# Resolving the Module of Derivations on an $n\times(n+1)$ Determinantal Ring

Henry Potts-Rubin  Mathematics Department, Syracuse University, Syracuse, NY
13244 U.S.A<EMAIL_ADDRESS>
###### Abstract.

We use the construction of the relative bar resolution via differential graded
structures to obtain the minimal graded free resolution of
$\operatorname{Der}_{R\mid k}$, where $R$ is a determinantal ring defined by
the maximal minors of an $n\times(n+1)$ generic matrix and $k$ is its
coefficient field. Along the way, we compute an explicit action of the
Hilbert-Burch differential graded algebra on a differential graded module
resolving the cokernel of the Jacobian matrix whose kernel is
$\operatorname{Der}_{R\mid k}$. As a consequence of the minimality of the
resulting relative bar resolution, we get a minimal generating set for
$\operatorname{Der}_{R\mid k}$ as an $R$-module, which, while already known,
has not been obtained via our methods.

##  1\. Introduction

This paper is concerned with showing that a truncation of a relative bar
resolution minimally resolves the module of $k$-linear derivations
$\operatorname{Der}_{R\mid k}$ on a determinantal ring $R$ defined by the
maximal minors of an $n\times(n+1)$ generic matrix $X$. The minimality of the
resolution yields a minimal generating set for $\operatorname{Der}_{R\mid k}$
over $R$, a result which may be found in [3], produced by a starkly different
method.

While minimal generators of modules of derivations on determinantal rings are
known, the minimal graded free resolution of $\operatorname{Der}_{R\mid k}$
over $R$ has not been computed. The first order differential operators
$D^{1}_{R\mid k}$ on $R$ decompose as $D^{1}_{R\mid
k}\cong\operatorname{Der}_{R\mid k}\oplus R$, and so after obtaining the
resolution of $\operatorname{Der}_{R\mid k}$, the resolution of $D^{1}_{R\mid
k}$ is immediate. Known resolutions of differential operators are few $-$ see
[6] for the case of low-order differential operators on an isolated
hypersurface singularity and [14] for derivations on a generic hyperplane
arrangement, as two examples. Differential operators need not behave the same
in positive characteristic as in characteristic zero, so it seems worthwhile
to point out that our approach to the minimal graded $R$-free resolution of
$\operatorname{Der}_{R\mid k}$ is characteristic-free.

The $R$-module $\operatorname{Der}_{R\mid k}$ is isomorphic to the kernel of
$J^{T}_{R}$, where $J^{T}_{R}$ is the transpose of the Jacobian matrix of the
maximal minors of $X$ tensored down to $R$. That is, the following sequence of
$R$-modules is exact:

$0\to\operatorname{Der}_{R\mid k}\to
R^{n(n+1)}\overset{J^{T}_{R}}{\to}R^{n+1}\to\operatorname{coker}J^{T}_{R}\to
0.$

Consequently, to obtain the minimal graded $R$-free resolution of
$\operatorname{Der}_{R\mid k}$, we may construct the minimal graded $R$-free
resolution of $\operatorname{coker}J^{T}_{R}$ and truncate it.

Our main result is the following. For the proof, we build the appropriate
differential graded (dg) structures to use the construction of the relative
bar resolution developed by Iyengar in [11].

######  Theorem (Theorem 6.3).

Let $\mathcal{A}$ be the minimal graded $Q$-free resolution of $R$ and
$\mathcal{U}$ be the minimal graded $Q$-free resolution of
$\operatorname{coker}J^{T}_{R}$. The minimal graded $R$-free resolution of
$\operatorname{Der}_{R\mid k}$ is the truncated relative bar resolution

$\ldots\longrightarrow
R^{n(n+1)(n+2)}\overset{\small\begin{bmatrix}J^{T}_{R}&\varphi_{2}\\\
M_{1,1}&M_{2,0}\end{bmatrix}}{\longrightarrow}R^{2n(n+1)}\overset{\small\begin{bmatrix}\partial_{2}&M_{1,0}\end{bmatrix}}{\longrightarrow}\operatorname{Der}_{R\mid
k}\longrightarrow 0,$

where $M_{i,j}$ comes from the action $\mathcal{A}_{i}\cdot\mathcal{U}_{j}$ of
the dg algebra $\mathcal{A}$ on the dg $\mathcal{A}$-module $\mathcal{U}$ and
$\varphi_{2}$ and $\partial_{2}$ come from the differentials of $\mathcal{A}$
and $\mathcal{U}$, respectively.

It is not always the case that this construction produces a minimal
resolution, but when the ring and module involved are Golod and the
resolutions of them used in the construction are minimal, it does. Our minimal
free resolutions are graded, and we refrain from saying so from here on. We do
not indicate shifts. In our setting, $\mathcal{A}$ is the minimal $Q$-free
resolution of $R$, which is Golod, and $\mathcal{U}$ is the minimal $Q$-free
resolution of $\operatorname{coker}J^{T}_{R}$, which is also Golod. The
resolution $\mathcal{A}$ is known to admit a dg algebra structure, and we show
that $\mathcal{U}$ admits a dg $\mathcal{A}$-module structure. The relative
bar resolution then yields the minimal $R$-free resolution of
$\operatorname{coker}J^{T}_{R}$, in which the first differential is
$J^{T}_{R}$. Truncating the relative bar resolution results in the minimal
$R$-free resolution of $\operatorname{Der}_{R\mid k}$.

The outline of sections is as follows. Section 2 introduces the necessary
background, including the dg algebra structure on the Hilbert-Burch complex
$\mathcal{A}$, and the specifics of this structure prove that $R$ is Golod.
Section 3, and in particular Lemma 3.8, discusses the relationship between
minors and partial derivatives, to be used in several subsequent proofs. In
Section 4, we give the minimal $Q$-free resolution $\mathcal{U}$ of
$\operatorname{coker}J^{T}_{R}$, where $J^{T}_{R}$ is the transpose of the
Jacobian of the maximal minors of $X$ tensored down to $R$. In Section 5, we
compute an explicit dg algebra action of $\mathcal{A}$ on $\mathcal{U}$ which
shows that $\operatorname{coker}J^{T}_{R}$ is Golod, so that in Section 6 we
may use the relative bar resolution via dg structures to obtain the minimal
$R$-free resolution of $\operatorname{Der}_{R\mid k}$.

##  2\. Background

In this section, we recall the background necessary to talk about the
ingredients going into the main result, Theorem 6.3. Throughout, let $k$ be a
field and $Q=k[x_{i,j}\mid 1\leq i\leq n,\,1\leq j\leq n+1]$, where $n$ is at
least 2. Take $X$ to be the $n\times(n+1)$ matrix of variables with
$(i,j)$-entry $x_{i,j}$. For $r\in\\{1,\ldots,n+1\\}$, denote by $F_{r}$ the
maximal minor of $X$ obtained by deleting the $r$th column. For lists of
positive integers $\alpha$ and $\beta$, write $X_{\alpha}^{\beta}$ for the
matrix obtained by deleting the columns with labels in $\alpha$ and rows with
labels in $\beta$, e.g., $F_{r}=\det X_{r}$.

Write $I_{n}$ for the ideal of $Q$ generated by the $n\times n$ (i.e.,
maximal) minors of $X$, i.e., $I_{n}=(F_{1},F_{2},\ldots,F_{n+1})$, and set
$R=Q/I_{n}$.

######  Definition 2.1.

The ring $R$ is called a determinantal ring, and $I_{n}$ is called a
determinantal ideal.

More generally, determinantal rings are defined by the $k\times k$ minors of
an $n\times m$ generic matrix; we simply focus on the case when $k=n$ and
$m=n+1$ (and note that our methods do not easily generalize). Much is known
about determinantal rings and ideals $-$ see, for example, [3]. One fact used
in this work is that the height of $I_{n}$ is 2 (see, for example, [2],
Theorem 7.3.1). Another relates to the minimal free resolution of $R$ over
$Q$, but before stating it, we introduce differential graded algebras and
differential graded modules.

######  Definition 2.2.

A differential graded (dg) algebra over a ring $S$ is a complex
$(A,\partial^{A})$ of free $S$-modules equipped with a unitary, associative
multiplication $A\otimes_{S}A\to A$ satisfying

  * (i)

$A_{i}A_{j}\subseteq A_{i+j}$,

  * (ii)

$\partial^{A}(a_{i}a_{j})=\partial^{A}(a_{i})a_{j}+(-1)^{i}a_{i}\partial^{A}(a_{j})$,

  * (iii)

$a_{i}a_{j}=(-1)^{ij}a_{j}a_{i}$,

  * (iv)

$a_{i}^{2}=0\text{ if $i$ is odd}$,

where $a_{\ell}\in A_{\ell}$.

######  Definition 2.3.

Let $(A,\partial^{A})$ be a dg algebra over a ring $S$. A differential graded
(dg) $A$-module is a complex $(U,\partial^{U})$ of free $S$-modules together
with a unitary, associative multiplication $A\otimes_{S}U\to U$ satisfying

  * (i)

$A_{i}U_{j}\subseteq U_{i+j}$,

  * (ii)

$\partial^{U}(a_{i}u_{j})=\partial^{A}(a_{i})u_{j}+(-1)^{i}a_{i}\partial^{U}(u_{j})$,

where $a_{i}\in A_{i}$ and $u_{j}\in U_{j}$.

A natural question is whether the minimal free resolution of a given module
has the structure of a dg algebra. For ideals of height 2 in a Cohen-Macaulay
ring, the answer is given by the Hilbert-Burch complex, shown to be a
resolution in the polynomial ring case by Hilbert in 1890 [10] and more
generally by Burch in 1968 [5]. The dg algebra structure is due to Herzog [9].
It is well-known that the height of $I_{n}$ is 2 (see, for example, [2],
Theorem 7.3.1), and so we state the next proposition in terms of $R$ and $Q$.

######  Proposition 2.4.

[5, 9, 10] The minimal $Q$-free resolution of $R$ is the Hilbert-Burch complex

$0\longrightarrow
Q^{n}\overset{\varphi_{2}}{\longrightarrow}Q^{n+1}\overset{\varphi_{1}}{\longrightarrow}Q\longrightarrow
R\longrightarrow 0,$

where
$\varphi_{1}=\begin{bmatrix}F_{1}&-F_{2}&\ldots&(-1)^{i+1}F_{i}&\ldots&(-1)^{n+2}F_{n+1}\end{bmatrix}$
and $\varphi_{2}=X^{T}$. Furthermore, the resolution $\mathcal{A}$ is a dg
algebra under the multiplication

$\displaystyle e_{i}\cdot e_{j}=-(e_{j}\cdot e_{i})$
$\displaystyle=\begin{cases}\displaystyle\sum_{k=1}^{n}(-1)^{i+j+k}\det
X^{k}_{i,j}T_{k},&\text{if }i<j,\\\ \\\ 0,&\text{if }i=j,\end{cases}$
$\displaystyle e_{i}\cdot T_{j}$ $\displaystyle=0,$ $\displaystyle T_{i}\cdot
T_{j}$ $\displaystyle=0,$

where $\\{1\\}$ is a basis for $\mathcal{A}_{0}$, $\\{e_{i}\\}$ is a basis for
$\mathcal{A}_{1}$, and $\\{T_{i}\\}$ is a basis for $\mathcal{A}_{2}$.

######  Remark 2.5.

More generally, the Eagon-Northcott complex resolves a determinantal ring
defined by the $k\times k$ minors of an $n\times m$ generic matrix over the
ambient polynomial ring [7]. In our case, the Hilbert-Burch and Eagon-
Northcott complexes coincide. The Eagon-Northcott complex admits the structure
of a dg algebra, computed by Srinivasan in [13].

######  Definition 2.6.

Given a surjective homomorphism of local rings
$(S,\mathfrak{m}_{S},k_{S})\to(S/J,\mathfrak{m}_{S/J},k_{S})$ and a finitely-
generated $S/J$-module $N$, there is a coefficientwise inequality of Poincaré
series

$P_{N}^{S/J}(t)\preceq\dfrac{P^{S}_{N}(t)}{1-t(P^{S}_{S/J}(t)-1)}.$

In the case that equality holds, $N$ is called Golod. We call the ring $S/J$
Golod if $k_{S}$ is Golod.

When the minimal free resolution of $S/J$ over $S$ is a dg algebra, the next
proposition provides an equivalent condition for $S/J$ to be Golod. We phrase
it in terms of $R$, $Q$, and $\mathcal{A}$.

