\(1) Then $\rank(\mc F) \le \beta^{\rm I}_{\bar
s}$, with equality if $\mb h = \mb h^{\rm I}_{\bar s}$.

\(2) Assume that $\rank(\mc F) = \beta^{\rm I}_{\bar s}$, that $\mc
A$ has integrity $s-b_1$, and that $\mc B^\dg$ is
$(1-b_1)$-stable.  Then $\mb h = \mb h^{\rm I}_{\bar
s}$ and $G_{\bu}\mc F \in \mb H\Psi_{\mc B}^{\mb h, \mb
t}$ and
$G_{\bu}(\mc F^*) \ox_{\mc A}\mc O_S = G_{\bu}(\mc F^* \ox_{\mc A}\mc
\end{proposition}

\begin{proof}
Each assertion holds if it holds on some neighborhood of each point of
$S$.  Now, $\mb t^*$ is the local generator type of $\mc
F^*$; so each point of $S$ has a neighborhood on which $\mc
F^*$ has a set of generators with precisely $\mb
t(q)$ generators in $F^{-q}\mc F^* - F^{1-q}\mc
F^*$.  Thus shrinking
$S$, we may assume these generators exist globally on $S$.

Set $\mc E := \bigoplus_q \mc A(q)^{\oplus \mb t(q)}$.  The generators
define a map $\ve\: \mc E \to \mc F^*$; it's surjective and filtered
of degree 0.  Let
$\?\ve\: \mc E/F^{1-b_1}\mc E \to \mc F^*/F^{1-b_1}\mc F^*$ be the
induced map.  It is surjective.  Plainly
$\rank (\mc E/F^{1-b_1}\mc E) = \sum_{p\ge b_1} \mb g_{\bar s}(p)$
owing to \eqref{eqsbImax}.  However,
$\mb g_{\bar s}(p) = \mb h^{\rm I}_{\bar s}(p)$ for $p\ge b_1$ by
\eqref{lehsI}(1).  Thus
$\rank(\mc F^* / F^{1-b_1}\mc F^*) \le \sum_{p\ge b_1} \mb h^{\rm
I}_{\bar s}(p)$.

On the other hand, $F^{1-b_1}\mc F^* \subset F^{1-b_1}\mc B^\dg$, and
$F^{1-b_1}\mc B^\dg / F^{1-b_1}\mc F^*$ is flat.  Hence
$\rank (F^{1-b_1}\mc F^*) \le \sum_{p< b_1} \mb b(p)$.  But
$\mb b(p) = \mb h^{\rm I}_{\bar s}(p)$ for $p< b_1$ by
\eqref{lehsI}(2).  Therefore,
$\rank (F^{1-b_1}\mc F^*) \le \sum_{p< b_1} \mb h^{\rm I}_{\bar
s}(p)$.  Hence $\rank(\mc F^*) \le \sum_p \mb h^{\rm I}_{\bar s}(p)$.
Thus $\rank(\mc F) \le \beta^{\rm I}_{\bar s}$.

Lastly, note $\rank(\mc F) = \rank(G_\bu\mc F)$.  But
$G_\bu\mc F \in \mb H\Psi_{\mc B}^{\mb h}$ since
$\mc F \in \mb F\Psi_{\mc B}^{\mb h, \mb t}$; hence,
$\rank(G_\bu\mc F) = \sum_p \mb h(p)$.  Thus
$\rank(\mc F) = \beta^{\rm I}_{\bar s}$ if
$\mb h = \mb h^{\rm I}_{\bar s}$.  Thus (1) holds.

{\bf For (2),} set
$\beta^{\rm I}_1 := \sum_{p\ge b_1} \mb h^{\rm I}_{\bar s}(p)$ and
$\beta^{\rm I}_2 := \sum_{p < b_1} \mb h^{\rm I}_{\bar s}(p)$.  Then
$\beta^{\rm I}_1 + \beta^{\rm I}_2 = \beta^{\rm I}_{\bar s}$.  And
$\rank(\mc F^* / F^{1-b_1}\mc F^*) \le \beta^{\rm I}_1$ and
$\rank (F^{1-b_1}\mc F^*) \le \beta^{\rm I}_2$; see the proof of (1).
But $\rank(\mc F^*) = \beta^{\rm I}_{\bar s}$ now.  Thus
$\rank(\mc F^* / F^{1-b_1}\mc F^*) = \beta^{\rm I}_1$ and
$\rank (F^{1-b_1}\mc F^*) = \beta^{\rm I}_2$.

Note $\rank(F^{1-b_1}\mc B^\dg) = \beta^{\rm I}_2$ and
$F^{1-b_1}\mc B^\dg / F^{1-b_1}\mc F^*$ is flat.  So
$F^{1-b_1}\mc F^* = F^{1-b_1}\mc B^\dg$.  But
$F^q\mc F^* = \mc F^* \cap F^q\mc B^\dg$ for all $q$; indeed, as
$\mc F$ has the induced filtration, the quotient map
$\mc B \onto \mc F$ is strict, and so its dual, the inclusion
$\mc F^* \into \mc B^*$, is strict too by \eqref{sbSm}.  Thus
$G_{-p}\mc F^* =\mc B^\dg_{-p}$ for $p < b_1$.  So
$\mb h(p) = \mb b(p)$ for $p < b$.  Recall
$\mb b(p) = \mb h^{\rm I}_{\bar s}(p)$ for $p< b_1$ by
\eqref{lehsI}(2).  Thus $\mb h(p) = \mb h^{\rm I}_{\bar s}(p)$ for
$p< b_1$.

By the above,
$\?\ve\: \mc E/F^{1-b_1}\mc E \onto \mc F^*/F^{1-b_1}\mc F^*$ is a
surjection of locally free sheaves of the same rank.  So $\?\ve$ is an
isomorphism.  So its restriction to $F^{-b_1}\mc E / F^{1-b_1}\mc E$
is injective.  Thus
$G_{-b_1}\ve \: G_{-b_1} \mc E \to G_{-b_1} \mc F^*$ is injective.

Proceed by induction: given $p\ge b_1$, assume
$G_{-p}\ve \: G_{-p} \mc E \to G_{-p} \mc F^*$ is injective.  Given
$\sigma\in S$ and $y \in G_{-p-1} \mc E_\sigma$ with
$(G_{-p-1} \ve) y = 0$, take any $x \in \mc A_{1,\sigma}$.  Then
$(G_{-p} \ve) (xy) = x((G_{-p-1} \ve) y) = 0$.  So $xy = 0$.  But
$\mc A$ has integrity $s-b_1$, and $p+1 > b_1$.  So $y=0$.  Thus
$G_{-p}\ve \: G_{-p} \mc E \to G_{-p} \mc F^*$ is injective for all
$p\ge b_1$.

