mathcal H"] & (S(\CEb'), \delta') \arrow[l, bend left=20, swap, "\Psi'", shift = {(0, -1mm)}]\arrow[out=-45, in=45, looseness=5, loop,swap, "\mathcal H'"]
\end{tikzcd}
\end{equation}
\vspace{-9mm}

\noindent
where $\Psi, \Psi' $ and each one of the components as in \eqref{eq:Ht} of the homotopies $\mathcal H_t=\Phi_t + \Delta_t \, dt , \mathcal H_t' =\Phi_t' + \Delta_t '\, dt $ preserve the ideals $S^{\geq2 } \CEb \oplus \CI   \CEb  $ and $S^{\geq 2} \CEb' \oplus \CI  \CEb' $.
\end{proposition}
\begin{proof}
Lemma \ref{lem:dga.morphisms}
implies the existence and uniqueness up to homotopy of morphisms of differential graded commutative algebras as in Equation \eqref{eq:existHomotopy}.
Remark \ref{rmk:aboutgreaterthan2} gives the conditions about the ideals  $ S^{\geq 2} \CEb \oplus \CI\CEb  $ and $S^{\geq 2} \CEb' \oplus \CI  \CEb' $. %\eqref{eq:condtionOnHomotopy}.
\end{proof}

\vspace{0.1cm}

\noindent
Having found such a homotopy we can make a following quotient construction: consider the quotient of a Koszul-Tate resolution $(S(\CEb), \delta)$ of $\CO/\CI $ by $\CI\CEb\oplus \Sdeg(\CEb)$.
This quotient is isomorphic to the tensor product $\CE_\bullet \otimes_{\CO} \CO/\CI$ as an $\CO$-module.
Since $\delta(\CI\CEb\oplus \Sdeg(\CEb)) \subset \CI\CEb\oplus \Sdeg(\CEb)$,  the quotient comes equipped with an induced differential $ \underline{\delta}$.

\begin{definition}
For a Koszul-Tate resolution $(S(\CEb), \delta)$ of $\CO/\CI$, the complex of $ \CO$-modules $(\CEb\otimes \CO/\CI, \underline{\delta})$ is called a reduced complex of $\CO/\CI$.
\end{definition}

\vspace{0.1cm}

\noindent
The following is an immediate corollary of Proposition \ref{prop:homotopicAlways}:

\begin{corollary}
\label{cor:reduced}
Two reduced complexes associated to two Koszul-Tate resolutions of $\CO/\CI $ are homotopy equivalent  and there is distinguished class of homotopies between them.
%in a unique up to homotopy manner.
In particular, its homology is canonically attached to the ideal $ \CI \subset \CO$.
\end{corollary}

\vspace{0.1cm}

\noindent
Consider an ideal $\CJ \subset \CO $ containing $ \CI$. The tensor product
$(\CE_\bullet \otimes \CO/\CJ, \underline \delta)$
of a reduced complex  of $ \CI $ with $ \CO/\CJ$
is a complex of $ \CO$-modules, on which $ \underline{\delta}$ induces a differential that we denote again by $ \underline{\delta}$.

\begin{definition}
For a Koszul-Tate resolution $(S(\CEb), \delta)$ of $\CO/\CI$ and an ideal $\CJ \subset \CO $ containing $ \CI$, the complex of $ \CO$-modules $(\CEb\otimes \CO/\CJ, \underline{\delta})$ is called a reduced complex of $\CO/\CI$ evaluated at $ \CJ$.
\end{definition}

\noindent
Corollary \ref{cor:reduced} implies that
the homology of this complex does not depend on the choice of a Koszul-Tate resolution of $ \CO/\CI$.
%tensor product of the reduced complex with $\CO/\CJ $ with $\CJ $ any ideal containing $\CI $.