######  Proposition 2.7 (Proposition 5.2.4, [1]).

The ring $R$ is Golod if and only if $\mathcal{A}_{\geq
1}\cdot\mathcal{A}_{\geq 1}\subseteq\mathfrak{m}_{Q}\mathcal{A}$, where
$\mathfrak{m}_{Q}$ is the (homogeneous) maximal ideal of $Q$. Similarly, an
$S/J$-module $N$ is Golod if and only if its minimal $Q$-free resolution
$\mathcal{U}$ is a dg module over $\mathcal{A}$ with structure satisfying that
the action of $\mathcal{A}_{\geq 1}$ on $\mathcal{U}$ lands in
$\mathfrak{m}_{Q}\mathcal{U}$.

######  Corollary 2.8.

The ring $R$ is Golod.

###### Proof.

The dg algebra structure of the Hilbert-Burch complex satisfies
$\mathcal{A}_{\geq 1}\cdot\mathcal{A}_{\geq
1}\subseteq\mathfrak{m}_{Q}\mathcal{A}$. ∎

The construction of the relative bar resolution via dg structures due to
Iyengar [11] is the main tool in this work. Given a dg algebra $A$ over a ring
$S$ resolving $S/J$ and a dg $A$-module $U$ over $S$ resolving an $S/J$-module
$N$, set

$B_{r}=\bigoplus_{i_{1}+\ldots+i_{p}+j+p=r}S/J\otimes_{S}\overline{A_{i_{1}}}\otimes_{S}\ldots\otimes_{S}\overline{A_{i_{p}}}\otimes_{S}U_{j},$

where the bars mean to shift by 1 homological degree.

Define $m_{n}:A^{\otimes n}\to A$ iteratively by $m_{1}=\partial^{A}$ and
$m_{n}$ ($n>1$) by the characterization that it satisfies the Stasheff
identity

$\sum_{s=1}^{n}\sum_{r+s+t=n}(-1)^{r+st}m_{r+1+t}(1^{\otimes r}\otimes
m_{s}\otimes 1^{\otimes t})=0,$

in which $r,t\geq 0$. Define $\mu_{n}:A^{\otimes n-1}\otimes U\to U$ such that
analogous identities hold.

######  Definition 2.9.

The relative bar resolution has $B_{r}$ as its $r$th term and differential

$\displaystyle\partial(s\otimes a_{1}\otimes\ldots\otimes a_{p}\otimes u)$
$\displaystyle=\sum_{i=1}^{p}\sum_{j=0}^{p-i}\pm s\otimes
a_{1}\otimes\ldots\otimes a_{j}\otimes m_{i}(a_{j+1}\otimes a_{j+i})\otimes
a_{j+i+1}\ldots\otimes a_{p}\otimes u$ $\displaystyle+\sum_{i=1}^{p+1}\pm
s\otimes a_{1}\otimes\ldots\otimes
a_{p-i+1}\otimes\mu_{i}(a_{p-i+2}\otimes\ldots\otimes a_{p}\otimes u).$

Exact signs are suppressed, and they are unimportant to the moral of the
story.

######  Remark 2.10.

The relative bar resolution is a resolution of $N$ over $S/J$. By considering
the construction of the differentials in the relative bar resolution via dg
structures and Proposition 2.7, one observes that if $S/J$ is Golod, $N$ is an
$S/J$-module which is Golod, and both $A$ and $U$ are minimal, then the
relative bar resolution resolving $N$ over $S/J$ is minimal. Thus, in this
setting, the Poincaré series for $N$ over $S/J$ may be realized as in
Definition 2.6 when equality holds. We will show that this is the case in our
setting, where $S=Q$, $S/J=R$, and $N$ is the cokernel of the Jacobian matrix
$J^{T}_{R}$ whose kernel is $\operatorname{Der}_{R\mid k}$ (to be defined in a
moment). The following will be expanded upon in Corollary 6.7:

$P^{R}_{N}(t)=\dfrac{P^{Q}_{N}(t)}{1-t(P^{Q}_{R}(t)-1)}.$

The module we will be resolving with a truncation of the relative bar
resolution is the module of $k$-linear derivations on $R$.

######  Definition 2.11.

A $k$-linear derivation on a $k$-algebra $S$ is a $k$-linear function
$\delta:S\to S$ satisfying the Leibniz rule:

$\delta(ab)=\delta(a)b+a\delta(b),$

for all $a,b\in S$. The collection of $k$-linear derivations on $S$ forms an
$S$-module under addition of functions and $S$-scalar multiplication. Denote
this $S$-module $\operatorname{Der}_{S\mid k}$.

######  Example 2.12.

A $k$-linear derivation $\delta$ on the polynomial $k$-algebra
$k[x_{1},\ldots,x_{s}]$ can be written

$\delta=\sum_{j=1}^{m}\alpha_{j}\dfrac{\partial}{\partial x_{j}},$

where $\alpha_{j}\in k$ and $\dfrac{\partial}{\partial x_{j}}$ is partial
differentiation with respect to $x_{j}$. The prototypical example of a
derivation that does not look like a derivation is the Euler derivation

$E(f):=\deg(f)\cdot f,$

where $f$ is a homogeneous polynomial. One can express
$E\in\operatorname{Der}_{Q\mid k}$ (or in $\operatorname{Der}_{R\mid k}$,
where $x_{k,\ell}$ is understood to be an equivalence class) as

$E=\displaystyle\sum_{k=1}^{n}\sum_{\ell=1}^{n+1}x_{k,\ell}\dfrac{\partial}{\partial
x_{k,\ell}}.$

Write $J^{T}_{Q}$ for the transpose of the Jacobian matrix of $I_{n}$ in $Q$.
That is, the $i$th row of $J^{T}_{Q}$ is indexed by $F_{i}$ and the columns of
$J^{T}_{Q}$ are indexed by the indeterminates $x_{i,j}$ according to the rule
$x_{i,j}\prec x_{i^{\prime},j^{\prime}}$ precisely when $i\leq i^{\prime}$ and
if $i=i^{\prime}$, $j\leq j^{\prime}$, i.e.,

$x_{1,1}\prec x_{1,2}\prec\ldots\prec x_{1,n+1}\prec x_{2,1}\prec
x_{2,2}\prec\ldots\prec x_{n,n+1},$

so that the $(i,(j,k))$-entry of $J^{T}_{Q}$ is $\dfrac{\partial
F_{i}}{\partial x_{j,k}}$. We define $J^{T}_{R}$ to be $J^{T}_{Q}\otimes_{Q}R$
and keep the above labeling convention throughout this work.

######  Example 2.13.

In the $2\times 3$ case,

$X=\begin{bmatrix}x_{1,1}&x_{1,2}&x_{1,3}\\\
x_{2,1}&x_{2,2}&x_{2,3}\end{bmatrix},$

$F_{1}=x_{1,2}x_{2,3}-x_{1,3}x_{2,2}$, $F_{2}=x_{1,1}x_{2,3}-x_{1,3}x_{2,1}$,
and $F_{3}=x_{1,1}x_{2,2}-x_{1,2}x_{2,1}$, so that

$J^{T}_{Q}=\begin{bmatrix}0&x_{2,3}&-x_{2,2}&0&-x_{1,3}&x_{1,2}\\\
x_{2,3}&0&-x_{2,1}&-x_{1,3}&0&x_{1,1}\\\
x_{2,2}&-x_{2,1}&0&-x_{1,2}&x_{1,1}&0\end{bmatrix}.$

######  Remark 2.14.

Notice that $\dfrac{\partial F_{i}}{\partial x_{k,i}}=0$ for all $i$ and $k$,
since to compute $F_{i}$, the $i$th column of $X$ is deleted.

##  3\. Useful Lemmas

In this section we further explore the relationship between minors and partial
derivatives, developing some useful identities. These identities are involved
primarily in the proofs of Lemmas 4.5 and 4.6, which combine with the
Buchsbaum-Eisenbud Acyclicity Theorem (Theorem 4.7) to produce the minimal
$Q$-free resolution $\mathcal{U}$ of $\operatorname{coker}J^{T}_{R}$, and in
the proof of the dg action of the Hilbert-Burch dg algebra $\mathcal{A}$
(Proposition 2.4) on $\mathcal{U}$ (Lemma 5.1).

######  Remark 3.1.

Recall that for lists of positive integers $\alpha$ and $\beta$, we write
$X_{\alpha}^{\beta}$ for the matrix obtained by deleting the columns with
labels in $\alpha$ and rows with labels in $\beta$. To compute $F_{i}$, one
starts with $X$ and deletes column $i$ to obtain the matrix $X_{i}$. The
determinant of $X_{i}$ is $F_{i}$, and one can compute this determinant via
Laplace expansion along the column of $X_{i}$ labeled $j$ (when $i\neq j$).
Finding $\dfrac{\partial F_{i}}{\partial x_{k,j}}$ thus amounts to finding the
determinants of $X_{i,j}^{k}$ for varying $k$, i.e., the polynomial
coefficient of $x_{k,j}$ in $F_{i}$. However, if $i<j$, the sign associated to
$x_{k,j}$ in $X$, which is $(-1)^{j+k}$, switches to $(-1)^{j+k+1}$ in
$X_{i}$. In light of this observation, we note that when $i\neq j$, we can
rewrite the multiplication in the dg algebra structure of the Hilbert-Burch
complex (Proposition 2.4) as

$e_{i}\cdot
e_{j}=\begin{cases}\sum_{k=1}^{n}(-1)^{i+j+k}(-1)^{j+k+1}\dfrac{\partial
F_{i}}{\partial x_{k,j}}T_{k}=\sum_{k=1}^{n}(-1)^{i+1}\dfrac{\partial
F_{i}}{\partial x_{k,j}}T_{k},&\text{if }i<j,\\\ \\\
\sum_{k=1}^{n}(-1)^{i+j+k}(-1)^{j+k}\dfrac{\partial F_{i}}{\partial
x_{k,j}}T_{k}=\sum_{k=1}^{n}(-1)^{i}\dfrac{\partial F_{i}}{\partial
x_{k,j}}T_{k},&\text{if }i>j.\end{cases}$

######  Remark 3.2.

We claim that

$\dfrac{\partial F_{i}}{\partial x_{k,j}}=(-1)^{i+j+1}\dfrac{\partial
F_{j}}{\partial x_{k,i}}.$

Without loss of generality, let $i>j$. We have

$\displaystyle(-1)^{k+j}\dfrac{\partial F_{i}}{\partial x_{k,j}}=\det
X_{i,j}^{k}=(-1)^{k+i+1}\dfrac{\partial F_{j}}{\partial x_{k,i}},$

from which the claim follows.

######  Definition 3.3.

The rank of a matrix $M$ (or of a map represented by a matrix $M$) is the size
of the largest non-vanishing minor of $M$.

######  Lemma 3.4.

The rank of $J^{T}_{Q}$ is $n+1$.

###### Proof.

Consider the maximal minor of $J^{T}_{Q}$ determined by the last column and
the $n$ columns with first entries $\dfrac{\partial F_{1}}{\partial x_{u,1}}$,
$u\in\\{1,\ldots,n\\}$. These entries are zero, and so the minor in question
is a product of $\pm\dfrac{\partial F_{1}}{\partial x_{n,n+1}}$ and the
determinant of a matrix $Y$ with entries $\dfrac{\partial F_{r\neq
1}}{\partial x_{u,1}}$, $u\in\\{1,\ldots,n\\}$.

Specialize by sending $x_{i,i+1}$ to itself for $i=1,\ldots,n-1$ and $x_{i,j}$
to zero for $i\neq j-1$. The matrix $Y$ becomes diagonal under this
specialization, since the entries that were $\dfrac{\partial F_{u+1}}{\partial
x_{u,1}}$ remain nonzero (since one summand is $\pm\displaystyle\sum_{i\neq
u}x_{i,i+1}$) and the entries that were $\dfrac{\partial F_{r}}{\partial
x_{u,1}}$ become zero for all $u$ and $r\neq u+1$. The entry $\dfrac{\partial
F_{1}}{\partial x_{n,n+1}}$ of $J^{T}_{Q}$ remains nonzero under this
specialization, as well.

The determinant of $Y$ under the specialization is thus a product of nonzero
entries in a domain, and so the minor we considered at the start of the proof
is nonzero. Thus, $J^{T}_{Q}$ has full rank $n+1$. ∎

######  Example 3.5.

In the $2\times 3$ case, i.e., the case when $n=2$, we have that
$\dfrac{\partial F_{1}}{\partial x_{2,3}}=x_{1,2}$. The matrix $Y$ from the
proof of Lemma 3.4 is

$Y=\begin{bmatrix}x_{2,3}&-x_{1,3}\\\ x_{2,2}&-x_{1,2}\end{bmatrix},$

and after specialization, we get

$\begin{bmatrix}x_{2,3}&0\\\ 0&-x_{1,2}\end{bmatrix}.$

######  Definition 3.6.

The grade against $Q$ of a $Q$-ideal $I$ is the length of the longest
$Q$-sequence in $I$, denoted $\operatorname{grade}(I,Q)$.