By \eqref{sbArt}, the functor $\mc M \mapsto G_\bu\mc M$ is biexact on
the filtered $\mc O_S$-Artinian modules $\mc M$.  Thus
$\?\ve\: \mc E / F^{1-b_1}\mc E \to \mc F^* / F^{1-b_1}\mc F^*$ is a
strict injection.

But $\?{\ve}$ is surjective. So $\?\ve$ is a strict isomorphism.  Thus
$G_\bu\?\ve$ is an isomorphism.

Hence $\mb g_{\bar s}(p) = \mb h(p)$ for $p\ge b_1$.  But
$\mb g_{\bar s}(p) = \mb h^{\rm I}_{\bar s}(p)$ for $p\ge b_1$ by
\eqref{lehsI}(1).  Thus $\mb h(p) = \mb h^{\rm I}_{\bar s}(p)$ for
$p\ge b_1$, so for all $p$ owing to the conclusion above, as desired.

Note $G_{\bu}\mc F \in \mb H\Psi_{\mc B}^{\mb h, \mb t}$ if
$G_{\bu}\mc F ^*$ is of local generator type $\mb t^*$, or $((G_\bu\mc F
^*) \ox_{\mc A}\mc O_S)_{-p}$ is locally free of rank $\mb t(p)$ for all
$p$.  The latter holds for $p \ge b_1$, as $G_\bu\?\ve$ is bijective and
as $\mc E$ is plainly of local generator type $\mb t^*$.  But $\mb t$ is
permissible; so $\mb t(p) = 0$ for $p < b_1$ by \eqref{leperm}.  Thus it
remains to show $((G_\bu\mc F ^*) \ox_{\mc A}\mc O_S)_{-p} = 0$ for $p <

Recall $G_{-p}\mc F^* = \mc B^\dg_{-p}$ for $p < b_1$.  But, by
hypothesis, $\mc B^\dg$ is $(1-b_1)$-stable; that is, $\mc A_1\cdot \mc
B^\dg_{-p} = \mc B^\dg_{1-p}$ for $p < b_1$.  Thus $((G_\bu\mc F^*)
\ox_{\mc A}\mc O_S)_{-p} = 0$ for $p < b_1-1$.

Recall $\?\ve\: \mc E / F^{1-b_1}\mc E \to \mc F^* / F^{1-b_1}\mc F^*$
is injective and $\ve\: \mc E \to \mc F^*$ is surjective.  Hence $\ve$
restricts to a surjection $F^{1-b_1}\mc E \onto F^{1-b_1}\mc F^*$ by the
Snake Lemma.  Thus $(G_{1-b_1}\ve) (\mc E_{1-b_1}) = G_{1-b_1}\mc F^*$.

Recall $\mc E$ is of local generator type $\mb t^*$ and $\mb t(b_1-1) =
0$.  So $(\mc E \ox_{\mc A}\mc O_S)_{1-b_1} = 0$.  Hence
$\bigoplus_{q>0}\mc A_q\cdot \mc E_{1-b_1-q} = G_{1-b_1}\mc E$.
$$\ts \bigoplus_{q>0}\mc A_q\cdot
\big((G_{1-b_1-q}\ve) (\mc E_{1-b_1-q})\big)
= (G_{1-b_1}\ve) (G_{1-b_1}\mc E).$$
But $(G_{1-b_1-q}\ve) (\mc E_{1-b_1-q})\subset G_{1-b_1-q}\mc F^*$,
and $(G_{1-b_1}\ve) (\mc E_{1-b_1}) = G_{1-b_1}\mc F^*$.  Hence
$\bigoplus_{q>0}\mc A_q\cdot G_{1-b_1-q}\mc F^* = G_{1-b_1}\mc F^*$.  So
$((G_\bu\mc F ^* )\ox_{\mc A}\mc O_S)_{1-b_1} = 0$.  Thus $G_{\bu}\mc F
\in \mb H\Psi_{\mc B}^{\mb h, \mb t}$.

Finally, by \eqref{leDiag} there's a canonical surjection $G_{\bu}(\mc
F^*) \ox_{\mc A}\mc O_S \onto G_{\bu}(\mc F^* \ox_{\mc A}\mc O_S)$.
It's an isomorphism as both its source and target are locally free with
the same Hilbert function, namely $\mb t^*$.  Thus (2) holds.
\end{proof}

\begin{proposition}\label{prIcprsd2}
Assume $\mb t$ is permissible.  Given $\mc F \in \mb F\Psi_{\mc
B}^{\mb h,\mb r}$ for some $\mb h$ and $\mb r$, assume $G_{\bu}\mc F \in
\mb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$.  Then $\mb h =
\mb h^{\rm I}_{\bar s}$ and $\mb r = \mb t$ and $\rank(\mc F) =
\beta^{\rm I}_{\bar s}$ .
\end{proposition}

\begin{proof}
As $\mc F \in \mb F\Psi_{\mc B}^{\mb h,\mb r}$ and $G_{\bu}\mc F
\in \mb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$, the
definitions yield $\mb h = \mb h^{\rm I}_{\bar s}$.

Note that $G_\bu \mc F \in \mb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s},
\mb t}$.  Hence $\rank(G_\bu\mc F) = \beta^{\rm I}_{\bar s}$ by
\eqref{prIcprsd1}(1).  However, $\rank(G_\bu \mc F) = \rank(\mc F)$.
Thus $\rank(\mc F) = \beta^{\rm I}_{\bar s}$.