\begin{corollary}
\label{cor:reducedWIthJ}
Two reduced complex of $\CO/\CI$ evaluated at $ \CJ$ associated to two Koszul-Tate resolutions of $\CO/\CI $ are homotopy equivalent  and there is a distinguished class of homotopies between them.
In particular, its homology is canonically attached to the ideals $ \CI \subset \CJ \subset \CO$.
\end{corollary}
\vspace{0cm}
\noindent
Let $\CJ$ be a maximal ideal of $\CO$. Since $\CO/\CJ \simeq \mathbb K $ is a vector space, any reduced complex of $\CO/\CI$ evaluated at $ \CJ$ is a complex of vector spaces over $\mathbb K $.  The homology of the reduced complex of $ \CO/\CI$ evaluated at $ \CJ$ is a vector space as well. We denote by $ b_i(\CO/\CI,\CJ) \in \mathbb N \cup \{+\infty\}$ the dimension of the homology in degree $i \in \mathbb N$. By the above discussion, $b_i(\CO/\CI,\CJ)$ does not depend on the choice of a Koszul-Tate resolution, hence the notation.
If the free $\CO $-module $\CE_i  $ is indexed by a set $ S_i$, then the chains in degree $i$ of the reduced complex of $ \CO/\CI$ evaluated at $ \CJ$ form a vector space indexed by the same set $S_i$. The $i$-th homology being a vector space of smaller dimension, the next statement follows:

\begin{theorem}
\label{thm:lowernboundonrank}
For any Koszul-Tate resolution $ (S(\CEb),\delta)$ of $ \CO/\CI$, any maximal ideal $ \CJ$ containing $ \CI$, and any $ i \in \mathbb N$, we have
${\mathrm{rk}}(\CE_i) \geq b_i(\CO/\CI,\CJ)$.
\end{theorem}

\section{Minimal Koszul-Tate resolutions}
\label{app:minimal}

\vspace{0.1cm}

\noindent
Although the arborescent Koszul-Tate resolutions have the advantages listed in the introduction, they may  not be the "smallest" Koszul-Tate resolution.  For instance, for $\CI \subset \CO$ a complete intersection with $r$ generators, the Koszul $\CO$-module resolution $((\FM_i = \wedge^i \CO^r)_{i \geq 1},d = \sum_{i=1}^{n}\phi_i \iota_{e_i}) $
of section \ref{sec:Koszul} is by construction an $\CO$-module resolution with finitely many generators. One can therefore consider the associated arborescent Koszul-Tate resolution. The latter is of course a huge symmetric algebra: for $ r \geq 2$, it has infinitely many generators.

\vspace{0.2cm}
\noindent
Theorem \ref{thm:lowernboundonrank} gives a lower bound to the possible ranks of the free $\CO$-modules $ \CEb$ entering a Koszul-Tate resolution of $ \CO/\CI$.
In this appendix, we give a context where this lower bound can be reached.
We consider two distinct cases, which may seem restrictive but quite a few examples exist, see section~\ref{sec:basicExamples}:
\begin{itemize}
\item[ ] Case i. $\CO$ is a local Noetherian ring with maximal ideal $\CJ$ and residue field $\mathbb K$.
\item[ ] Case ii. $\CO$ is the polynomial ring $\mathbb K[x_1, \dots, x_n]$, $\CJ$ is the maximal ideal $\CJ:=\langle x_1, \dots, x_n \rangle$ and  $\CI$ is an ideal with homeogeneous generators.
\end{itemize}
Here is a classical theorem, see e.g. [13] for case i and  Theorem 2.12 of [31] for case ii. Let $\CJ$ be a maximal ideal containing $\CI $.

\begin{theorem}
\label{thm:basis.theorem}
Let $M$ be a finitely generated $\CO$-module and $ \CJ$ a maximal ideal of $\CO $ containing $\CI $. In Case ii., in addition, $M$ is graded with respect to the polynomial degree $\mathrm{pol}(.)$.
\begin{itemize}
\item[1.] If $\bar e = \lbrace \bar e_1, \dots , \bar e_j\rbrace $ is a basis of the vector space $M/\CJ M$, then its preimage in $M$ is a minimal set of generators of $M$.
\item[2.] Every minimal set of generators is obtained through 1.
\item[3.] In Case i., if $e = \lbrace e_1, \dots , e_j\rbrace$ and $f = \lbrace f_1, \dots, f_j \rbrace$ are two minimals sets of generators of $M$ and  $e_i = \sum_{k=1}^j A_{ij}f_j$ for some matrix $(A_{ij})$, then the matrix $(A_{ij})$ is invertible. In Case ii., one asks additionally that $e$ and $f$ are homogeneous and that the matrix  $(A_{ij})$  has homogeneous entries.
\end{itemize}
\end{theorem}