######  Remark 3.7.

The grade against $Q$ of the $Q$-ideal generated by the rank-size minors of
$J^{T}_{Q}$ is at least 1, since this ideal is nonzero by Lemma 3.4 and $Q$ is
a domain.

The next lemma provides some identities involving partials and minors that
will be used in developing the minimal $Q$-free resolution of
$\operatorname{coker}J^{T}_{R}$ in Section 4 and the differential graded
structure appearing in Section 5.

######  Lemma 3.8.

There are equalities

(1) $\displaystyle\sum_{u=1}^{n+1}x_{i,u}\dfrac{\partial F_{r}}{\partial
x_{s,u}}=\begin{cases}0,&i\neq s,\\\ F_{r},&i=s,\end{cases}$ (2)
$\displaystyle\sum_{k=1}^{n}x_{k,i}\dfrac{\partial F_{i}}{\partial
x_{k,j}}=(-1)^{i+j+1}F_{j}\,\text{ when }i\neq j,$ (3)
$\displaystyle\sum_{r\neq i}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{j,r}=(-1)^{i+1}F_{i},$ (4)
$\displaystyle\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,t}=0\text{ when $i$, $k$, and $t$ are distinct}.$

###### Proof.

To see the case of (1) when $i\neq s$, let $Y$ be the matrix obtained from
$X_{r}$ by replacing the row indexed $s$ with a second row indexed $i$. The
determinant of $Y$ is zero, and one way to compute $\det Y$ is via Laplace
expansion along one of the rows indexed by $i$ in $Y$. Explicitly, this
computation is

$0=\det Y=\pm\sum_{u=1}^{n+1}x_{i,u}\dfrac{\partial F_{r}}{\partial x_{s,u}}.$

This yields the case of (1) in which $i\neq s$.

When $i=s$, the result follows from the observation that the sum in (1) is the
calculation of $F_{r}$ via Laplace expansion along the row of $X_{r}$ indexed
$s$.

For equality (2), we examine $x_{k,i}\dfrac{\partial F_{i}}{\partial
x_{k,j}}$:

$x_{k,i}\dfrac{\partial F_{i}}{\partial
x_{k,j}}=\begin{cases}(-1)^{j+k+1}x_{k,i}\det X_{i,j}^{k},&i<j,\\\
(-1)^{j+k}x_{k,i}\det X_{i,j}^{k},&i>j.\end{cases}$

By Laplace expansion to compute $F_{j}$ along the column of $X_{j}$ indexed
$i$,

$\sum_{k=1}^{n}(-1)^{i+k+1}x_{k,i}\det X_{i,j}^{k}=F_{j}$

when $i>j$, and

$\sum_{k=1}^{n}(-1)^{i+k}x_{k,i}\det X_{i,j}^{k}=F_{j}$

when $i<j$. Equality (2) follows:

$\sum_{k=1}^{n}x_{k,i}\dfrac{\partial F_{i}}{\partial
x_{k,j}}=(-1)^{i+j+1}F_{j}.$

Alternatively, equation (2) follows from Remark 3.2 and Laplace expansion as
in the $i=s$ case of equation (1).

To see (3), consider the determinant of $X_{i}$ as computed via Laplace
expansion along the row indexed by $j$:

$\displaystyle F_{i}=\det X_{i}$
$\displaystyle=\sum_{r=1}^{i-1}(-1)^{j+r}x_{j,r}\det
X_{i,r}^{j}+\sum_{r=i+1}^{n}(-1)^{j+r+1}x_{j,r}\det X_{i,r}^{j}.$

When $i>r$,

$\dfrac{\partial F_{r}}{\partial x_{j,i}}=(-1)^{i+j+1}\det X_{i,r}^{j},$

and when $i<r$,

$\dfrac{\partial F_{r}}{\partial x_{j,i}}=(-1)^{i+j}\det X_{i,r}^{j}.$

Thus,

$\displaystyle F_{i}$
$\displaystyle=\sum_{r=1}^{i-1}(-1)^{i+r+1}x_{j,r}\dfrac{\partial
F_{r}}{\partial x_{j,i}}+\sum_{r=i+1}^{n}(-1)^{i+r+1}x_{j,r}\dfrac{\partial
F_{r}}{\partial x_{j,i}}$ $\displaystyle=(-1)^{i+1}\sum_{r\neq
i}(-1)^{r}\dfrac{\partial F_{r}}{\partial x_{j,i}}x_{j,r},$

from which (3) follows.

To see (4), let $Y$ be the matrix obtained from $X_{i}$ by replacing the
column indexed $k$ by a second column indexed $t$. By Laplace expansion along
one of the columns indexed $t$ in $Y$,

$0=\det Y=\pm\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,t}.$

∎

######  Example 3.9.

In the $2\times 3$ case with $i=2$ and $j=1$, equation (3) in Lemma 3.8 reads

$\displaystyle\sum_{r\neq 2}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{1,2}}x_{1,r}$ $\displaystyle=-\dfrac{\partial F_{1}}{\partial
x_{1,2}}x_{1,1}-\dfrac{\partial F_{3}}{\partial x_{1,2}}x_{1,3}$
$\displaystyle=-x_{2,3}x_{1,1}+x_{2,1}x_{1,3}$ $\displaystyle=-F_{2}.$

##  4\. Resolving the Cokernel over $Q$

In the introduction, we noted that $\operatorname{Der}_{R\mid k}$ is
isomorphic to the kernel of $J^{T}_{R}$, i.e., there is an exact sequence of
$R$-modules

$0\to\operatorname{Der}_{R\mid k}\to
R^{n(n+1)}\overset{J^{T}_{R}}{\to}R^{n+1}\to\operatorname{coker}J^{T}_{R}\to
0.$

Resolving $\operatorname{Der}_{R\mid k}$ over $R$ thus amounts to resolving
$\operatorname{coker}J^{T}_{R}$ over $R$. To do so, we first resolve
$\operatorname{coker}J^{T}_{R}$ over $Q$, and that is the content of this
section. Using the resolution we develop in this section (see Theorem 6.2)
together with the Hilbert-Burch complex from Proposition 2.4, we construct the
relative bar resolution of $\operatorname{coker}J^{T}_{R}$ over $R$ in Section
6. “Chopping off” the first term of this resolution results in the minimal
$R$-free resolution of $\operatorname{Der}_{R\mid k}$.

We start by examining what will be the second differential in the $Q$-free
resolution of $\operatorname{coker}J^{T}_{R}$. Let $A_{k}$ be the
$(n+1)\times(n-1)$ matrix obtained from the $(n+1)\times n$ matrix

$X^{T}=\begin{bmatrix}x_{1,1}&x_{2,1}&\ldots&x_{n,1}\\\
x_{1,2}&x_{2,2}&\ldots&x_{n,2}\\\ \vdots&\vdots&\vdots&\vdots\\\
x_{1,n+1}&x_{2,n+1}&\ldots&x_{n,n+1}\\\ \end{bmatrix}$

by removing the $k$th column. Let $\vec{e}_{\ell}$ be the $\ell$th
$n(n+1)\times 1$ standard basis vector and let $B_{\ell}$ be the $n(n+1)\times
1$ column matrix

$\sum_{s=1}^{n+1}\left(-x_{1,s}\vec{e}_{s}+x_{\ell,s}\vec{e}_{(\ell-1)(n+1)+s}\right).$

Let $\partial_{2}:Q^{(n-1)(n+1)}\to Q^{n(n+1)}$ be the map of free $Q$-modules
be given by the block matrix

$\begin{bmatrix}A_{1}&0&\ldots&0&0&|&|&|&|\\\ 0&A_{2}&\ldots&0&0&|&|&|&|\\\
\vdots&\vdots&\ddots&\vdots&\vdots&B_{2}&B_{3}&\ldots&B_{n}\\\
0&0&\ldots&A_{n-1}&0&|&|&|&|\\\ 0&0&\ldots&0&A_{n}&|&|&|&|\end{bmatrix}$

Let $I$ be the ideal of $(n-1)(n+1)\times(n-1)(n+1)$ (i.e., maximal) minors of
$\partial_{2}$.

######  Lemma 4.1.

We have that $\operatorname{grade}(I,Q)$ is at least 2.

###### Proof.

For a matrix $Y$ with entries in $Q$, denote by $I(Y)$ the $Q$-ideal generated
by the maximal minors of $Y$.

Take an $(n-1)(n+1)$ minor of $\partial_{2}$ by choosing $n-1$ rows from the
collection of $n+1$ rows that intersects $A_{1}$ and $n$ rows from each of the
$n-1$ collections of $n+1$ rows that intersect each $A_{k}$, $k\geq 2$. Using
column operations, one sees that the matrix corresponding to this minor can be
calculated along a block diagonal as an element of $I(X^{1})\cdot I(X)^{n-1}$.
The collection of minors taken this way generates the ideal $I(X^{1})\cdot
I(X)^{n-1}$, so that $I(X^{1})\cdot I(X)^{n-1}\subseteq I$. Using that
$I(X)^{n-1}\subseteq I(X)\subseteq I(X^{1})$, we have

$\sqrt{I(X^{1})\cdot I(X)^{n-1}}=\sqrt{I(X^{1})\cap
I(X)^{n-1}}=\sqrt{I(X)^{n-1}}=\sqrt{I(X)}=I(X).$

We use above that $I(X)$ is prime and thus radical. We note that
$\operatorname{ht}\left(\sqrt{I(X^{1})\cdot
I(X)^{n-1}}\right)=\operatorname{ht}\left(I(X^{1})\cdot I(X)^{n-1}\right)$ and
that $\operatorname{grade}(I(X),Q)=2$ (see, for example, [2], Theorem 7.3.1).
Since height and grade coincide in the case of a Cohen-Macaulay ring,
$\operatorname{grade}(I,Q)\geq 2$. ∎

######  Example 4.2.

In the $3\times 4$ case, the matrix $\partial_{2}$ is

$\begin{bmatrix}{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}-x_{1,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}-x_{1,1}}\\\
x_{2,2}&x_{3,2}&0&0&0&0&-x_{1,2}&-x_{1,2}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}-x_{1,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}-x_{1,3}}\\\
x_{2,4}&x_{3,4}&0&0&0&0&-x_{1,4}&-x_{1,4}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,2}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,2}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,2}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}\\\
0&0&x_{1,4}&x_{3,4}&0&0&x_{2,4}&0\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,1}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,1}}\\\
0&0&0&0&x_{1,2}&x_{2,2}&0&x_{3,2}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,3}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,3}}\\\
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{1,4}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{2,4}}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}0}&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}x_{3,4}}\\\
\end{bmatrix},$

in which we have highlighted a maximal minor described in the proof of Lemma
4.1.

######  Remark 4.3.

In Lemma 4.1, the minors we take are products of nonzero elements of $Q$,
which is a domain (the factors are determinants of matrices of distinct
indeterminates, hence nonzero). Thus, the map $\partial_{2}$ has full rank
$(n-1)(n+1)$.

Having said something about the map we will claim as the second differential
in the resolution of $\operatorname{coker}J^{T}_{R}$ over $Q$, we turn now to
the rest of the resolution. We begin with a lemma that will aid in showing
that the resolution is indeed a complex.

######  Lemma 4.4.

Let $S=k[x_{1},\ldots,x_{m}]$ and $f\in S$. Let
$\delta\in\operatorname{Der}_{S\mid k}$ have coordinate vector $\vec{\alpha}$,
and let $J^{T}_{S}(f)$ be the transpose of the Jacobian of $f$ over $S$. The
composition $J^{T}_{S}(f)\circ\vec{\alpha}=\vec{0}$ over $S$ if and only if
$\delta(f)=0$.

###### Proof.

Say $\delta=\sum_{j=1}^{m}\alpha_{j}\dfrac{\partial}{\partial x_{j}}$. We have
that $J^{T}_{S}(f)\circ\vec{\alpha}$ is equal to

$\sum_{j=1}^{m}\dfrac{\partial f}{\partial
x_{j}}\alpha_{j}=\sum_{j=1}^{m}\alpha_{j}\dfrac{\partial f}{\partial
x_{j}}=\delta(f).$

∎

######  Lemma 4.5.

The sequence of maps and free modules

$Q^{(n-1)(n+1)}\overset{\partial_{2}}{\longrightarrow}Q^{n(n+1)}\overset{J^{T}_{Q}}{\longrightarrow}Q^{n+1}$

is a complex.

###### Proof.

We want to show that $J^{T}_{Q}\circ\partial_{2}=0$. By Lemma 4.4, it suffices
to show that the columns of $\partial_{2}$ correspond to $k$-linear
derivations on $Q$ that send the polynomials $F_{r}$ (the generators of
$I_{n})$ to zero. Define

$V_{i,s}:=\sum_{u=1}^{n+1}x_{i,u}\dfrac{\partial}{\partial x_{s,u}}.$

For $i\neq s$, $V_{i,s}$ is the derivation corresponding to the column of
$\partial_{2}$ having first nonzero entry $x_{i,1}$. We will use $B_{\ell}$ to
denote the derivation corresponding to the column $B_{\ell}$. Notice that
$B_{\ell}=-V_{1,1}+V_{\ell,\ell}$.