By definition, $\mb r^*$ is the generating function of $G_\bu (\mc F^*
\ox_{\mc A}\mc O_S)$, and $\mb t^*$ is that of $(G_\bu\mc F^*)\ox_{\mc
A}\mc O_S$.  But there's a canonical surjection $G_\bu (\mc F^*)\ox_{\mc
A}\mc O_S \onto G_\bu (\mc F^* \ox_{\mc A}\mc O_S)$ by \eqref{leDiag}.
Thus $\mb r(p) \le \mb t(p)$ for all $p$.  Define $\alpha^{\rm I}$ like
$\beta^{\rm I}_{\bar s}$ in \eqref{eqbetaI} using $\mb r$ in place of
$\mb t$.  Thus $\alpha^{\rm I} \le \beta^{\rm I}_{\bar s}$, and
$\alpha^{\rm I} = \beta^{\rm I}_{\bar s}$ iff $\mb r = \mb t$.  But
$\rank(\mc F) \le \alpha^{\rm I}$ by \eqref{prIcprsd1}(1).  So
$\beta^{\rm I}_{\bar s} = \rank(\mc F) \le \alpha^{\rm I} \le \beta^{\rm
I}_{\bar s}$.  So $\alpha^{\rm I} = \beta^{\rm I}_{\bar s}$.  Thus $\mb
r = \mb t$.
\end{proof}

\begin{lemma}\label{lePotLift}
Fix $\mb h$ and $\mc C\in \mb H\Psi_{\mc B}^{\mb h, \mb t}$.  Set
$\mc N := \mc C^*\ox_{\mc A}\mc O_S$ and $\mc W := F^{-s}\mc B^\dg/\mc
C^*$.  Fix an affine $S$-scheme $T$.  Then given a homogeneous $\mc
O_T$-section $\nu\:\mc N_T \into \mc C_T^*$ of $\mc C_T^*\onto \mc N_T$,
and one $\omega \: \mc W_T \into F^{-s}\mc B_T^\dg$ of $F^{-s}\mc
B_T^\dg \onto \mc W$ and given $\mc F \in \mb F\Psi_{\mc B_T}^{\mb
h}$ with $G_\bu \mc F = \mc C_T$, there's a unique $\gamma \in
F^1\Hom_{\mc O_T}(\mc N_T,\,\mc W_T)$ with $\mc F^* := \mc A_T \cdot(\nu
+\omega\gamma)(\mc N_T)$.
\end{lemma}

\begin{proof}
Note that $G_\bu \mc F^*$ is a finitely generated and flat $\mc
O_T$-module.  So, as $T$ is affine, each projection $F^p\mc F^* \onto
G_p\mc F^*$ splits over $\mc O_T$; say $F^p\mc F^* = \mc F'_p \oplus
F^{p+1}\mc F^*$.  But $G_\bu \mc F^* = \mc C_T^*$.  Thus $F^p\mc B_T^\dg
= \mc F'_p \oplus \omega(\mc W_T)_p \oplus F^{p+1}\mc B_T^\dg$.

Recall from \eqref{se6} that $F^q\mc B = \mc B$ for all $q\le 0$.  Hence
$F^q\mc B^\dg = 0$ for all $q\ge 1$.  It follows that $F^p\mc F^* =
\bigoplus_{q\ge p} \mc F'_q$ and that $F^p\mc B_T^\dg = F^p\mc F^*
\oplus \omega(F^p\mc W_T)$.  Denote the corresponding projections by
$\vf_p\: F^p\mc B_T^\dg \to F^p\mc F^*$ and $\pi_p\: F^p\mc B_T^\dg \to
F^p\mc W_T$.

Recall that $G_\bu \mc F^* = \mc C_T^*$.  So for every $p$, there's a
corresponding projection $F^p\mc F^* \onto {(\mc C_T^*)}_p$.  Fix a
lifting $\nu'_p \: (\mc N_T)_p \into F^p\mc F^*$ of $\nu_p\: ({\mc
N_T})_p \into (\mc C_T^*)_p$.  Then $\nu'_p - \nu_p \in \Hom_{\mc
O_S}(\mc N_p,\, F^{p+1}\mc B^\dg)$.  Set $\gamma_p := \pi_{p+1}(
\nu'_p - \nu_p) \in \Hom_{\mc O_S}( ({\mc N_T})_p,\,F^{p+1}\mc W_T)$
and $\gamma := \sum \gamma_p \in F^1\Hom_{\mc O_S}( \mc N_T,\,\mc W_T)$.

Note that $\vf_p + \omega_p\pi_p = 1_{F^p\mc B_T^\dg }$.  Therefore,
$\omega_p\gamma_p = (\nu'_p - \nu_p) - \vf_{p+1}(\nu'_p - \nu_p)$
in $\Hom_{\mc O_S}( (\mc N_T)_p ,\, F^{p+1}\mc B_T^\dg)$.  Furthermore,
$$\nu_p + \omega_p\gamma_p  =  \nu'_p
- \vf_{p+1}(\nu'_p - \nu_p) \text{\quad in\quad}
\Hom_{\mc O_S}( (\mc N_T)_p ,\, F^p\mc F^*).$$
Hence $(\nu +\omega\gamma)(\mc N_T) \subset \mc F^*$ and $G_\bu (\nu
+\omega\gamma)(\mc N_T) = \mc N_T$.  But $G_\bu \mc F^* = \mc C_T^*$ and
$\mc A_T\cdot \mc N_T = \mc C_T^*$.  Apply the last paragraph of
\eqref{sbArt}.  Thus $\mc A_T\cdot (\nu +\omega\gamma)(\mc N_T) = \mc

Lastly, given $\gamma^\ft\in \Hom_{\mc O_T}(\mc N_T,\,\mc W_T)$, set
$\delta := \gamma^\ft- \gamma$.  Note $\delta \in \Hom_{\mc O_T}(\mc
N_T,\,\mc W_T)$.  So $\delta(\mc N_T) \subset \mc W_T$.  So
$(\omega\delta)(\mc N_T) \subset \omega(\mc W_T)$.  Now, assume $(\nu
+\omega\gamma^\ft)(\mc N_T) \subset \mc F^*$.  Note $(\nu
+\omega\gamma^\ft) - (\nu +\omega\gamma) = \omega\delta$.  Hence
$(\omega\delta)(\mc N_T) \subset \mc F^*$.  Thus $(\omega\delta)(\mc N_T)
\subset \mc F^*\cap \omega(\mc W_T)$.  But $F^p\mc F^* \cap
\omega(F^p\mc W_T) = 0$ for all $p$.  So $(\omega\delta)(\mc N_T) = 0$.
So $\omega\delta = 0$.  But $\omega$ is injective.  So $\delta = 0$.  Thus
$\gamma^\ft = \gamma$, as desired.
\end{proof}

\begin{definition}\label{deEff}
Fix $\mc C\in \mb H\Psi_{\mc B}^{\mb h}$ for some $\mb h$.  Given a
filtered quotient $\mc F$ of $\mc B_T$ for some $T/S$, call $\mc F$ a
{\it lifting\/} of $\mc C$ if $G_\bu \mc F = \mc C_T$ as quotients of
\index[terminology]{lifting}
$\mc B_T$ (so $\mc F \in \mb F\Psi_{\mc B_T}^{\mb h}$).