\vspace{0.1cm}

\noindent
Recall that a free or projective resolution $(\FM_{\bullet}, d)$ of $\CO/\CI $ is said to be \emph{minimal at $\CJ $}  if $d(\FM_{i+1}) \subset \CJ \FM_{i}$ for all $i \geq 1 $.
Theorem \ref{thm:basis.theorem} implies that in  case i. the $  \CO$-module resolution $(\FM_{\bullet}, d)$ obtained as in Construction \ref{const:free.resolution} is minimal at $\CJ $ if and only if at each step
%of Construction \ref{const:free.resolution}
we choose a minimal set of generators of $\mathrm{Ker}(d_i)$.
Also, in this case, a minimal free $ \CO$-module resolution is unique up to isomorphisms, and any free  $ \CO$-module resolution can be decomposed as a direct sum of a trivial complex and a minimal free  $ \CO$-module resolution.
In case ii., the same conclusion holds provided that the generators of $\FMb$ are chosen to be homogeneous at every step.
By construction, all free $\CO$-modules $\FMb$ are graded with respect to the polynomial degree $\mathrm{pol}(.)$, and  $\mathrm{pol}(d) = 0$.

%$ \CI$ has homogeneous generators.

\vspace{0.1cm}

\noindent
We give the following definition of minimality at $\CJ $ of a Koszul-Tate resolution of $\CO/\CI$:
\begin{definition}
Let $ \CJ$ be a maximal ideal containing $\CI $. A Koszul-Tate resolution $(S(\CE), \delta)$ is said to be minimal at $\CJ $ if $\delta(\CE_{i+1})  \subset \CJ \CE_{i} \oplus \Sdeg (\CE)_i$ for all $i\geq 1$.
\end{definition}

\vspace{0.2cm}
\noindent
The following result is straightforward.
\begin{lemma}
\label{lem;asbi}
Let $ (S(\CEb), \delta)$ be a Koszul-Tate resolution of $ \CO/\CI$ such that $\CE_i $ is a free module of finite rank for all $ i \geq 1$.
Then $ (S(\CEb), \delta)$ is minimal  at $ \CJ$ if and only if $ {\mathrm{rk}}(\CE_i)=b_i(\CO/\CI,\CJ)$.
\end{lemma}
\begin{proof}
Let $ i\in \mathbb N$.
By Theorem \ref{thm:basis.theorem}, the rank of $ \CE_i$ coincides with the dimension of $ \CE_i/\CJ \CE_i$.
The minimality condition means that the differential of the reduced complex  of $ \CO/\CI$ evaluated at $\CJ$ is zero, so that the dimension of $ \CE_i/\CJ \CE_i$ is $b_i(\CO/\CI,\CJ)$.
\end{proof}

%Sometimes to specify the maximal ideal $\CJ$ we say a Koszul-Tate resolution is minimal at $\CJ$.
\vspace{0.1cm}
\noindent
For Koszul-Tate resolution obtained through Tate algorithm, i.e. Construction \ref{const:tate.algorithm}, this notion of minimality coincides with the naive one.

\begin{proposition}
\label{prop:Tateisminimal}
Assume $\CO,\CI $ and $\CJ $ are as in Cases (i) and (ii) in the beginning of this section. A Koszul-Tate resolution $(S(\CE), \delta)$ obtained out of the Tate algorithm, i.e. Construction \ref{const:tate.algorithm}, is minimal at $\CJ $ if and only if for every $i \geq 1$, in the $i$-th step of the construction, a minimal set of generators of $H_i(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_i) , \delta_i) $ is chosen.
\end{proposition}
\begin{proof} We will present the proof in the case of local ring, i.e. in case (\emph{i}). For the case  (\emph{ii}), the invertible elements of $\CO$ are simply non-zero constant polynomials, and all families considered have to be made of homogeneous elements.