That $V_{i,s}(F_{r})=0$ when $i\neq s$ and $V_{s,s}(F_{r})=F_{r}$ is the
content of Lemma 3.8, equation (1). Since $V_{s,s}(F_{r})=F_{r}$, we get that
$B_{\ell}(F_{r})=0$.

By Lemma 4.4, $J^{T}_{Q}\circ\partial_{2}=0$. ∎

We could extend the complex from Lemma 4.5 to the right to include a
surjection onto $\operatorname{coker}J^{T}_{Q}$. However, we would like to
resolve $\operatorname{coker}J^{T}_{R}$ over $Q$. The next lemma states that
this is not an issue.

######  Lemma 4.6.

As $Q$-modules,
$\operatorname{coker}J^{T}_{Q}\cong\operatorname{coker}J^{T}_{R}$.

###### Proof.

We have that

$\operatorname{coker}J^{T}_{R}=\frac{R^{n+1}}{\hskip 2.84526pt\text{im}\hskip
2.84526ptJ^{T}_{R}}\cong\frac{(Q/I)^{n+1}}{\hskip 2.84526pt\text{im}\hskip
2.84526ptJ^{T}_{R}}\cong\frac{Q^{n+1}}{\langle F_{i}\vec{e}_{j}\rangle+\hskip
2.84526pt\text{im}\hskip 2.84526ptJ^{T}_{Q}},$

where $\\{\vec{e}_{j}\\}$ is a basis for the $j$th copy of $Q$ and $\langle
F_{i}\vec{e}_{j}\rangle$ is the ideal of $Q^{n+1}$ generated by the
$F_{i}\vec{e}_{j}$, $i,j=1,\ldots,n+1$. The module $\dfrac{Q^{n+1}}{\langle
F_{i}\vec{e}_{j}\rangle+\hskip 2.84526pt\text{im}\hskip 2.84526ptJ^{T}_{Q}}$
is isomorphic to the cokernel of a matrix equal to $J^{T}_{Q}$ with adjoined
columns that are standard basis vectors scaled by the $F_{i}$. Call this
matrix $D$. The lemma follows if we can show that the columns of $D$ are
$Q$-linear combinations of the columns of $J^{T}_{Q}$.

Write $b_{i,j}$ for the column of $J^{T}_{Q}$ corresponding to the partial
derivative with respect to $x_{i,j}$. When $i\neq j$, we claim that

$\sum_{k=1}^{n}x_{k,i}b_{k,j}=(-1)^{i+j+1}F_{j}\vec{e}_{i}.$

The $r$th entry of the vector $\sum_{k=1}^{n}x_{k,i}b_{k,j}$ is

$\sum_{k=1}^{n}x_{k,i}\dfrac{\partial F_{r}}{\partial
x_{k,j}}=\begin{cases}(-1)^{i+j+1}F_{j},&i=r,\\\ 0,&i\neq r.\end{cases}$

The $i=r$ case is Lemma 3.8, equation (2). The $i\neq r$ case is Lemma 3.8,
equation (4).

When $i=j$, we claim that

$\displaystyle\sum_{u\neq i}x_{i,u}b_{i,u}-\sum_{k\neq
i}x_{k,i}b_{k,i}=F_{i}\vec{e}_{i}.$

The $r$th entry of $\sum_{u\neq i}x_{i,u}b_{i,u}-\sum_{k\neq i}x_{k,i}b_{k,i}$
is

$\displaystyle\sum_{u\neq i}x_{i,u}\dfrac{\partial F_{r}}{\partial
x_{i,u}}-\sum_{k\neq i}x_{k,i}\dfrac{\partial F_{r}}{\partial x_{k,i}}.$

When $r=i$, this sum is $F_{i}$ by Laplace expansion computing $F_{i}$ along
the row of $X_{i}$ indexed by $i$. When $r\neq i$,

$\displaystyle\sum_{u\neq i}x_{i,u}\dfrac{\partial F_{r}}{\partial
x_{i,u}}-\sum_{k\neq i}x_{k,i}\dfrac{\partial F_{r}}{\partial x_{k,i}}$
$\displaystyle=\sum_{u=1}^{n+1}x_{i,u}\dfrac{\partial F_{r}}{\partial
x_{i,u}}-x_{i,i}\dfrac{\partial F_{r}}{\partial
x_{i,i}}-\left(\sum_{k=1}^{n}x_{k,i}\dfrac{\partial F_{r}}{\partial
x_{k,i}}-x_{i,i}\dfrac{\partial F_{r}}{\partial x_{i,i}}\right)$
$\displaystyle=F_{r}-x_{i,i}\dfrac{\partial F_{r}}{\partial
x_{i,i}}-F_{r}+x_{i,i}\dfrac{\partial F_{r}}{\partial x_{i,i}}$
$\displaystyle=0$

∎

To show that what we claim is the $Q$-free resolution of
$\operatorname{coker}J^{T}_{R}$ is in fact acyclic, we call upon the
Buchsbaum-Eisenbud Acyclicity Theorem.

######  Theorem 4.7 (Buchsbaum-Eisenbud Acyclicity [4]).

Let

$0\longrightarrow
F_{s}\overset{d_{s}}{\longrightarrow}F_{s-1}\longrightarrow\ldots\longrightarrow
F_{1}\overset{d_{1}}{\longrightarrow}F_{0}$

be a complex of free $Q$-modules in which $F_{i}=Q^{f_{i}}$. Let $r_{i}$ be
the rank of the matrix representing $d_{i}$. By convention, $r_{s+1}=0$. Then,
$\mathcal{F}$ is acyclic if and only if the following two conditions hold

  * (i)

$f_{i}=r_{i}+r_{i+1}$ for all $1\leq i\leq s$,

  * (ii)

for all $1\leq i\leq s$, we have that $\operatorname{grade}(I(d_{i}),Q)\geq
i$, where $I(d_{i})$ is the ideal of $Q$ generated by to $r_{i}\times r_{i}$
minors of $d_{i}$.

To end this section, we bring its contents together to obtain the minimal
$Q$-free resolution of $\operatorname{coker}J^{T}_{R}$.

######  Theorem 4.8.

The minimal $Q$-free resolution of $\operatorname{coker}J^{T}_{R}$ is

$0\longrightarrow
Q^{(n-1)(n+1)}\overset{\partial_{2}}{\longrightarrow}Q^{n(n+1)}\overset{J^{T}_{Q}}{\longrightarrow}Q^{n+1}\longrightarrow\operatorname{coker}J^{T}_{R}\longrightarrow
0$

###### Proof.

By Lemma 4.5 and Lemma 4.6, $\mathcal{U}$ is a complex. By Lemma 3.4 and
Remark 4.3, the ranks of $J^{T}$ and $\partial_{2}$ are maximal. By Lemma 4.1
and Remark 3.7, the grades of the ideals of rank-sized minors of the
differentials of $\mathcal{U}$ are large enough so that by the Buchsbaum-
Eisenbud Acyclicity Theorem (Theorem 4.7), we may conclude that $\mathcal{U}$
is a resolution. The entries of the matrices representing the differentials
are in the maximal ideal of $Q$, and so $\mathcal{U}$ is indeed minimal.

∎

##  5\. Differential Graded Structure

Given a dg algebra $\mathcal{A}$ resolving $R$ over $Q$ and a dg
$\mathcal{A}$-module $\mathcal{U}$ resolving an $R$-module $N$ over $Q$, one
can produce the relative bar resolution resolving $N$ over $R$ (Theorem 1.2,
[11]). This section develops the dg module structure required when the module
$N$ is $\operatorname{coker}J^{T}_{R}$.

We restate the resolutions $\mathcal{A}$ and $\mathcal{U}$ developed earlier,
with the free modules in them labeled by their bases. Recall that
$\mathcal{A}$ is the Hilbert-Burch complex (Proposition 2.4):

$0\longrightarrow\underset{\\{T_{i}\\}}{Q^{n}}\overset{\varphi_{2}}{\longrightarrow}\underset{\\{e_{i}\\}}{Q^{n+1}}\overset{\varphi_{1}}{\longrightarrow}\underset{\\{1\\}}{Q}\longrightarrow
R\longrightarrow 0,$

where
$\varphi_{1}=\begin{bmatrix}F_{1}&-F_{2}&\ldots&(-1)^{i+1}F_{i}&\ldots&(-1)^{n+2}F_{n+1}\end{bmatrix}$
and $\varphi_{2}=X^{T}$. Furthermore, the resolution $\mathcal{A}$ is a dg
algebra under the multiplication

$\displaystyle e_{i}\cdot e_{j}=-(e_{j}\cdot e_{i})$
$\displaystyle=\begin{cases}\displaystyle\sum_{k=1}^{n}(-1)^{i+j+k}\det
X^{k}_{i,j}T_{k},&\text{if }i<j,\\\ \\\ 0,&\text{if }i=j,\end{cases}$
$\displaystyle e_{i}\cdot T_{j}$ $\displaystyle=0,$ $\displaystyle T_{i}\cdot
T_{j}$ $\displaystyle=0.$

Also recall that $\mathcal{U}$ is the minimal $Q$-free resolution of
$\operatorname{coker}J^{T}_{R}$ (Theorem 4.8):

$0\longrightarrow\underset{\\{c_{i,j}\\}}{Q^{(n-1)(n+1)}}\overset{\partial_{2}}{\longrightarrow}\underset{\\{b_{j,k}\\}}{Q^{n(n+1)}}\overset{J^{T}_{Q}}{\longrightarrow}\underset{\\{a_{i}\\}}{Q^{n+1}}\longrightarrow\operatorname{coker}J^{T}_{R}\longrightarrow
0.$

Restated in more detail, we have

$\bullet$ $\\{1\\}$ is a basis for $\mathcal{A}_{0}$

$\bullet$ $\\{e_{i}\\}$ is a basis for $\mathcal{A}_{1}$

$\bullet$ $\\{T_{i}\\}$ is a basis for $\mathcal{A}_{2}$

$\bullet$ $\\{a_{i}\\}$ is a basis for $\mathcal{U}_{0}$

$\bullet$ $\\{b_{j,k}\\}$, $1\leq j\leq n$, $1\leq k\leq n+1$, is a basis for
$\mathcal{U}_{1}$

$\bullet$ $\partial(b_{i,j})=\displaystyle\sum_{\ell=1}^{n+1}\dfrac{\partial
F_{\ell}}{\partial x_{i,j}}a_{\ell}$

$\bullet$ $\\{c_{i,j}\\}$, $1\leq i,j\leq n$, $(i,j)\neq(1,1)$, is a basis for
$\mathcal{U}_{2}$; by convention, we take $c_{1,1}=0$

$\bullet$
$\partial(c_{i,j})=\displaystyle\sum_{\ell=1}^{n+1}x_{i,\ell}b_{j,\ell}-\delta_{i,j}\sum_{k=1}^{n+1}x_{1,k}b_{1,k}$,
that is, $c_{i,j}$ corresponds

to the column of $\partial_{2}$ in which $x_{i,k}$ is in the $j$th block of
$n+1$ entries

######  Lemma 5.1.

The resolution $\mathcal{U}$ admits the structure of a dg module over the dg
algebra $\mathcal{A}$. Such a structure is given by

$\displaystyle e_{i}\cdot a_{j}$
$\displaystyle=\begin{cases}(-1)^{j}\left(\displaystyle\sum_{k=1}^{n}x_{k,j}b_{k,i}\right),&\text{if
}i\neq j,\\\ \\\ (-1)^{i+1}\left(\displaystyle\sum_{k\neq
i}x_{1,k}b_{1,k}-\displaystyle\sum_{\ell\neq
1}x_{\ell,i}b_{\ell,i}\right),&\text{if }i=j,\end{cases}$ $\displaystyle
e_{i}\cdot b_{j,k}$
$\displaystyle=(-1)^{i+1}\displaystyle\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}c_{\ell,j},$ $\displaystyle e_{i}\cdot c_{j,k}$
$\displaystyle=0,$ $\displaystyle T_{i}\cdot a_{j}$
$\displaystyle=(-1)^{j}\sum_{k=1}^{n}x_{k,j}c_{i,k},$ $\displaystyle
T_{i}\cdot b_{j,k}$ $\displaystyle=0,$ $\displaystyle T_{i}\cdot c_{j,k}$
$\displaystyle=0.$

Before the proof, we make a remark and give an example.