Set $\mc W := F^{-s}\mc B^\dg/\mc C^*$ and $\mc N := \mc C^*\ox_{\mc
A}\mc O_S$.  Call $\mc F$ a {\it potential lifting\/} of $\mc C$ if
\indt{potential lifting@potential lifting of $\mc C$}
there exist homogeneous sections $\nu\:\mc N_T \into \mc C_T^*$ of $\mc
C_T^*\onto \mc N_T$ and $\omega \: \mc W_T \into F^{-s}\mc B_T^\dg$ of
$F^{-s}\mc B_T^\dg \onto \mc W$ and a map $\gamma \in F^1\Hom_{\mc
O_T}(\mc N_T,\,\mc W_T)$ with $\mc F^* := \mc A_T \cdot(\nu
+\omega\gamma)(\mc N_T)$.

Call $\mc C$ {\it effective\/} if any potential lifting $\mc F$ on any
\indt{effective@effective $\mc C$}
(Noetherian) affine $T$ is, in fact, a lifting.  Call $\mc C$ {\it very
effective\/} if, in addition, $\mc F$ has the same socle type $\mb t$ as
$\mc C$.
\indt{very effective@very effective $\mc C$}
\end{definition}

\begin{proposition}\label{prIveryeff}
Every $\mc C\in \mb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$
with $\mb t$ permissible is very effective.
\end{proposition}

\begin{proof}
Fix a potential lifting $\mc F$ of $\mc C$, and use the setup of
\eqref{deEff}.  Set $\mc E := \mc A\ox_{\mc O_S}\mc N$.  Define $\mu \:
\mc E_T \to \mc B_T^\dg$ by $\mu(a\ox f) := a\cdot (\nu
+\omega\gamma)(f)$.  Then $\mu(\mc E_T) = \mc F^*$.  And $G_\bu\mu\: \mc
E_T \to B_T^\dg$ is given by $G_\bu\mu(a\ox f) := a\cdot \nu(f)$.  But
$\mc A_T\cdot \nu(\mc N_T) = \mc C_T^*$.  Thus $G_{-p}\mu$ restricts to
a surjection $(\mc E_T)_{-p} \onto (\mc C_T^*)_{-p}$ for all $p$; also
$\mc C_T^*\subset G_\bu \mc F^*$, but equality cannot yet be asserted as
it's not obvious $\mu$ is strict.

For all $p$, note $\mc E_{-p} = \bigoplus_{q+r=p}\mc A_{-q}\ox_{\mc
O_S}\mc N_{-r}$.  Thus \eqref{eqsbImax} gives $\rank\mc E_{-p} = \mb
g_{\bar s}(p)$.

For $p\ge b_1$, note $\mb h^{\rm I}_{\bar s}(p) = \mb g_{\bar s}(p)$ by
\eqref{lehsI}(1) as $\mb t$ is permissible.  But $\mc C^* \in \mb
H\Delta_{\mc B^\dg}^{\mb h^*}$ where $\mb h := \mb h^{\rm I}_{\bar
s}$.  So $\mb h^*(-p) = \mb g_{\bar s}(p)$.  Thus $(\mc
E_T)_{-p} \onto (\mc C_T^*)_{-p}$ is an isomorphism.

Form the induced map $\bar\mu\: \mc E_T/ F^{1-b_1}\mc E_T \to\mc B_T^\dg
/ F^{1-b_1}\mc B_T^\dg$.  Then $G_{\bu}\bar\mu$ is injective, because it
factors as $\bigoplus_{p\ge b}(\mc E_T)_{-p} \risom \bigoplus_{p\ge
b}(C^*_T)_{-p} \into \mc B_T^\dg$.  But owing to \eqref{sbArt}, the
functor $\mc M \mapsto G_\bu \mc M$ is biexact on the filtered $\mc
O_T$-Artinian modules $\mc M$.  Thus $\?{\mu}$ is a strict injection.

Note $\bar\mu$ factors as $\mc E_T/ F^{1-b_1}\mc E_T \onto \mc F^*/
F^{1-b_1}\mc F^* \into \mc B_T^\dg / F^{1-b_1}\mc B_T^\dg$ since $\mc
F^*$ has the induced filtration.  So $\bar\mu$ gives a strict
isomorphism $\mc E_T/ F^{1-b_1}\mc E_T \risom \mc F^*/ F^{1-b_1}\mc
F^*$.  But $\?{\mu}$ is a strict injection.  So $G_{\bu}\bar\mu$ gives
an isomorphism $(\mc E_T)_{-p} \risom G_{-p}\mc F^*$ for $p\ge b_1$.
Thus $(\mc C_T^*)_{-p} = G_{-p}\mc F^* \subset (\mc B_T^\dg)_{-p}$ for
$p\ge b_1$.

For $p< b_1$, note $\mb h^{\rm I}_{\bar s}(p) = \mb b(p)$ by
\eqref{lehsI}(2).  But $\mc C^* \in \mb H\Delta_{\mc B^\dg}^{\mb
h^*}$ where $\mb h := \mb h^{\rm I}_{\bar s}$.  So $\mb h^*(-p) = \mb
b(p)$.  But $(\mc C^*_T)_{-p} \subset G_{-p}\mc F^* \subset (\mc
B_T^\dg)_{-p}$.  Further, $\mc B^\dg_{-p} \big/ \mc C^*_{-p}$ is flat,
because $\mc C^* \in \mb H\Delta_{\mc B^\dg}^{\mb h^*}$.  Thus $(\mc
C_T^*)_{-p} = G_{-p}\mc F^*= (\mc B_T^\dg)_{-p}$.  Thus $\mc C_T^* =
G_\bu \mc F^*$.

Finally, form the following commutative diagram, which is induced by
\begin{equation}\label{eqFHGH}
\begin{CD}
(F^{1-b_1}\mc E_T)\ox_{\mc A_T}\mc O_T  @>>>  \mc E_T\ox_{\mc A_T}\mc O_T
@>>> \bigl(\mc E_T \big/ F^{1-b_1}\mc E_T\bigr)\ox_{\mc A_T}\mc O_T
@>>>  0 \\
@VVV                  @VVV
@VVV \\
(F^{1-b_1}\mc F^*)\ox_{\mc A_T}\mc O_T   @>>> \mc F^*\ox_{\mc A_T}\mc O_T
@>>> \bigl( \mc F^* \big/ F^{1-b_1}\mc F^*\bigr) \ox_{\mc A_T}\mc O_T
@>>> 0
\end{CD}
\end{equation}
Its rows are right exact.  Its vertical maps are surjective; in fact,
the right map is an isomorphism, as the isomorphism $\mc E_T/
F^{1-b_1}\mc E_T \risom \mc F^*/ F^{1-b_1}\mc F^*$ was proved above.