\vspace{0.1cm}

\noindent
Recall that in case (\emph{i}), all elements not in the maximal ideal $\CJ $ are invertible.
Let as assume that the Koszul-Tate resolution $(S(\CE), \delta)$ obtained out of the Tate algorithm is not minimal. This means that $\exists i\in \mathbb N$ such that $\delta_{i+1}(\CE_{i+1}) \not\subset \CJ\CE_{i}\oplus \Sdeg(\CE)_i$.
%Since  $i= -1$ is impossible since we always have $\delta(\CE_1)\subset \CI \subset \CJ$).
Recall that when applying the Tate algorithm at step $i$, one needs to choose  $\chi = \lbrace \chi_1, \dots, \chi_k \rbrace$ a set of generators of $H_{i-1}(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i-1}) , \delta_{i-1})$  and to associate to it a basis  $e_\chi = \lbrace e_{1}, \dots, e_{k}  \rbrace$ of $\CE_{i}$. We then extend $ \delta_i$ to $ S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i})$  by imposing $\delta_i (e_i)= \chi_i$ for all $ i=1,\dots, k$.
If $\delta_{i+1}(\CE_{i+1}) \not\subset \CJ\CE_{i}\oplus \Sdeg(\CE)_i$, then there exists $f\in \CE_{i+1}$ such that
\delta_{i+1} f = \sum_{j=1}^k a_j e_j\, \, \mathrm{mod}\,\, \Sdeg(\CE)_i
where at least one of the elements $a_1, \dots, a_k$ does not belong to $ \CJ$,  hence is invertible.
Without any loss of generality, we may assume  $a_1$ is invertible.
After multiplying the previous equality by $a_1^{-1}$, we obtain
$$\delta_{i+1} \left(  a^{-1}_1 f \right) = e_{1} + \sum_{j=2}^k \frac{a_j}{a_1} e_j\, \, \mathrm{mod}\,\, \Sdeg (\CE)_i.$$
%    xxx what does that mean? yyy
% On the level of homology at degree $i-1$, this translates into:
By applying $ \delta_{i}$ to the previous equality, one obtains
\delta_i(e_1) + \frac{a_2}{a_1}\delta_i(e_2) + \dots + \frac{a_k}{a_1}\delta_i (e_k) = \delta_i(g).
for some $g \in S^{\geq 2} S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i-1}) $. In particular, $ \delta_i(g)= \delta_{i-1}(g) $ and the previous relation means that in $H_{i-1}(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i-1}) , \delta_{i-1})$ we have
\chi_1 + \frac{\overline{a_2}}{\overline{a_1}}\chi_2 + \dots + \frac{\overline{a_k}}{\overline{a_1}}\chi_k = 0.
Therefore, $\chi$ is not a minimal set of generators of $H_{i-1}(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i-1}) , \delta_{i-1})$.

\vspace{0.1cm}

\noindent
Let us prove the converse implication.
Assume that the Tate algorithm is applied, and that at the $i$-th step one chooses a set $\chi$ which is not a minimal set of generators of $H_{i-1}(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i-1}) , \delta_{i-1})$. Then, after relabeling the generators, we have a relation of the form
\chi_1 + \overline{b_2}\chi_2 + \dots + \overline{b_k}\chi_k = 0
saying that $\chi$ is redundant. On the level of generators of $\CE_{i}$ this equality reads as
\left \{e_{1} + \sum_{2}^k b_j e_j + \dots \right \}\hbox{ is $ \delta$-closed.}
Here $\dots$ stands for terms in $\Sdeg_{i+1}(\CE)$ which are governed by a choice of representatives of $H_i(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_i) , \delta_i) $. Since $\delta \Sdeg(\CE)\cap\CE \subset \CI\CE$, this cycle is not exact in $(S(\mathcal E_1 \oplus \cdots \oplus \mathcal E_{i}) , \delta_{i})$. Hence there exists $f\in \CE_{i+2}$ such that $\delta f = e_{1} + \sum_{j=2}^k b_j e_j + \dots$. Therefore, the Koszul-Tate resolution $(S(\CE), \delta)$ is not a minimal Koszul-Tate resolution.
\end{proof}

%In section \ref{}, the notions of "unnecessary and necessary families" was defined.

%Let $\CI \subset \CJ \subset \CO $ be as above.

\vspace{0.1cm}

\noindent
Let us now explain how a Koszul-Tate resolution of $\CO/\CI $ minimal at  $ \CJ$ can be obtained as a quotient of an arbitrary Koszul-Tate resolution $(S(\mathcal E), \delta) $ of $\CO/\CI $ in cases i. and ii.
%Let $(S(\mathcal E), \delta) $ be a Koszul-Tate resolution of $\CO/\CI $.
Let us first interpret the minimality condition in terms of the \emph{reduced complex} $(\CE_\bullet \otimes \CO/\CI, \underline{\delta}) $ of Appendix \ref{app:reduced}), and more precisely in terms of the complex
$  ( \mathcal E_\bullet \otimes \CO / \CJ , \underline{\delta} )$
as in Corollary \ref{cor:reducedWIthJ}.