######  Remark 5.2.

Notice that in the action $e_{i}\cdot a_{i}$, one could take the sums to be
over all $k$ and all $\ell$, respectively, since
$x_{1,i}b_{1,i}-x_{1,i}b_{1,i}=0$. Additionally, there is nothing special
about the number one here. We could have instead defined

$e_{i}\cdot a_{i}=(-1)^{i+1}\left(\left(\displaystyle\sum_{k\neq
i}x_{s,k}b_{s,k}\right)-\left(\displaystyle\sum_{\ell\neq
s}x_{\ell,i}b_{\ell,i}\right)\right)$

for any fixed $s\in\\{1,\ldots,n\\}$. This is because the difference

$\displaystyle(-1)^{i+1}\left(\left(\displaystyle\sum_{k\neq
i}x_{s,k}b_{s,k}\right)-\left(\displaystyle\sum_{\ell\neq
s}x_{\ell,i}b_{\ell,i}\right)\right)-(-1)^{i+1}\left(\left(\displaystyle\sum_{k\neq
i}x_{1,k}b_{1,k}\right)-\left(\displaystyle\sum_{\ell\neq
1}x_{\ell,i}b_{\ell,i}\right)\right)$
$\displaystyle=(-1)^{i+1}\left[\left(\displaystyle\sum_{k\neq
i}x_{s,k}b_{s,k}\right)-\left(\displaystyle\sum_{k\neq
i}x_{1,k}b_{1,k}\right)-x_{1,i}b_{1,i}+x_{s,i}b_{s,i}\right]$
$\displaystyle=(-1)^{i+1}\left[\left(\displaystyle\sum_{k=1}^{n+1}x_{s,k}b_{s,k}\right)-\left(\displaystyle\sum_{k=1}^{n+1}x_{1,k}b_{1,k}\right)\right]$
$\displaystyle=\partial((-1)^{i+1}c_{s,s})$

is a cycle.

######  Example 5.3.

Let $n=3$, $i=1$, and $j=k=2$. Then,

$\displaystyle e_{1}\cdot b_{2,2}$
$\displaystyle=(-1)^{2}\sum_{\ell=1}^{n}\dfrac{\partial F_{1}}{\partial
x_{\ell,2}}c_{\ell,2}$ $\displaystyle=\dfrac{\partial F_{1}}{\partial
x_{1,2}}c_{1,2}+\dfrac{\partial F_{1}}{\partial
x_{2,2}}c_{2,2}+\dfrac{\partial F_{1}}{\partial x_{3,2}}c_{3,2}$
$\displaystyle=(x_{2,3}x_{3,4}-x_{2,4}x_{3,3})c_{1,2}-(x_{1,3}x_{3,4}-x_{1,4}x_{3,3})c_{2,2}+(x_{1,3}x_{2,4}-x_{1,4}x_{2,3})c_{3,2}.$

###### Proof of Lemma 5.1.

We show a number of computations in the proof. While similar in flavor, we
include as much as we do since different cases require different tricks.

The fact that $\mathcal{U}$ is a dg module over $\mathcal{A}$ follows from
Proposition 2.2.5 in [1], but the explicit structure does not. That the
products $e_{i}\cdot c_{j,k}$, $T_{i}\cdot b_{j,k}$, and $T_{i}\cdot c_{j,k}$
are all equal to zero comes from the fact that $\mathcal{U}_{\geq 3}=0$.

In what follows, an equality is tagged $(n)$ if it follows from Lemma 3.8,
equation $(n)$.

Begin $e_{i}a_{j}$ action:

We have that $\partial(e_{i}a_{j})=(-1)^{i+1}F_{i}a_{j}$. When $i\neq j$,

$\displaystyle\partial\left[(-1)^{j}\left(\sum_{k=1}^{n}x_{k,j}b_{k,i}\right)\right]$
$\displaystyle=(-1)^{j}\left(\sum_{k=1}^{n}x_{k,j}\partial(b_{k,i})\right)$
$\displaystyle=(-1)^{j}\left(\sum_{k=1}^{n}x_{k,j}\left(\sum_{\ell=1}^{n+1}\dfrac{\partial
F_{\ell}}{\partial x_{k,i}}a_{\ell}\right)\right)$
$\displaystyle=(-1)^{j}\left(\sum_{\ell=1}^{n+1}\sum_{k=1}^{n}x_{k,j}\dfrac{\partial
F_{\ell}}{\partial x_{k,i}}a_{\ell}\right)$
$\displaystyle=(-1)^{j}\left(\sum_{k=1}^{n}x_{k,j}\dfrac{\partial
F_{j}}{\partial x_{k,i}}a_{j}+\sum_{\ell\neq
j}\sum_{k=1}^{n}x_{k,j}\dfrac{\partial F_{\ell}}{\partial
x_{k,i}}a_{\ell}\right)$ (4)
$\displaystyle=(-1)^{j}\left(\sum_{k=1}^{n}x_{k,j}\dfrac{\partial
F_{j}}{\partial x_{k,i}}a_{j}\right)$ (2)
$\displaystyle=(-1)^{j}(-1)^{i+j+1}F_{i}a_{j}$
$\displaystyle=(-1)^{i+1}F_{i}a_{j}.$

When $i=j$,

$\displaystyle\partial\left[(-1)^{i+1}\left(\left(\sum_{k\neq
i}x_{1,k}b_{1,k}\right)-\left(\sum_{\ell\neq
1}x_{\ell,i}b_{\ell,i}\right)\right)\right]$
$\displaystyle=(-1)^{i+1}\left(\sum_{k\neq
i}x_{1,k}\partial(b_{1,k})\right)-\left(\sum_{\ell\neq
1}x_{\ell,i}\partial(b_{\ell,i})\right)$
$\displaystyle=(-1)^{i+1}\left(\sum_{k\neq
i}x_{1,k}\left(\sum_{u=1}^{n+1}\dfrac{\partial F_{u}}{\partial
x_{1,k}}a_{u}\right)-\sum_{\ell\neq
1}x_{\ell,i}\left(\sum_{v=1}^{n+1}\dfrac{\partial F_{v}}{\partial
x_{\ell,i}}a_{v}\right)\right)$ (1)
$\displaystyle=(-1)^{i+1}\left(\sum_{u=1}^{n+1}\left(F_{u}a_{u}-x_{1,i}\frac{\partial
F_{u}}{\partial x_{1,i}}a_{u}\right)-\sum_{\ell\neq
1}x_{\ell,i}\left(\sum_{v\neq i}\dfrac{\partial F_{v}}{\partial
x_{\ell,i}}a_{v}\right)\right)$
$\displaystyle=(-1)^{i+1}\left(F_{i}a_{i}+\sum_{u\neq
i}\left(F_{u}a_{u}-x_{1,i}\frac{\partial F_{u}}{\partial
x_{1,i}}a_{u}\right)-\sum_{\ell\neq 1}x_{\ell,i}\left(\sum_{v\neq
i}\dfrac{\partial F_{v}}{\partial x_{\ell,i}}a_{v}\right)\right)$ (1)
$\displaystyle=(-1)^{i+1}\left(F_{i}a_{i}+\sum_{u\neq
i}\left(F_{u}a_{u}-x_{1,i}\frac{\partial F_{u}}{\partial
x_{1,i}}a_{u}\right)-\sum_{v\neq i}\left(F_{v}a_{v}-x_{1,i}\frac{\partial
F_{v}}{\partial x_{1,i}}a_{v}\right)\right)$
$\displaystyle=(-1)^{i+1}F_{i}a_{i}.$

End $e_{i}a_{j}$ action.

Begin $e_{i}b_{j,k}$ action:

We have that

$\displaystyle\partial(e_{i}b_{j,k})$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-e_{i}\left(\sum_{r=1}^{n+1}\dfrac{\partial
F_{r}}{\partial x_{j,k}}a_{r}\right)$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r=1}^{n+1}\dfrac{\partial
F_{r}}{\partial x_{j,k}}e_{i}a_{r}$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq i}\dfrac{\partial
F_{r}}{\partial x_{j,k}}e_{i}a_{r}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}e_{i}a_{i}$ $\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq
i}\left(\dfrac{\partial F_{r}}{\partial
x_{j,k}}(-1)^{r}\left(\sum_{\ell=1}^{n}x_{\ell,r}b_{\ell,i}\right)\right)-\dfrac{\partial
F_{i}}{\partial x_{j,k}}(-1)^{i+1}\left(\sum_{t\neq
i}x_{1,t}b_{1,t}-\sum_{u\neq 1}x_{u,i}b_{u,i}\right)$

Case: $i=k$.

We claim that we may take $e_{i}\cdot
b_{j,i}=(-1)^{i+1}\displaystyle\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,i}}c_{\ell,j}=0$. When $i=k$, $\dfrac{\partial
F_{i}}{\partial x_{j,k}}e_{i}a_{i}=0$, and so

$\displaystyle\partial(e_{i}b_{j,i})$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-\sum_{r\neq i}\dfrac{\partial
F_{r}}{\partial x_{j,i}}e_{i}a_{r}$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-\sum_{r\neq i}\left(\dfrac{\partial
F_{r}}{\partial
x_{j,i}}(-1)^{r}\left(\sum_{\ell=1}^{n}x_{\ell,r}b_{\ell,i}\right)\right)$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-\sum_{r\neq
i}\sum_{\ell=1}^{n}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-\sum_{\ell\neq j}\sum_{r\neq
i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)-\sum_{r\neq
i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial x_{j,i}}x_{j,r}b_{j,i}\right)$
(3) $\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-\sum_{\ell\neq j}\sum_{r\neq
i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)-(-1)^{i+1}F_{i}b_{j,i}$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,i}-0-(-1)^{i+1}F_{i}b_{j,i}$
$\displaystyle=0.$

To see that

$\sum_{\ell\neq j}\sum_{r\neq i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)=0,$

we examine

$\sum_{r\neq i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)$

for a fixed $\ell\neq j$. Let $Y$ be the matrix obtained from $X_{i}$ by
replacing the row indexed by $j$ with a second row indexed by $\ell$. By
Laplace expansion along the row in $Y$ originally indexed by $\ell$, we get

$\displaystyle 0=\det Y$
$\displaystyle=\sum_{r=1}^{i-1}(-1)^{\ell+r}x_{\ell,r}\det
X_{i,r}^{j}+\sum_{r=i+1}^{n}(-1)^{\ell+r+1}x_{\ell,r}\det X_{i,r}^{j}.$

When $i>r$,

$\dfrac{\partial F_{r}}{\partial x_{j,i}}=(-1)^{i+j+1}\det X_{i,r}^{j},$

and when $i<r$,

$\dfrac{\partial F_{r}}{\partial x_{j,i}}=(-1)^{i+j}\det X_{i,r}^{j}.$

So,

$\displaystyle\det Y$
$\displaystyle=\sum_{r=1}^{i-1}(-1)^{\ell+r+i+j+1}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial
x_{j,i}}+\sum_{r=i+1}^{n}(-1)^{\ell+r+i+j+1}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial x_{j,i}}$ $\displaystyle=(-1)^{\ell+i+j+1}\sum_{r\neq
i}(-1)^{r}x_{\ell,r}\dfrac{\partial F_{r}}{\partial x_{j,i}}.$

Thus,

$\sum_{\ell\neq j}\sum_{r\neq i}\left((-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,i}}x_{\ell,r}b_{\ell,i}\right)=0.$

It follows that we may take $e_{i}\cdot b_{j,i}=0$.

Case: $j=1$, $i\neq k$.

When $j=1$, we claim that

$e_{i}b_{1,k}=(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}c_{\ell,1}.$

We may take $i\neq k$. Consider

$\displaystyle\partial(e_{i}b_{1,k})$
$\displaystyle=(-1)^{i+1}F_{i}b_{1,k}-\sum_{r\neq i}\left(\dfrac{\partial
F_{r}}{\partial
x_{1,k}}(-1)^{r}\left(\sum_{\ell=1}^{n}x_{\ell,r}b_{\ell,i}\right)\right)-\dfrac{\partial
F_{i}}{\partial x_{1,k}}(-1)^{i+1}\left(\sum_{t\neq
i}x_{1,t}b_{1,t}-\sum_{u\neq 1}x_{u,i}b_{u,i}\right)$
$\displaystyle=(-1)^{i+1}F_{i}b_{1,k}-\sum_{r\neq i}\left(\dfrac{\partial
F_{r}}{\partial
x_{1,k}}(-1)^{r}\left(\sum_{\ell=1}^{n}x_{\ell,r}b_{\ell,i}\right)\right)$
$\displaystyle\hskip 72.55452pt-(-1)^{i+1}\dfrac{\partial F_{i}}{\partial
x_{1,k}}b_{1,k}+\dfrac{\partial F_{i}}{\partial
x_{1,k}}(-1)^{i+1}\left(\sum_{t\neq i,k}x_{1,t}b_{1,t}-\sum_{u\neq
1}x_{u,i}b_{u,i}\right).$

We first examine the coefficients of the basis elements $b_{m,n}$ in
$\partial(e_{i}b_{1,k})$.