Note $\mc E\ox_{\mc A}\mc O_S = (\mc A\ox_{\mc O_S}\mc N)\ox_{\mc A}\mc
O_S = \mc N\ox_{\mc O_S}\mc O_S = \mc N$.  But $F^{1-b_1}\mc N = 0$ as
$\mc C^*$ has local generator type $\mb t^*$ and as $\mb t$ is
permissible.  In \eqref{eqFHGH}, the upper left map preserves the
filtrations.  So it's 0.  So the lower left map is 0 too, as the left
vertical map is surjective.  Thus \eqref{eqFHGH} yields $\mc N_T \risom
\mc F^*\ox_{\mc A_T} \mc O_T$.  But this isomorphism is strict, as $\mc
E/ F^{1-b_1}\mc E \risom \mc F^*/ F^{1-b_1}\mc F^*$ is strict.  Thus
$\mc N_T \risom G_\bu (\mc F^*\ox_{\mc A_T} \mc O_T)$.  Thus $\mc F$ has
the same socle type $\mb t$ as $\mc C$.  Thus $\mc C$ is very effective.
\end{proof}

\begin{proposition}\label{prAffBdl}
Fix $\mb h$.  Set $r_p := \sum_{q<p}(\mb b(q) - \mb h(q))$ and $\mb R :=
\sum_p \mb t(p)r_p$.  Given an effective $\mc C\in \mb H\Psi_{\mc
B_T}^{\mb h, \mb t}$ for some $T/S$, form the associated map $T\to \bb
H\Psi_{\mc B}^{\mb h}$.  Then $T\times_{\bb H\Psi_{\mc B}^{\mb h}}
\bb F\Psi_{\mc B}^{\mb h}$ is an affine-space bundle over $T$ of fiber
dimension $\mb R$.
\end{proposition}

\newbox\FHbox \setbox\FHbox=\hbox{$F^1\Hom_{\mc O_R}(\mc N_R,\,\mc W_R)$}
\newbox\gambox \setbox\gambox=\hbox to \wd\FHbox{\hfil$\gamma$\hfil}
\begin{proof}
Set $\bb F := T\times_{\bb H\Psi_{\mc B}^{\mb h}} \bb F\Psi_{\mc
B}^{\mb h}$.  We must prove each point $x \in T$ has a neighborhood
$U_x$ equipped with an isomorphism $\vf_x \: \bb F_{U_x} \risom \bb
A^{\mb R}_{U_x}$ such that, on each overlap $U_{xy} := U_x\cap U_y$, the map
$(\vf_y | U_{xy})(\vf_x | U_{xy}) ^{-1}$ is an affine transformation of
$\bb A^{\mb R}_{U_{xy}}$.  Notice that the cocycle condition automatically

Set $\mc N := \mc C^*\ox_{\mc A_T} \mc O_T$ and set $\mc W := \mc
B_T^\dg / \mc C_T^*$.  Fix $p\le 0$.  For any $R/T$, set $\mb V_p(R) :=
\Hom_{\mc O_R}\bigl( (\mc N_p)_R,\ (F^{p+1}\mc W)_R\bigr)$.  As $R$
varies, the $\mb V_p(R) $ form a functor; by \cite{EGAI}*{(9.6.1),
p.\,377}, it's representable, say by $\bb V_p / T$.  Also, note $\rank
(\mc N_p) = \mb t(p)$ and $\rank (F^{p+1}\mc W) = r_p$.  Thus $\dim (\bb
V_p / T) =\mb t(p)r_p$.

Set $\mb V(R) := F^1\Hom_{\mc O_R}( \mc N_R,\,\mc W_R)$.  Note $\mb V(R)
= \bigoplus_p \mb V_p(R)$.  Thus, as $R$ varies, the $\mb V(R)$ form a
functor, and it's representable by the fiber product, say $\bb V / T$,
of the $\bb V_p / T$.  Furthermore, $\dim (\bb V / T) = \mb R$.

Given $x \in T$, choose a neighborhood $U_x$ such that $\mc N_{U_x}$ and
$\mc W_{U_x}$ are free over $\mc O_{U_x}$.  Choosing bases yields an
isomorphism $\psi_x \: \bb V_{U_x} \risom \bb A^{\mb R}_{U_x}$.

Also, then there exist homogeneous sections $\nu\:\mc N_{U_x} \into \mc
C_{U_x}^*$ of $\mc C_{U_x}^*\onto \mc N_{U_x}$ and $\omega \: \mc
W_{U_x} \into F^{-s}\mc B_{U_x}^\dg$ of $F^{-s}\mc B_{U_x}^\dg \onto \mc
W_R$.  But $\mc C$ is effective for $T$.  So for any affine $R/U_x$,
there exists a map of sets
\begin{align*}%\label{eqFHG1}
\copy\FHbox &\longrightarrow \big\{\,\mc F \in
\mb F\Psi_{B_R}^{\mb h}\ \big| \ G_\bu \mc F = \mc C_R \,\big\} \\
\copy\gambox & \longmapsto  \hphantom{\big\{\,}
\mc F^* := \mc A_R\cdot(\nu_R +\omega_R\gamma)(\mc N_R).
\end{align*}
Owing to \eqref{lePotLift}, this map is bijective.  Plainly, it's
functorial in $R$.  So it defines an isomorphism $\theta_x\: \bb F_{U_x}
\risom \bb V_{U_x}$.  Set $\vf_x := \psi_x\theta_x$, so that $\vf_x \:
\bb F_{U_x} \risom \bb A^{\mb R}_{U_x}$.