\vspace{0.1cm}
\noindent
A direct consequence of Proposition  \ref{prop:Tateisminimal} and Lemma \ref{lem;asbi} is that the lower bounds of Theorem \ref{thm:lowernboundonrank} can be always reached:

\begin{corollary}
Assume $\CO,\CI $ and $\CJ $ are as in Cases (i) and (ii) in the beginning of this section. There exists a Koszul-Tate resolution $(S(\CE), \delta)$ of $ \CO/\CI$ such that for all $ i \in \mathbb N$, the equality $ {\mathrm{rk}}(\CE_i)=b_i(\CO/\CI,\CJ)$ holds.
\end{corollary}

\vspace{0.1cm}

\noindent
By construction, the evaluation at $\CJ $ of the reduced complex is a complex of $\CO/\CJ \simeq \mathbb K $-modules, i.e. a complex of vector spaces. Now, for complex of vector spaces, a normal form exists [13]. To be more precise, there exist sequences $ (C_i)_{i \geq 0}$ and $ (H_i)_{i \geq 1}$ of vector spaces with $ C_0=0$ such that for all $i \geq 1 $, we have
$$  \CE_i \otimes \CO/\CJ = C_{i} \oplus H_i \oplus C_{i-1}  $$
and the differential is given by $\underline{\delta} (c',h,c) = (c , 0 , 0) $ for any $ i \geq 1$ and any $c' \in C_{i} \, , \, h \in H_i \, , \, c \in C_{i-1} $.
In particular, $H_i $ is isomorphic to the $i$-th homology of the reduced complex evaluated at $\CJ $.
%which is canonical by Corollary  \ref{cor:reducedWIthJ}.
We have for all $i \geq 1 $, in view of Theorem \ref{thm:basis.theorem}, a decomposition of $\CE_i$ as follows:
\begin{equation}
\label{eq:decomponE}
\begin{array}{rcl}\CE_i  &\, = \,& \CO \otimesk C_{i}  \, \oplus \, \CO \otimesk H_i \, \oplus \, \CO \otimesk C_{i-1}  \\
& \, =\, &   \CC_{i} \oplus \CE'_\bullet \oplus  \CC[-1]_{i} \end{array}
\end{equation}
where we have introduced the following free graded  $\CO$-submodules of $\CE_{\bullet}$:  $\CC[-1]_{\bullet} = \oplus_{i=1}^{\infty}\CO\otimesk C_{i}$, whose $i$-th degree component is $\CO\otimesk C_{i-1}$,  $\CC_{\bullet} = \oplus_{i=1}^{\infty}\CO\otimesk C_{i}$
, whose $i$-th degree component is $\CO\otimesk C_{i}$, and $ \mathcal E_\bullet' = \oplus_{i \geq 0} \CO \otimesk H_i$.

\begin{proposition}
\label{prop:quotientisKT}
Assume $\CO,\CI $ and $\CJ $ are as in Cases (i) and (ii) in the beginning of this section.
The quotient of a Koszul-Tate resolution $ (S (\mathcal E), \delta) $ by the ideal  generated by $\CC[-1]_{\bullet}\oplus \delta \left( \CC[-1]_{\bullet}\right) $
is a minimal Koszul-Tate resolution of $\CO/\CI $ isomorphic to $ S(\CE')$.
\end{proposition}

\noindent
First, let us show that $ \CC[-1]_{\bullet}\cap \delta \left(  \CC[-1]_{\bullet} \right) = 0$, so the direct sum is well-defined. If the intersection is not zero, then any $a \in  \CC[-1]_{\bullet}\cap \delta \left( \CC[-1]_{\bullet} \right)$ belongs as well to the kernel of $\delta$. Let $s:\CC[-1]_{\bullet} \longrightarrow \CC_{\bullet}$ and $s^{-1}: \CC_{\bullet} \longrightarrow \CC[-1]_{\bullet}$ be the suspension and the desuspension maps. Let $\delta_s$ be the degree $-1$ derivation  of  $S(\CE_{\bullet})$ whose restriction to $\CE$ maps $(a, b, c)$ to $(c, 0, 0)$. Here we used  decomposition \eqref{eq:decomponE}.