The coefficient of $b_{\ell,i}$ for $\ell\neq 1$ is

$\displaystyle-\sum_{r\neq i}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{1,k}}x_{\ell,r}+(-1)^{i+1}\dfrac{\partial F_{i}}{\partial
x_{1,k}}x_{\ell,i}=-\left(\sum_{r=1}^{n+1}(-1)^{r}\dfrac{\partial
F_{r}}{\partial x_{1,k}}x_{\ell,r}\right).$

Let $Y$ be the matrix obtained from $X_{k}$ by replacing the row indexed 1 by
a second row indexed $\ell$. The determinant of $Y$ is zero, and one way to
compute $\det Y$ is via Laplace expansion along the first row:

$\displaystyle 0=\det Y$
$\displaystyle=(-1)^{\ell}\left(\sum_{r<k}(-1)^{r}x_{\ell,r}\det
X^{1}_{r,k}+\sum_{r>k}(-1)^{r+1}x_{\ell,r}\det X^{1}_{r,k}\right)$
$\displaystyle=(-1)^{\ell}\left(\sum_{r<k}(-1)^{r+1}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial x_{1,k}}+\sum_{r>k}(-1)^{r+1}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial x_{1,k}}\right)$
$\displaystyle=(-1)^{\ell+1}\left(\sum_{r<k}(-1)^{r}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial x_{1,k}}+\sum_{r>k}(-1)^{r}x_{\ell,r}\dfrac{\partial
F_{r}}{\partial x_{1,k}}\right)$
$\displaystyle=\sum_{r=1}^{n+1}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{1,k}}x_{\ell,r}.$

Thus, the coefficient of $b_{\ell,i}$ for $\ell\neq 1$ in
$\partial(e_{i}b_{1,k})$ is zero.

The coefficient of $b_{1,i}$ is

$-\sum_{r\neq i}\dfrac{\partial F_{r}}{\partial x_{1,k}}(-1)^{r}x_{1,r}.$

The coefficient of $b_{1,k}$ is

$(-1)^{i+1}F_{i}-(-1)^{i+1}\dfrac{\partial F_{i}}{\partial x_{1,k}}x_{1,k}.$

The coefficient of $b_{1,t}$ for $t\neq i,k$ is

$\dfrac{\partial F_{i}}{\partial x_{1,k}}(-1)^{i}x_{1,t}.$

We now examine the coefficients of the basis elements $b_{m,n}$ in
$\partial\left((-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}c_{\ell,1}\right)$. We have

$\displaystyle\partial\left((-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}c_{\ell,1}\right)$
$\displaystyle=(-1)^{i+1}\sum_{\ell=2}^{n}\left(\dfrac{\partial
F_{i}}{\partial
x_{\ell,k}}\left(\sum_{s=1}^{n+1}x_{\ell,s}b_{1,s}\right)\right)$
$\displaystyle=(-1)^{i+1}\sum_{\ell=2}^{n}\sum_{s=1}^{n+1}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,s}b_{1,s}$
$\displaystyle=(-1)^{i+1}\sum_{\ell=2}^{n}\sum_{s\neq k}\dfrac{\partial
F_{i}}{\partial
x_{\ell,k}}x_{\ell,s}b_{1,s}+(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,k}b_{1,k}$
$\displaystyle=(-1)^{i+1}\sum_{\ell=2}^{n}\sum_{s\neq k}\dfrac{\partial
F_{i}}{\partial
x_{\ell,k}}x_{\ell,s}b_{1,s}+(-1)^{i+1}\left(F_{i}-\dfrac{\partial
F_{i}}{\partial x_{1,k}}x_{1,k}\right)b_{1,k}.$

The coefficient of $b_{\ell,i}$ for $\ell\neq 1$ is zero.

The coefficient of $b_{1,i}$ is

$(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,i}$

We claim that

$(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,i}=-\sum_{r\neq i}\dfrac{\partial F_{r}}{\partial
x_{1,k}}(-1)^{r}x_{1,r}.$

By Lemma 3.8, equation (2),

$\displaystyle(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,i}$
$\displaystyle=(-1)^{i+1}\left((-1)^{i+k+1}F_{k}-\dfrac{\partial
F_{i}}{\partial x_{1,k}}x_{1,i}\right)$
$\displaystyle=(-1)^{k}F_{k}+(-1)^{i}\dfrac{\partial F_{i}}{\partial
x_{1,k}}x_{1,i}.$

Also,

$\displaystyle-\sum_{r\neq i}\dfrac{\partial F_{r}}{\partial
x_{1,k}}(-1)^{r}x_{1,r}$ $\displaystyle=-\sum_{r\neq i,k}\dfrac{\partial
F_{r}}{\partial x_{1,k}}(-1)^{r}x_{1,r}$ $\displaystyle=-\sum_{r\neq
k}\dfrac{\partial F_{r}}{\partial
x_{1,k}}(-1)^{r}x_{1,r}+(-1)^{i}\dfrac{\partial F_{i}}{\partial
x_{1,k}}x_{1,i}$ (3) $\displaystyle=(-1)^{k+1+1}F_{k}+(-1)^{i}\dfrac{\partial
F_{i}}{\partial x_{1,k}}x_{1,i}$
$\displaystyle=(-1)^{k}F_{k}+(-1)^{i}\dfrac{\partial F_{i}}{\partial
x_{1,k}}x_{1,i}.$

The coefficient of $b_{1,k}$ is

$(-1)^{i+1}\left(F_{i}-\dfrac{\partial F_{i}}{\partial
x_{1,k}}x_{1,k}\right).$

The coefficient of $b_{1,t}$ for $t\neq i,k$ is

$(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,t}$

By Lemma 4.4, equation (4),

$(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,t}=\dfrac{\partial F_{i}}{\partial
x_{1,k}}(-1)^{i}x_{1,t}.$

Since all coefficients of the basis elements $b_{m,n}$ match, we may take

$e_{i}b_{1,k}=(-1)^{i+1}\sum_{\ell=2}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}c_{\ell,1}.$

Case: $j\neq 1$, $i\neq k$.

Assume $j\neq 1$, and take $i\neq k$. We claim that

$e_{i}b_{j,k}=(-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}c_{\ell,j}.$

Consider

$\displaystyle\partial(e_{i}b_{j,k})$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq i}\left(\dfrac{\partial
F_{r}}{\partial
x_{j,k}}(-1)^{r}\left(\sum_{\ell=1}^{n}x_{\ell,r}b_{\ell,i}\right)\right)-\dfrac{\partial
F_{i}}{\partial x_{j,k}}(-1)^{i+1}\left(\sum_{t\neq
i}x_{1,t}b_{1,t}-\sum_{u\neq 1}x_{u,i}b_{u,i}\right)$
$\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq
i}\sum_{\ell=1}^{n}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,k}}x_{\ell,r}b_{\ell,i}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}(-1)^{i+1}\left(\sum_{t\neq i}x_{1,t}b_{1,t}-\sum_{u\neq
1}x_{u,i}b_{u,i}\right)$ $\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq
i}\left(\sum_{\ell=1}^{j-1}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,k}}x_{\ell,r}b_{\ell,i}+\sum_{\ell=j+1}^{n}(-1)^{r}\dfrac{\partial
F_{r}}{\partial x_{j,k}}x_{\ell,r}b_{\ell,i}\right)$
$\displaystyle-\sum_{r\neq i}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,k}}x_{j,r}b_{j,i}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}(-1)^{i+1}\left(\sum_{t\neq i}x_{1,t}b_{1,t}-\sum_{u\neq
1}x_{u,i}b_{u,i}\right)$ (3) $\displaystyle=(-1)^{i+1}F_{i}b_{j,k}-\sum_{r\neq
i}\left(\sum_{\ell=1}^{j-1}(-1)^{r}\dfrac{\partial F_{r}}{\partial
x_{j,k}}x_{\ell,r}b_{\ell,i}+\sum_{\ell=j+1}^{n}(-1)^{r}\dfrac{\partial
F_{r}}{\partial x_{j,k}}x_{\ell,r}b_{\ell,i}\right)$
$\displaystyle-\left((-1)^{k+1}F_{k}-(-1)^{i}\dfrac{\partial F_{i}}{\partial
x_{j,i}}\right)b_{j,i}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}(-1)^{i+1}\left(\sum_{t\neq i}x_{1,t}b_{1,t}-\sum_{u\neq
1}x_{u,i}b_{u,i}\right).$

We first examine the coefficients of the basis elements $b_{m,n}$ in
$\partial(e_{i}b_{j,k})$.

The coefficient of $b_{j,k}$ is

$(-1)^{i+1}F_{i}.$

The coefficient of $b_{1,v}$ for $v\neq i$ is

$(-1)^{i}\dfrac{\partial F_{i}}{\partial x_{j,k}}x_{1,v}.$

The coefficient in $b_{1,i}$ is

$-\sum_{r\neq i}\dfrac{\partial F_{r}}{\partial x_{j,k}}(-1)^{r}x_{1,r}.$

When $v=i$, we claim no discrepancy, i.e.,

$(-1)^{i}\dfrac{\partial F_{i}}{\partial x_{j,k}}x_{1,i}=-\sum_{r\neq
i}\dfrac{\partial F_{r}}{\partial x_{j,k}}(-1)^{r}x_{1,r}.$

To see this, let $Y$ be the matrix obtained from $X_{k}$ by replacing the row
indexed by $j$ by a second row indexed by $1$. The determinant of $Y$ is zero,
and one way to compute $\det Y$ is via Laplace expansion along the first row:

$\displaystyle 0=\det Y$
$\displaystyle=\pm\left(\sum_{r<k}(-1)^{1+r}x_{1,r}\det
X_{k,r}^{1}+\sum_{r>k}(-1)^{1+r+1}x_{1,r}\det X_{k,r}^{1}\right)$ (Remark 3.2)
$\displaystyle=\pm\left(\sum_{r<k}(-1)^{1+r}x_{1,r}(-1)^{1+k+1}\dfrac{\partial
F_{r}}{\partial
x_{1,k}}+\sum_{r>k}(-1)^{1+r+1}x_{1,r}(-1)^{1+k}\dfrac{\partial
F_{r}}{\partial x_{1,k}}\right)$
$\displaystyle=\pm(-1)^{k+1}\sum_{r=1}^{n+1}(-1)^{r}x_{1,r}\dfrac{\partial
F_{r}}{\partial x_{j,k}}.$

The coefficient of $b_{u,i}$ for $u\neq 1$ is

$\displaystyle-\sum_{r\neq i}\left(\dfrac{\partial F_{r}}{\partial
x_{j,k}}(-1)^{r}x_{u,r}\right)+(-1)^{i+1}\dfrac{\partial F_{i}}{\partial
x_{j,k}}x_{u,i}=-\sum_{r=1}^{n+1}\dfrac{\partial F_{r}}{\partial
x_{j,k}}(-1)^{r}x_{u,r}.$

When $j=u$, Lemma 3.8, equation (3), tell us that

$-\sum_{r=1}^{n+1}\dfrac{\partial F_{r}}{\partial
x_{j,k}}(-1)^{r}x_{u,r}=(-1)^{k}F_{k}.$

When $j\neq u$,

$\sum_{r=1}^{n+1}\dfrac{\partial F_{r}}{\partial x_{j,k}}(-1)^{r}x_{u,r}=0.$

To see this, let $Y$ be the matrix obtained from $X_{k}$ by replacing the row
indexed by $j$ by a second row indexed by $u$. The determinant of $Y$ is zero,
and one way to compute $\det Y$ is via Laplace expansion along one of the rows
indexed by $u$:

$\displaystyle 0=\det Y$
$\displaystyle=\pm\left(\sum_{r<k}(-1)^{u+r}x_{u,r}\det
X_{k,r}^{u}+\sum_{r>k}(-1)^{u+r+1}x_{u,r}\det X_{k,r}^{u}\right)$ (Remark 3.2)
$\displaystyle=\pm\left(\sum_{r<k}(-1)^{u+r}x_{u,r}(-1)^{u+k+1}\dfrac{\partial
F_{r}}{\partial
x_{u,k}}+\sum_{r>k}(-1)^{u+r+1}x_{u,r}(-1)^{u+k}\dfrac{\partial
F_{r}}{\partial x_{u,k}}\right)$
$\displaystyle=\pm(-1)^{k+1}\sum_{r=1}^{n+1}(-1)^{r}x_{u,r}\dfrac{\partial
F_{r}}{\partial x_{j,k}}.$

Thus, the coefficient of $b_{u,i}$ for $u\neq 1$ and $j\neq u$ is zero.

The coefficients of all other $b_{m,n}$ are zero.