The isomorphisms $\psi_x$ and $\theta_x$ depend on the choices of
several $\mc O_{U_x}$-maps, but two different choices of the same map
differ by an appropriate $\mc O_{U_x}$-isomorphism.  Thus although
$\vf_y | U_{xy}$ and $\vf_x | U_{xy}$ needn't be the same isomorphism
from $\bb F_{U_{xy}}$ to $\bb A^{\mb R}_{U_{xy}}$, nevertheless $(\vf_y
| U_{xy})(\vf_x | U_{xy}) ^{-1}$ is an affine transformation of $\bb
A^{\mb R}_{U_{xy}}$.
\end{proof}

\begin{theorem}\label{thFHdim}
Set $r_p := \sum_{q<p}(\mb b(q) - \mb h^{\rm I}_{\bar s}(q))$ and $\mb
R := \sum_p \mb t(p)r_p$.  Assume that $\mc A$ has integrity $s-b_1$,
and $\mc B^\dg$ is $(1-b_1)$-stable\emdash for example, that $\mc A$ is
the symmetric algebra on a locally free $\mc O_S$-module and that $\mc B
= \mc A$.  Assume $\mb t$ is permissible and $\bb H\Psi_{\mc B}^{\mb
h^{\rm I}_{\bar s}, \mb t} \neq \emptyset$.  Then $\bb F\Psi_{\mc
B}^{\mb h^{\rm I}_{\bar s}, \mb t}$ is an affine-space bundle over $\bb
H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$ of fiber dimension $\mb
R$, and $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is one over $\bb
H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$.
\end{theorem}

\begin{proof}
Set $\mb h := \mb h^{\rm I}_{\bar s}$.  Set $T := \bb H\Psi_{\mc
B}^{\mb h, \mb t}$ and $\bb F := T\times_{\bb H\Psi_{\mc B}^{\mb h}} \bb
F\Psi_{\mc B}^{\mb h}$.  Let $\mc C$ be the universal quotient of $\mc
B_T$.  Then $\mc C$ is (very) effective by \eqref{prIveryeff}.  So $\bb
F$ is an affine-space bundle over $T$ of fiber dimension $r$ by
\eqref{prAffBdl}.  Thus we must prove $\bb F = \bb F\Psi_{\mc B}^{\mb h,
\mb t}$.

Fix $R/S$.  An $R$-point of $\bb F$ is given by an $\mc F \in \mb
F\Psi_{\mc B_R}^{\mb h}$ with $G_\bu \mc F \in \mb H\mb \Psi_{\mc
B_R}^{\mb h, \mb t}$.  So $\mc F \in \mb F \Psi_{\mc B_R}^{\mb h, \mb
t}$ by \eqref{prIcprsd2} or \eqref{prIveryeff}.  Thus $\mc F$ gives an
$R$-point of $\bb F\Psi_{\mc B}^{\mb h, \mb t}$.  Conversely, an
$R$-point of $\bb F\Psi_{\mc B}^{\mb h, \mb t}$ is given by an $\mc F
\in  F\Psi_{\mc B_R}^{\mb h, \mb t}$.  Then $G_\bu \mc F \in \mb H\mb
\Psi_{\mc B_R}^{\mb h, \mb t}$ by \eqref{prIcprsd1}(1),\,(2) and
\eqref{sbIS}.  Thus $\bb F$ and $\bb F\Psi_{\mc B}^{\mb h, \mb t}$ have
the same $R$-points, so they're equal as schemes.

The last assertion follows because $F\Lambda_{\mc B}^{\{\mb h^{\rm
I}_m\}} = \bb H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}\times_{\bb
H\Psi_{\mc B}^{\mb h}} \bb F\Psi_{\mc B}^{\mb h}$ essentially by
defintion; see \eqref{sbMLev}.
\end{proof}

\begin{corollary}\label{coFIrr}
Assume $\mb t$ is permissible, and $S$ is irreducible.

\(1) Assume $\bb F\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$ is nonempty.
Then it is irreducible.

\(2)  Assume  $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is nonempty. Then
it is open and dense in $\bb F\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb
\end{corollary}

\begin{proof}
{\bf For (1),} recall that $\bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s},
\mb t}$ is irreducible by \eqref{prIrr}(1).  So $\bb F\Psi_{\mc B}^{\mb
h^{\rm I}_{\bar s}, \mb t}$ is irreducible, as it's an
affine space bundle over $\bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s},
\mb t}$ by \eqref{thFHdim}.  Thus (1) holds

The proof of (2) is similar with \eqref{prIrr}(2) in place of
\eqref{prIrr}(1).
\end{proof}

\begin{corollary}\label{coFdim}
Set $\mb F := \sum_p \mb t(p)\big(\sum_{q\le p}(\mb b(q) - \mb h^{\rm
I}_{\bar s}(q))\bigr)$.  Assume that $\mc A$ has integrity $s-b_1$ and
that $\mc B^\dg$ is $(1-b_1)$-stable\emdash for example, that $\mc A$ is
the symmetric algebra on a locally free $\mc O_S$-module and that $\mc B
= \mc A$.  Assume that $\mb t$ is permissible%
%  and that $\bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$ is
% nonempty.

\(1) Then $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is covered by
(possibly empty) open subschemes, each one isomorphic to an open
subscheme of the affine space over $S$ of fiber dimension $\mb F$.

\(2) If $S$ is irreducible, then so are $\bb H\Lambda_{\mc B}^{\{\mb
h^{\rm I}_m\}}$ and $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$, but
they may be empty.
\end{corollary}

\begin{proof}
First, note that $\bb H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is
nonempty iff $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is so.
Moreover, both (1) and (2) are trivial if these schemes are empty.  So
assume they're nonempty.

{\bf For (1),} note $\bb H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ and
$\bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$ are both subschemes
of $\bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}}$ by \eqref{prMLevS} and
\eqref{prLevS}.  In fact, $\bb H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}
\subset \bb H\Psi_{\mc B}^{\mb h^{\rm I}_{\bar s}, \mb t}$, since the
universal quotient on $\bb H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ is of
$S$-socle type $ t$ by \eqref{leperm}(6).  So $\bb F\Lambda_{\mc
B}^{\{\mb h^{\rm I}_m\}}$ is an affine-space bundle over $\bb
H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ of fiber dimension $\mb R$ by
\eqref{thFHdim}.  Thus (1) \eqref{prSmOp}(2) yields (1).

{\bf For (2),} note $\{\mb h^{\rm I}_m\}$ is recursively maximal by
\eqref{coImax}.  Assume $S$ is irreducible. Then $\bb H\Lambda_{\mc
B}^{\{\mb h^{\rm I}_m\}}$ is irreducible by \eqref{thMax} applied with
$S_{\rm red}$ for $S$.  Hence $\bb F\Lambda_{\mc B}^{\{\mb h^{\rm
I}_m\}}$ is irreducible, as it's an affine-space bundle over $\bb
H\Lambda_{\mc B}^{\{\mb h^{\rm I}_m\}}$ by \eqref{thFHdim}.  Thus (2)

Alternatively, (2) follows immediately from \eqref{prIrr} and
\eqref{coFIrr}.
\end{proof}

\begin{example}[smoothable quotients, elementary components]
\label{exElemCpts}
Fix a polynomial ring $A := k[X_1,\dotsc,X_r]$ with $k$ a field.  Let's
see how \eqref{coFdim} can help in finding examples of $k$-Artinian
quotients $C$ of $A$ that aren't {\it smoothable\/} (deformable to
$k$-etale quotients) and examples of {\it elementary\/} components of
\indt{smoothable quotient}
\indt{elementary component} $\Hilb_{\bb A_{k}^r/k}^d$ where $d :=
\dim_kC$ (absolutely irreducible components whose $K$-points for every
field $K\supset k$ are quotients $C'$ of $A_K$ with $\Spec(C')$
absolutely irreducible).  Notably, we are going to find a new {\it
small\/} elementary component (one of dimension
\indt{small elementary component}
smaller than $dr$, which is the dimension of the {\it principal
component\/} of $\Hilb_{\bb A_k^r/k}^d$, which parameterizes all
smoothable $C$).