\begin{lemma}
\label{lem:ker=0}
$\ker \delta|_{\CC[-1]_{\bullet}} = 0$.
\end{lemma}
\begin{proof}
From the definition of $\delta_s$ it follows that the difference $\mathcal D\coloneq (\delta - \delta_s)|_{\CC[-1]_i}$ is valued in $\CJ  \CE_{i-1}  \oplus  S^{\geq 2}(\CE)_{i-1}$.
Let $a\in \ker \delta|_{\CC[-1]_i}$, then $\mathrm{p}_{\CC_{i-1}} \circ \delta(a) = \delta_s (a) + \mathrm{p}_{\CC_{i-1}} \circ \mathcal D (a) = (s + \mathrm{p}_{\CC_{i-1}}\circ \mathcal D)(a) = 0$, where $\mathrm{p}_{\CC_{i-1}}$ is the projection on $\CC_{i-1}$. In Case i. the map $s + \mathrm{p}_{\CC_{i-1}}\circ \mathcal D: \CC[-1]_i \longrightarrow \CC_{i-1}$ has an invertible determinant, therefore it is invertible, so we deduce $a$ = 0. In Case ii. we decompose $a$ into a finite sum of its homogeneous components with respect to $\mathrm{pol}(.)$: $sa = \sum_{i\in B} a^{(i)}$ for some ordered set $B$, such that $\mathrm{pol}(a^{(i)}) < \mathrm{pol}(a^{(j)})$ if $i < j$. Note that $s$ does not change the polynomial degree. Let $a^{(0)}$ be the first component of this decomposition. Then $(sa)^{(0)} = -\mathrm{p}_{\CC_{i-1}}\circ \mathcal D(a^{(0)}) = Xa^{(0)}$, where $X$ is a matrix with components belonging to $\CJ$. Therefore $a^{(0)} = 0$. By repeating the argument, we establish $a = 0$.
\end{proof}
\noindent
From Lemma \ref{lem:ker=0} it follows that $ \CC[-1]_{\bullet} \simeq \delta \left( \CC[-1]_{\bullet} \right)$. Therefore, $\delta \left( \CC[-1]_{\bullet} \right)$ is a free $\CO$-module. Note that $\delta \left( \CC[-1]_{\bullet} \right) \cap \CEb' = 0$, since $\mathrm{p}_{\CC_{i-1}} \circ \delta$ is invertible as established in Lemma  \ref{lem:ker=0}.
\begin{lemma}
$S(\CEb) = S(\delta (\CC[-1]_{\bullet})\oplus \CEb' \oplus{\mathcal C[-1]}_{\bullet} )$.
\end{lemma}
\begin{proof}
It is enough to establish that the generators of $\CEb$ can be expressed through the combinations of the generators $\delta (\CC[-1]_{\bullet})\oplus \CEb' \oplus{\mathcal C[-1]}_{\bullet}$. Let $e_1, \dots, e_n$ be a basis of $\CE_i$ and $f_1, \dots, f_m$ be a basis of $\delta (\CC[-1]_{\bullet})\oplus \CEb' \oplus{\mathcal C[-1]}_{\bullet}$. Then it is clear from the previous discussion that $m = n$ and $(f_1, \dots, f_n) = A(e_1, \dots, e_n) + R(e)$, where $A$ is an invertible matrix and $R(e)$ is an $n$-tuple where each component is at least quadratic in $e_i$. By induction in the homological degree we establish the statement of the lemma.
\end{proof}
\vspace{1mm}
\noindent
Let us introduce a map $\mathfrak j \colon S^{\geq 1}(\mathcal C_{\bullet} \oplus {\mathcal C}[-1]_{\bullet}) \longrightarrow S(\CE)$, a graded algebra morphism that extends a map $j\colon \mathcal C_{\bullet} \oplus {\mathcal C}[-1]_{\bullet} \longrightarrow \delta\left (\CC[-1]_{\bullet} \right) \oplus \CEb' \oplus{\mathcal C[-1]}_{\bullet}\,\,\,$, $(a, b) \mapsto ( \delta \circ s^{-1}(a), 0,  b)$. Denote by $\mathcal R$ the image of $\mathfrak j$.