We now examine the coefficients of the basis elements $b_{m,n}$ in
$\partial\left((-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}c_{\ell,j}\right)$. We have

$\displaystyle\partial\left((-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}c_{\ell,j}\right)$
$\displaystyle=(-1)^{i+1}\sum_{\ell=1}^{n}\left(\dfrac{\partial
F_{i}}{\partial
x_{\ell,k}}\left(\sum_{u=1}^{n+1}x_{\ell,u}b_{j,u}-\delta_{\ell,j}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)\right)$
$\displaystyle=(-1)^{i+1}\left(\sum_{\ell=1}^{n}\sum_{u=1}^{n+1}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,u}b_{j,u}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)$ (4)
$\displaystyle=(-1)^{i+1}\left(\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,k}b_{j,k}+\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,i}b_{j,i}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)$ (1)
$\displaystyle=(-1)^{i+1}\left(F_{i}b_{j,k}+\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,k}}x_{\ell,i}b_{j,i}-\dfrac{\partial F_{i}}{\partial
x_{j,k}}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right).$

The coefficient of $b_{j,k}$ is $(-1)^{i+1}F_{i}$.

The coefficient of $b_{1,v}$ is

$(-1)^{i}\dfrac{\partial F_{i}}{\partial x_{j,k}}x_{1,v}.$

The coefficient of $b_{j,i}$ is

$(-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,k}}x_{\ell,i},$

which, by Lemma 3.8, equation 2, is equal to

$(-1)^{i+1}(1)^{i+k+1}F_{k}=(-1)^{k}F_{k}.$

The coefficients of all other $b_{m,n}$ are zero.

End $e_{i}b_{j,k}$ action.

Begin $T_{i}a_{j}$ action:

We claim that

$T_{i}\cdot a_{j}=(-1)^{j}\sum_{k=1}^{n}x_{k,j}c_{i,k}.$

We have

$\displaystyle\partial(T_{i}a_{j})$
$\displaystyle=\sum_{\ell=1}^{n+1}x_{i,\ell}e_{\ell}a_{j}$
$\displaystyle=\sum_{\ell\neq
j}\left(x_{i,\ell}\left((-1)^{j}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)\right)+(-1)^{j+1}\left(x_{i,j}\sum_{u\neq
j}x_{1,u}b_{1,u}-\sum_{v\neq 1}x_{v,j}b_{v,j}\right)$
$\displaystyle=\sum_{\ell\neq
j}\left(x_{i,\ell}\left((-1)^{j}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)\right)+(-1)^{j+1}x_{i,j}\sum_{u=1}^{n+1}x_{1,u}b_{1,u}$
$\displaystyle+(-1)^{j+1}x_{i,j}x_{1,j}b_{1,j}-(-1)^{j+1}x_{i,j}x_{1,j}b_{1,j}-(-1)^{j+1}x_{i,j}\sum_{v=1}^{n}x_{v,j}b_{v,j}$
$\displaystyle=\sum_{\ell\neq
j}\left(x_{i,\ell}\left((-1)^{j}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)\right)+(-1)^{j+1}x_{i,j}\sum_{u=1}^{n+1}x_{1,u}b_{1,u}-(-1)^{j+1}x_{i,j}\sum_{v=1}^{n}x_{v,j}b_{v,j}.$

Also,

$\displaystyle\partial\left((-1)^{j}\sum_{k=1}^{n}x_{k,j}c_{i,k}\right)$
$\displaystyle=(-1)^{j}\sum_{k=1}^{n}\left(x_{k,j}\left(\sum_{\ell=1}^{n+1}x_{i,\ell}b_{k,\ell}-\delta_{i,k}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)\right)$
$\displaystyle=(-1)^{j}\sum_{k\neq
i}\sum_{\ell=1}^{n+1}x_{k,j}x_{i,\ell}b_{k,\ell}+(-1)^{j}x_{i,j}\left(\sum_{\ell=1}^{n+1}x_{i,\ell}b_{i,\ell}-\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)$
$\displaystyle=(-1)^{j}\sum_{\ell=1}^{n+1}\left(x_{i,\ell}\left(\sum_{k=1}^{n}x_{k,j}b_{k,\ell}-x_{i,j}b_{i,\ell}\right)\right)+(-1)^{j}x_{i,j}\left(\sum_{\ell=1}^{n+1}x_{i,\ell}b_{i,\ell}-\sum_{v=1}^{n+1}x_{1,v}b_{1,v}\right)$
$\displaystyle=(-1)^{j}\sum_{\ell=1}^{n+1}\left(x_{i,\ell}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)-(-1)^{j}\sum_{\ell=1}^{n+1}x_{i,\ell}x_{i,j}b_{i,\ell}$
$\displaystyle+(-1)^{j}\sum_{\ell=1}^{n+1}x_{i,j}x_{i,\ell}b_{i,\ell}-(-1)^{j}\sum_{v=1}^{n+1}x_{i,j}x_{1,v}b_{1,v}$
$\displaystyle=(-1)^{j}\sum_{\ell\neq
j}\left(x_{i,\ell}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)+(-1)^{j}x_{i,j}\sum_{k=1}^{n}x_{k,j}b_{k,j}-(-1)^{j}\sum_{\ell=1}^{n+1}x_{i,\ell}x_{i,j}b_{i,\ell}$
$\displaystyle+(-1)^{j}\sum_{\ell=1}^{n+1}x_{i,j}x_{i,\ell}b_{i,\ell}-(-1)^{j}\sum_{v=1}^{n+1}x_{i,j}x_{1,v}b_{1,v}$
$\displaystyle=(-1)^{j}\sum_{\ell\neq
j}\left(x_{i,\ell}\sum_{k=1}^{n}x_{k,j}b_{k,\ell}\right)+(-1)^{j+1}x_{i,j}\sum_{v=1}^{n+1}x_{1,v}b_{1,v}-(-1)^{j+1}x_{i,j}\sum_{k=1}^{n}x_{k,j}b_{k,j},$

which is $\partial(T_{i}a_{j})$.

Thus,

$T_{i}\cdot a_{j}=(-1)^{j}\sum_{k=1}^{n}x_{k,j}c_{i,k}.$

End $T_{i}a_{j}$ action.

It remains to show that the multiplication in the dg structure we have defined
is associative. For degree reasons, most triple products are zero. It is not
yet clear that $(e_{i}e_{j})a_{k}=e_{i}(e_{j}a_{k})$.

Case: $i=j$.

When $i=j$, $(e_{i}e_{j})a_{k}=(e_{i}e_{i})a_{k}=0$. If we further assume
$i\neq k$, then

$\displaystyle e_{i}(e_{i}a_{k})$
$\displaystyle=e_{i}\left((-1)^{k}\sum_{u=1}^{n}x_{u,k}b_{u,i}\right)$
$\displaystyle=(-1)^{k}\sum_{u=1}^{n}x_{u,k}e_{i}b_{u,i}$ $\displaystyle=0.$

If instead we assume $i=k$, then

$\displaystyle e_{i}(e_{i}a_{k})$
$\displaystyle=e_{i}\left((-1)^{i+1}\left(\sum_{u\neq
i}x_{1,u}b_{1,u}-\sum_{v\neq 1}x_{v,i}b_{v,i}\right)\right)$
$\displaystyle=(-1)^{i+1}\sum_{u\neq i}x_{1,u}e_{i}b_{1,u}$
$\displaystyle=(-1)^{i+1}\sum_{u=1}^{n+1}\left(x_{1,u}\left((-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{\ell,u}}c_{\ell,1}\right)\right)$
$\displaystyle=\sum_{u=1}^{n+1}\sum_{\ell=1}^{n}x_{1,u}\dfrac{\partial
F_{i}}{\partial x_{\ell,u}}c_{\ell,1}.$

The coefficient of $c_{\ell,1}$ for $\ell\neq 1$ above is

$\sum_{u=1}^{n+1}x_{1,u}\dfrac{\partial F_{i}}{\partial x_{\ell,u}}=0,$

by Lemma 3.8, equation (1). Thus,

$e_{i}(e_{i}a_{k})=0.$

Case: $i\neq j$.

When $i<j$,

$\displaystyle(e_{i}e_{j})a_{k}$ $\displaystyle=\sum_{u=1}^{n}(-1)^{i+j+u}\det
X^{u}_{i,j}T_{u}a_{k}$ $\displaystyle=\sum_{u=1}^{n}\left((-1)^{i+j+u}\det
X^{u}_{i,j}(-1)^{k}\sum_{v=1}^{n}x_{v,k}c_{u,v}\right)$
$\displaystyle=(-1)^{k}\sum_{v=1}^{n}\sum_{u=1}^{n}(-1)^{i+j+u}\det
X^{u}_{i,j}x_{v,k}c_{u,v}$ (a)
$\displaystyle=(-1)^{k}\sum_{v=1}^{n}\sum_{u=1}^{n}(-1)^{i+1}\dfrac{\partial
F_{i}}{\partial x_{u,j}}x_{v,k}c_{u,v}.$

When $i>j$,

$\displaystyle(e_{i}e_{j})a_{k}$
$\displaystyle=-\sum_{u=1}^{n}(-1)^{i+j+u}\det X^{u}_{i,j}T_{u}a_{k}$
$\displaystyle=-\sum_{u=1}^{n}\left((-1)^{i+j+u}\det
X^{u}_{i,j}(-1)^{k+1}\sum_{v=1}^{n}x_{v,k}c_{u,v}\right)$
$\displaystyle=(-1)^{k+1}\sum_{v=1}^{n}\sum_{u=1}^{n}(-1)^{i+j+u}\det
X^{u}_{i,j}x_{v,k}c_{u,v}$ (b)
$\displaystyle=(-1)^{k+1}\sum_{v=1}^{n}\sum_{u=1}^{n}(-1)^{i}\dfrac{\partial
F_{i}}{\partial x_{u,j}}x_{v,k}c_{u,v}.$

If we further assume that $j\neq k$, then

$\displaystyle e_{i}(e_{j}a_{k})$
$\displaystyle=e_{i}\left((-1)^{k}\displaystyle\sum_{v=1}^{n}x_{v,k}b_{v,j}\right)$
$\displaystyle=(-1)^{k}\displaystyle\sum_{v=1}^{n}x_{v,k}e_{i}b_{v,j}$
$\displaystyle=(-1)^{k}\displaystyle\sum_{v=1}^{n}\left(x_{v,k}\left((-1)^{i+1}\displaystyle\sum_{u=1}^{n}\dfrac{\partial
F_{i}}{\partial x_{u,j}}c_{u,v}\right)\right)$ (c)
$\displaystyle=(-1)^{i+k+1}\displaystyle\sum_{v=1}^{n}\sum_{u=1}^{n}x_{v,k}\dfrac{\partial
F_{i}}{\partial x_{u,j}}c_{u,v}.$

If instead we assume $j=k$, then

$\displaystyle e_{i}(e_{j}a_{k})=e_{i}(e_{j}a_{j})$
$\displaystyle=e_{i}\left((-1)^{j+1}\left(\sum_{u\neq
j}x_{1,u}b_{1,u}-\sum_{v\neq 1}x_{v,j}b_{v,j}\right)\right)$
$\displaystyle=(-1)^{j+1}\left(\sum_{u\neq j}x_{1,u}e_{i}b_{1,u}-\sum_{v\neq
1}x_{v,j}e_{i}b_{v,j}\right)$ $\displaystyle=(-1)^{j+1}\left(\sum_{u\neq
j}x_{1,u}\left((-1)^{i+1}\sum_{\ell=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{\ell,u}}c_{\ell,1}\right)-\sum_{v\neq
1}x_{v,j}\left((-1)^{i+1}\sum_{s=1}^{n}\dfrac{\partial F_{i}}{\partial
x_{s,j}}c_{s,v}\right)\right)$
$\displaystyle=(-1)^{j+1}\left((-1)^{i+1}\sum_{u\neq
j}\sum_{\ell=1}^{n}x_{1,u}\dfrac{\partial F_{i}}{\partial
x_{\ell,u}}c_{\ell,1}-(-1)^{i+1}\sum_{v\neq
1}\sum_{s=1}^{n}x_{v,j}\dfrac{\partial F_{i}}{\partial x_{s,j}}c_{s,v}\right)$
$\displaystyle=(-1)^{j+1}\left((-1)^{i}\sum_{\ell=1}^{n}x_{1,j}\dfrac{\partial
F_{i}}{\partial x_{\ell,j}}c_{\ell,1}-(-1)^{i+1}\sum_{v\neq
1}\sum_{s=1}^{n}x_{v,j}\dfrac{\partial F_{i}}{\partial x_{s,j}}c_{s,v}\right)$
$\displaystyle=(-1)^{i+j+1}\left(\sum_{\ell=1}^{n}x_{1,j}\dfrac{\partial
F_{i}}{\partial x_{\ell,j}}c_{\ell,1}+\sum_{v\neq
1}\sum_{s=1}^{n}x_{v,j}\dfrac{\partial F_{i}}{\partial x_{s,j}}c_{s,v}\right)$
$\displaystyle=(-1)^{i+j+1}\sum_{v=1}^{n}\sum_{s=1}^{n}x_{v,j}\dfrac{\partial
F_{i}}{\partial x_{s,j}}c_{s,v}$ (d)
$\displaystyle=(-1)^{i+k+1}\sum_{v=1}^{n}\sum_{s=1}^{n}x_{v,k}\dfrac{\partial
F_{i}}{\partial x_{s,j}}c_{s,v}.$

It remains to note that (a), (b), (c), and (d) are all equal.