To simplify notation, set $\mb h_m := \mb
h^{\rm I}_m$ for all $m$.  Define the {\it translation map}
\indt{translation map}
\begin{equation}\label{eqTr}\ts
\theta\:\bb F\Lambda_A^{\{\mb h_m\}}\x\bb A_k^r \to \Hilb_{\bb A_k^r/k}
\end{equation}
as follows (cf.\ Jelisiejew's definition \cite{JJ-JLMS2019}*{p.\,258}).
It suffices to define the action of $\theta$ on $K$-points for an
arbitrary $k$-algebra $K$.  A $K$-point of $\bb A_k^r$ is given by a
vector $\mb x := (x_1,\dotsc,x_r)$ with $x_i \in K$.  Define a
$K$-automorphism $\alpha_{\mb x}\: A_K \risom A_K$ by $\alpha_{\mb
x}(X_i) := X_i - x_i$.  Given $A_K/I \in \bb F\Lambda_{A_K}^{\{\mb
h_m\}}$, set $\theta((A_K/I),\,\mb x) := A_K/\alpha_{\mb x}(I)$.  This
assignment is functorial in $K$.  Thus it defines the desired map of
schemes $\theta$.

Note that $\theta$ is injective on $K$-points if $K$ is a field, as $\mb
x = \Supp(\alpha_{\mb x}(I))$ and $I = \alpha_{\mb -x}(\alpha_{\mb x}(I)
Note that the definition of $\theta$ and its injectivity remain valid
if $k$ is replaced by an arbitrary Noetherian ring; this added
generality is used in \eqref{prLift}.

Iarrobino and Emsalem, in their pioneering paper \cite{IE78}, proved the
first version of \eqref{coFdim}.  It treats I-compressed filtered
Gorenstein $C$ over an arbitrary algebraically closed field. In
particular, when the Hilbert vector of these $C$ is $(1,r,r,1)$, then
$\Im(\theta)$ is of dimension $r^3/6 + r^2 + 5r/6 -1$.  But the
principal component of $\Hilb_{\bb A_k^r/k}^{2r+2}$ is of dimension
$r(2r+2)$; so it can't contain $\Im(\theta)$ for $r\ge 8$.  Thus, they
proved (see \cite{IE78}*{p.\,152, mid}) that most such $C$ aren't
smoothable if $r\ge 8$.

Similarly, suppose the socle vector is $(0,0,1,1)$.  Then the nonzero
Hilbert vectors of an I-compressed $C$ are $(1,r,r+1,1)$ and
$(1,r,r,1)$; see \eqref{se9}.  So \eqref{coFdim} yields
\begin{align*}\label{eq}\ts
\mb F &= \mb t(3)\Bigl(\Bigl(\tbinom{r+2}{3} -1\Bigr)
+ \Bigl(\tbinom{r+1}{2} -(r+1)\Bigr)\Big)
+  \mb t(2)\Bigl(\tbinom{r+1}{2} -(r+1)\Bigr)\\
&= r^3/6 + 3r^2/2 - 2r/3 - 3.
\end{align*}
The principal component is of dimension $(2r+3)r$.  But $\mb F+r >
(2r+3)r$ if $r \ge 7$, and $\mb F+r < (2r+3)r$ if $r\le 6$.  Thus most
$C$ in $\mb F\Lambda_A^{\{h_m\}}$ aren't smoothable if $r\ge 7$.

Iarrobino and Emsalem devoted most of \cite{IE78} to another novel
method, known as the {\it method of  small tangent spaces} or {\it
\indt{method of  small tangent spaces}
\indt{method of trivial negative tangents}
trivial negative tangents\/}.  It grew out of Schlessinger's work with
the induced grading on the tangent space $T = \Hom_A(I,C)$ to
$\Hilb_{\bb A_k^r/k}^d$ at a homogeneous quotient $C$ with ideal $I$
(his setting is more general); see p.\,151, top.  Iarrobino and Emsalem
described the method in terms of $T$ and the functor $T^1(\bu)$ of
deformation theory.  In essence, see \cite{IE78}*{p.\,155 f.}, they
showed that, under mild restrictions on $\car(k)$, if $I$ is generated
by several general quadrics and if $T^1(C)_{-1} = 0$, then $T$ is small
in the sense that $\dim_kT_{-1} \le r$ and $T_v = 0$ for $v \le -2$, and
so $C$ belongs to an elementary component.  Unfortunately, they falsely
concluded, due to a computational error, see \cite{LNM1721}*{p.\.221,
bot.}, that there's an elementary component when the Hilbert vector is
$(1,r,r,1)$ for $r = 4,5$.

However, the method itself is sound, and was put in definitive form for
filtered $C$ by Jelisiejew \cite{JJ-JLMS2019}*{Thm.\,4.5, p.\,259 and
Thm.\,4.9, p.\,260}.  At the top of p.\,250, he made this definition:
$C$ has {\it trivial negative tangents\/} if $T^1(C)\big/F^0T^1(C) = 0$.
\indt{trivial negative tangents}
Moreover, by Cor.\,4.7 on p.\,260, if $C$ is homogeneous, then it has
trivial negative tangents iff $\sum_{v<0} \dim_k T_v = r$ (caution,
there's a typo: $\Hom(I_R,O_R)$ should be $\Hom(I_R,O_R)_{<0}$ in the
statement and in l.\,2 of the proof; cf.\ arXiv:1710.06124v5.)