\begin{corollary}
\label{cor:iso.red}
The graded algebra morphism $\mathfrak j $ is injective, and therefore it is a graded algebra isomorphism
$\mathcal R \simeq  S^{\geq 1}(\mathcal C_{\bullet} \oplus {\mathcal C[-1]}_{\bullet}) $. The quotient $ S(\CE)/\mathcal R$ is isomorphic to  $S(\CE')$.
\end{corollary}

\noindent
By construction, $\delta $ restricts to $ \mathcal R$, and we still denote by $ \delta$ this restriction.

\begin{lemma}
The complex $(\mathcal R, \delta)$ is acyclic.
\end{lemma}
\begin{proof}
Let us equip $ S^{\geq 1}(\mathcal C_{\bullet} \oplus {\mathcal C[-1]}_{\bullet}) $ with the differential $d_s$ given by $d_s(a, b) = (sb, 0)$ for all $ a \in \mathcal C_{\bullet}, b \in {\mathcal C[-1]}_{\bullet}$, then extended to a graded derivation. By construction, $\mathfrak j$ is a chain map.
Consider the degree $ +1$ $\CO$-linear derivation ${h} $ on $S^{\geq 1}(\mathcal C_{\bullet} \oplus {\mathcal C[-1]}_{\bullet})$
which is defined on $\mathcal C_{\bullet} \oplus {\mathcal C[-1]}_{\bullet}$ by
$$ h (a,b) = (0, s^{-1}a) $$
for all $a \in \CC[-1]_{\bullet}    $ and $ b \in {\mathcal C}_{\bullet}$.
We have $ h \circ \delta_s + \delta_s \circ h= {\mathrm{id}} $ on $\mathcal C_{\bullet} \oplus \widehat{\mathcal C}_{\bullet} $.
Hence for any $a \in S^{k}(\mathcal C_{\bullet} \oplus \widehat{\mathcal C}_{\bullet})$ we have $(\delta_s \circ h + h \circ \delta_s)a = ka$. The acyclicity of $(S^{\geq 1}(\mathcal C_{\bullet} \oplus \widehat{\mathcal C}_{\bullet}), \delta_s)$

\end{proof}

\begin{proof}[Proof of Proposition \ref{prop:quotientisKT}]
By Corollary \ref{cor:iso.red},
there is a natural graded algebra isomorphism
\begin{equation}\label{eq:desirediso} S(\CEb) \simeq S(\CEb') \otimes\left( \CO \oplus \mathcal R \right) ,\end{equation}
under which the differential $\delta $ decomposes as follows,
\xymatrix{
S(\CEb')_{i-3} \otimes \mathcal R_{j+1} & & & \\
& S(\CEb')_{i-2} \otimes \mathcal R_{j+1}& & \\
& & S(\CEb')_{i-1} \otimes \mathcal R_j& S(\CEb')_i \otimes \mathcal R_j \ar[d]^{{\mathrm{id}} \otimes \delta}
\ar[l]^{ \, \delta' \otimes {\mathrm{id}}
} \ar[ull]\ar[uulll]
\\
& & & S(\CEb')_i \otimes \mathcal R_{j-1}\\}
with the understanding that $ \mathcal R_{0} = \CO  $. Above, the horizontal and vertical components of the differential  $\delta$ are $ \delta'\otimes {\mathrm{id}}$ and ${\mathrm{id}} \otimes \delta $ respectively, where $ \delta'$ is the differential induced by $ \delta$ on  $ S(\CEb')=S(\CEb)/\mathcal R$.
\vspace{0.1cm}
\noindent
Since the homology of all vertical lines is zero in every degree except in degree $0$ where it is $\CO $, the homology of $(S(\CEb),\delta) $ coincides with the homology of
$(S(\CEb'), \delta')$.
%in every degree.
Hence $(S(\CE'), \delta') $ is again a Koszul-Tate resolution of $ \CO/\CI$. Last, the reduced complex of $(S(\CEb'),\delta') $ is by construction equipped with a differential valued in $\CJ \CEb' $, so that the reduced complex tensored with $\CO/\CJ$ has a trivial differential. Since its computes a complex whose homology in degree $i$ is $ b_i(\CO/\CI,\CJ)$, this completes the proof.
\end{proof}

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\end{document}