∎

######  Corollary 5.4.

The $R$-module $\operatorname{Der}_{R\mid k}$ is Golod.

###### Proof.

By Lemma 5.1, the action of $\mathcal{A}_{\geq 1}$ on $\mathcal{U}$ lands in
$\mathfrak{m}_{Q}\mathcal{U}$, and so applying Proposition 2.7 yields that the
$R$-module $\operatorname{coker}J^{T}_{R}$ is Golod. By a theorem of Levin
([12], Theorem 1.1), the first syzygy of a Golod module is Golod. ∎

##  6\. The Resolution of $\operatorname{Der}_{R\mid k}$

Having established the ingredients needed to produce the relative bar
resolution of $\operatorname{coker}J^{T}_{R}$ over $R$, we do so after
introducing notation for the derivations coming from the dg action defined in
the previous section. Truncating, we obtain the minimal $R$-free resolution of
$\operatorname{Der}_{R\mid k}$.

######  Notation 6.1.

Let $L_{i,j}$ be the derivation corresponding to $e_{i}\cdot a_{j}$, i.e.,

$L_{i,j}=\begin{cases}(-1)^{j}\displaystyle\sum_{k=1}^{n}x_{k,j}\dfrac{\partial}{\partial
x_{k,i}},&i\neq j,\\\ \\\ (-1)^{i+1}\left(\displaystyle\sum_{k\neq
i}x_{1,k}\dfrac{\partial}{\partial x_{1,k}}-\displaystyle\sum_{\ell\neq
1}x_{\ell,i}\dfrac{\partial}{\partial x_{\ell,i}}\right),&i=j.\end{cases}$

Let $M$ be the $n(n+1)\times(n+1)^{2}$ matrix with rows indexed by

$b_{1,1},\,b_{1,2},\ldots,b_{1,n+1},\,b_{2,1},\ldots,b_{n+1,n+1}$

and columns corresponding (via $\frac{\partial}{\partial
x_{i,j}}\leftrightarrow b_{i,j}$) to

$L_{1,1},\,L_{1,2},\ldots,L_{1,n+1},\,L_{2,1},\ldots,L_{n+1,n+1}.$

Set $M=M_{1,0}$, and define $M_{i,j}$ similarly, as the matrix coming from the
multiplication $\mathcal{A}_{i}\cdot\mathcal{U}_{j}$.

We now combine the ingredients from the previous sections to produce the
minimal $R$-free resolution of $\operatorname{coker}J^{T}_{R}$. This is done
using the construction of the relative bar resolution via dg structures due to
Iyengar [11]. For us, the construction requires a dg algebra resolution of $R$
over $Q$ (the Hilbert-Burch complex $\mathcal{A}$ from Proposition 2.4) and a
dg $\mathcal{A}$-module resolution of $\operatorname{coker}J^{T}_{R}$ over $Q$
(the resolution $\mathcal{U}$ from Theorem 4.8). We saw that $R$ is a Golod
ring (Corollary 2.8) and that $\operatorname{coker}J^{T}_{R}$ is a Golod
module (see the proof of Corollary 5.4). The Golod property together with the
fact that $\mathcal{A}$ and $\mathcal{U}$ are minimal implies that the
resulting relative bar resolution of $\operatorname{coker}J^{T}_{R}$ over $R$
is minimal.

######  Proposition 6.2.

The minimal $R$-free resolution of $\operatorname{coker}J^{T}_{R}$ is the
relative bar resolution

$\dots\longrightarrow
R^{n(n+1)(n+2)}\overset{\small\begin{bmatrix}J^{T}_{R}&\varphi_{2}\\\
M_{1,1}&M_{2,0}\end{bmatrix}}{\longrightarrow}R^{2n(n+1)}\overset{\small\begin{bmatrix}\partial_{2}&M_{1,0}\end{bmatrix}}{\longrightarrow}\normalsize
R^{n(n+1)}\overset{J^{T}_{R}}{\longrightarrow}R^{n+1}\longrightarrow\operatorname{coker}J^{T}_{R}\longrightarrow
0.$

We recall that $\operatorname{Der}_{R\mid k}$ is isomorphic to the kernel of
$J^{T}_{R}$. It follows that by truncating the relative bar resolution from
Proposition 6.2, we get the minimal $R$-free resolution of
$\operatorname{Der}_{R\mid k}$.

######  Theorem 6.3.

The minimal $R$-free resolution of $\operatorname{Der}_{R\mid k}$ is the
truncated relative bar resolution

$\dots\longrightarrow
R^{n(n+1)(n+2)}\overset{\small\begin{bmatrix}J^{T}_{R}&\varphi_{2}\\\
M_{1,1}&M_{2,0}\end{bmatrix}}{\longrightarrow}R^{2n(n+1)}\overset{\small\begin{bmatrix}\partial_{2}&M_{1,0}\end{bmatrix}}{\longrightarrow}\operatorname{Der}_{R\mid
k}\longrightarrow 0.$

Theorem 6.3 recovers a minimal generating set for $\operatorname{Der}_{R\mid
k}$ (see [3] for another minimal generating set of $\operatorname{Der}_{R\mid
k}$, obtained using a different approach).

######  Corollary 6.4.

A minimal generating set for $\operatorname{Der}_{R\mid k}$ over $R$ is

$\mathcal{M}:=\left\\{V_{r,s},B_{\ell},L_{i,j}\right\\},$

where $r,s,\in\\{1,\ldots,n\\}$, $r\neq s$, $\ell\in\\{2,\ldots,n\\}$, and
$i,j\in\\{1,\ldots,n+1\\}$.

For the sake of visualization, we give an example of what the first
differential in the resolution from Theorem 6.3 looks like in unabbreviated
form.

######  Example 6.5.

When $n=2$, the matrix $\begin{bmatrix}\partial_{2}&M_{1,0}\end{bmatrix}$ is
equal to

$\left[\begin{tabular}[]{cccccccccccc}$x_{2,1}$&$0$&$-x_{1,1}$&$0$&$x_{1,2}$&$-x_{1,3}$&$0$&$-x_{1,1}$&$0$&$0$&$0$&$x_{1,1}$\\\
$x_{2,2}$&$0$&$-x_{1,2}$&$x_{1,2}$&$0$&$0$&$-x_{1,1}$&$0$&$-x_{1,3}$&$0$&$0$&$x_{1,2}$\\\
$x_{2,3}$&$0$&$-x_{1,3}$&$x_{1,3}$&$0$&$0$&$0$&$-x_{1,3}$&$0$&$-x_{1,1}$&$x_{1,2}$&$0$\\\
$0$&$x_{1,1}$&$x_{2,1}$&$-x_{2,1}$&$x_{2,2}$&$-x_{2,3}$&$0$&$0$&$0$&$0$&$0$&$0$\\\
$0$&$x_{1,2}$&$x_{2,2}$&$0$&$0$&$0$&$-x_{2,1}$&$x_{2,2}$&$-x_{2,3}$&$0$&$0$&$0$\\\
$0$&$x_{1,3}$&$x_{2,3}$&$0$&$0$&$0$&$0$&$0$&$0$&$-x_{2,1}$&$x_{2,2}$&$-x_{2,3}$\\\
\end{tabular}\right],$

in which the columns are indexed

$c_{2,1},\,c_{1,2},\,c_{2,2},\,L_{1,1},\,L_{1,2},\,L_{1,3},\,L_{2,1},\,L_{2,2},\,L_{2,3},\,L_{3,1},\,L_{3,2},\,L_{3,3}.$

######  Remark 6.6.

One may wonder how the Euler derivation (Example 2.12) arises from the minimal
set of generators from Corollary 6.4. In terms of the elements of
$\mathcal{M}$, the Euler derivation $E$ can be expressed as

$E=n\left(\sum_{k=1}^{n+1}(-1)^{k+1}L_{k,k}\right)+(n+1)\left(\sum_{\ell=2}^{n}B_{\ell}\right).$

To see this, examine the coefficients of $\dfrac{\partial}{\partial x_{u,v}}$
in

$\displaystyle
n\left(\sum_{k=1}^{n+1}(-1)^{k+1}L_{k,k}\right)+(n+1)\left(\sum_{\ell=2}^{n}B_{\ell}\right)$
$\displaystyle=n\left(\sum_{k=1}^{n+1}(-1)^{k+1}(-1)^{k+1}\left(\sum_{u\neq
k}x_{1,u}\dfrac{\partial}{\partial x_{1,u}}-\sum_{v\neq
1}x_{v,k}\dfrac{\partial}{\partial x_{v,k}}\right)\right)$
$\displaystyle+(n+1)\left(\sum_{\ell=2}^{n}\left(\sum_{s=1}^{n+1}x_{\ell,s}\dfrac{\partial}{\partial
x_{\ell,s}}-\sum_{t=1}^{n+1}x_{1,t}\dfrac{\partial}{\partial
x_{1,t}}\right)\right)$
$\displaystyle=n\left(\sum_{k=1}^{n+1}\left(\sum_{u\neq
k}x_{1,u}\dfrac{\partial}{\partial x_{1,u}}-\sum_{v\neq
1}x_{v,k}\dfrac{\partial}{\partial x_{v,k}}\right)\right)$
$\displaystyle+(n+1)\left(\sum_{\ell=2}^{n}\left(\sum_{s=1}^{n+1}x_{\ell,s}\dfrac{\partial}{\partial
x_{\ell,s}}-\sum_{t=1}^{n+1}x_{1,t}\dfrac{\partial}{\partial
x_{1,t}}\right)\right).$

The coefficient of $\dfrac{\partial}{\partial x_{1,v}}$ is
$n^{2}x_{1,v}-(n+1)(n-1)x_{1,v}=x_{1,v}$, and the coefficient of
$\dfrac{\partial}{\partial x_{u,v}}$ for $u\neq 1$ is
$-nx_{u,v}+(n+1)x_{u,v}=x_{u,v}$.

The construction of the relative bar resolution via dg structures implies that
the Poincaré series of $\operatorname{Der}_{R\mid k}$ is rational.

######  Corollary 6.7.

The Poincaré series of $\operatorname{Der}_{R\mid k}$ over $R$ is

$P^{R}_{\operatorname{Der}_{R\mid
k}}(t)=\dfrac{(n+1)(1+(n-1)t)}{(1-t-nt^{2})t^{2}}.$

###### Proof.

Since $\operatorname{coker}J^{T}_{R}$ is Golod,

$\displaystyle P^{R}_{\operatorname{coker}J^{T}_{R}}(t)$
$\displaystyle=\dfrac{P^{Q}_{\operatorname{coker}J^{T}_{R}}(t)}{1-t\left(P^{Q}_{R}(t)-1\right)}$
$\displaystyle=\dfrac{n+1+n(n+1)t+(n-1)(n+1)t^{2}}{1-t(1+(n+1)t+nt^{2}-1)}$
$\displaystyle=\dfrac{(n+1)(1+nt+(n-1)t^{2})}{1-(n+1)t^{2}-nt^{3}}$
$\displaystyle=\dfrac{(n+1)(1+t)(1+(n-1)t)}{(1+t)(1-t-nt^{2})}$
$\displaystyle=\dfrac{(n+1)(1+(n-1)t)}{1-t-nt^{2}}.$

To obtain $P^{R}_{\operatorname{Der}_{R\mid k}}(t)$, multiply by
$\dfrac{1}{t^{2}}$. ∎

As a final corollary, we note the linearity of the resolution of
$\operatorname{Der}_{R\mid k}$ when $n=2$.

######  Corollary 6.8.

When $n=2$, the minimal $R$-free resolution of $\operatorname{Der}_{R\mid k}$
is linear, i.e., the entries of the matrices representing the differentials in
the minimal $R$-free resolution of $\operatorname{Der}_{R\mid k}$ are either
zero or variables.

###### Proof.

This is because partials of minors in the $2\times 3$ case are just variables.
∎

## Acknowledgements

Special thanks to Claudia Miller for all of her guidance on this project.
Thanks also to Josh Pollitz and Jack Jeffries for helpful discussions. Several
results in this paper I first conjectured based on examples I computed using
Macaulay2 [8].

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