Further, Thm.\,4.5 asserts that, if $C$ has trivial negative tangents,
then $\theta$ restricts to an isomorphism between neighborhoods of
$(C,0)$ and $C$.  Hence, cf.\ \cite{JJ-JLMS2019}*{ Cor.\,4.6, proof,
p.\.259}, if $C$ is also I-compressed, then \eqref{coFdim} implies $C$
is a simple point of $\Hilb_{\bb A_k^r/k}^d$, so lies on a unique
elementary component, which is of dimension $\mb F+r$.  Conversely, by
\cite{JJ-JLMS2019}*{Thm.\,4.9, p.\,260}, if $\car(k) = 0$ and $C$ is a
simple point of $\Hilb_{\bb A_k^r/k}^d$ lying on an elementary
component, then $C$ has trivial negative tangents.  Versions of these
important results also hold by \eqref{prLift}, which is a self-contained
alternative to \cite{JJ-JLMS2019}, but was greatly inspired by it; these
versions suffice here.

In the case of Hilbert vector $(1,r,r,1)$, Iarrobino and Kanev
\cite{LNM1721}*{Lem.\,6.21, p.\,221} asserted that an elementary
component exists for $6 \le r \le 12$, but $r \neq 7$.  To prove it,
they reformulated the condition to have trivial negative tangents in
terms of the Hilbert function of $A/I^2$; this reformulation is valid
essentially because $\Hom_A(I,C) = (I/I^2)^*(-3)$ as $C$ is Gorenstein.
Then they used Macaulay 2 to verify that, over several finite fields,
specific $C$ have trivial negative tangents; they said that therefore
they ``consider'' the existence of such $C$ true when $\car(k) = 0$.

Recently, Szafarczyk \cite{RSz} about settled the case of $(1,r,r,1)$.
He showed that a certain well-chosen $C$ has a small tangent space.  He
did so in Section 3 for $r\ge 18$ and arbitrary $k$ via a lengthy
elementary hand-calculation.  He did so in Section~4 with Macaulay~2 for
$6 \le r \le 17$, but $r \neq 7$, and $k = \bb Q,\ \bb Z/2,\ \bb Z/3$.
He concluded in Proposition 2.20 on p.\,7 that this $C$ is a simple
point of an elementary component.

Independently, the present authors used Macaulay 2 to show that, for
$8\le r \le 19$ and $k = \bb Z/101$, a randomly generated homogeneous
$C$ has trivial negative tangents, so is a simple point of $\Hilb_{\bb
A_k^r/k}^d$ and lies on a unique elementary component, which is of
dimension $\mb E := r^3/6 + r^2 + 5r/6 -1$.
Hence, over $\bb Q$ too, for $8\le r \le 19$, there exists a
generically smooth elementary component of dimension $\mb E$, by virtue
of \eqref{prLift} with $k := \bb Z$.

Notice that it suffices to show $\dim T \le \mb E$ in one positive
characteristic.  Notice also that \eqref{prLift} provides a general
method for establishing the existence of an elementary component with
given permissible socle type in characteristic 0 by producing one
example of a $C$ with trivial negative tangents in positive
characteristic (and its base field needn't be prime or even finite).
Thus \eqref{prLift} illustrates an advantage gained from having
\eqref{coFdim} with a base $S$ that's not the Spec of a field.

In the case of socle vector $(0,0,a,1)$, the authors' computations with
Macaulay 2 have yielded randomly generated I-compressed homogeneous $C$
with trivial negative tangents for $r = 5,7,8,9$ and $a=1$, for $r = 6,
7, 8$ and $2 \le a \le 5$, for $r = 7, 8$ and $6 \le a \le 8$, and for
$r = 8$ and $9 \le a \le 10$, all over the prime field $k$ of
characteristic 101 and several others (31, 701, 3001) for good measure
when $a=1$.  Similarly, they found such $C$ with Hilbert vector
$(1,r,2r,2)$ for $r = 6,7,8$ and $(1,r,2r+1,2)$ for $r = 7,8$ in
characteristic 101.  More precisely, the computations show that $\dim
T_{-1} = r$ and $T_{-2} = 0$ and that the ideal $I$ of $C$ is generated
by quadrics, say $n$ in number, giving a surjection $A(-2)^{\oplus n}
\onto I$; but $T := \Hom_A(I,C)$, so $T\into C(2)^{\oplus n}$, thus $T_v
= 0$ for $v < -2$.

Thus in those cases, each $C$ lies in the smooth locus of a unique
elementary component, which is of dimension $\mb F + r$.
Hence, \eqref{prLift} yields, in each case, a generically smooth
elementary component of dimension $\mb F + r$ over $\bb Q$.
Independently, in infinitely many related cases, Satriano and Staal
\cite{SS2}*{Thm.\,1.3, p.\,3} produced on 25 Oct.\ 2022 a generically
smooth elementary component over an algebraically closed field of
characteristic 0 by conceptual means.

The case of socle vector $(0,0,1,1)$ and $r = 5$ is of particular
interest.  Indeed, $\mb F+r = 57$ and $(2r+3)r = 65$; thus the component
is small.  But it's not in Jelisiejew's \cite{JJ-JLMS2019}*{Thm.\,1.4,
p.\,251, and Ex.\,6.9 and Rmk.\ 6.10, p.\.269}, which include all small
elementary components known in 2019; also, it's not in Satriano and
Staal's preprint \cite{SS1} of 2 Dec.\,2021.  For $r = 5$, moreover,
Macaulay 2 worked over $\bb Q$, thus corroborating \eqref{prLift}.

In the case of socle vector $(0,0,1,1)$ and $r = 6$ and $k = \bb Q$,
most I-compressed $C$ are smoothable, even though this case is
sandwiched between the cases $r = 5$ and $r =7$, where, as noted above,
most aren't.  This smoothability was just proved via a Macaulay 2
calculation by Jelisiejew, who kindly informed the authors about it in
an email.  Thus the situation is analogous to that of Gorenstein $C$
with Hilbert vector $(1,r,r,1)$.  For $r=6$ and $r = 8$ most such $C$
aren't smoothable as noted above.  But, for $r = 7$, all such $C$ are
smoothable by Bertone, Cioffi, and Roggero's \cite{BCR}*{Thm.\,3.7};
also, Jelisiejew said he had recovered this result via a Macaulay~2
calculation like the one for $(1,6,7,1)$.

The authors found no small elementary component among the examples with
socle vector $(0,0,a,1)$ computed above with $a \ge 1$ and $r \ge 6$.
Similarly, they found none for $(0,0,0,2)$ and $(0,0,1,2)$.
Furthermore, the computations suggest that a general enough $C$ with
Hilbert vector $(1,r,br+a,b)$ lies in a unique elementary component for
any $a \ge 0$, any $b\ge1$, and any $r\gg0$.

Iarrobino and Kanev conjectured in \cite{LNM1721}*{Conj.\,6.30, p.\,226}
that a general homogeneous Gorenstein $C$ of odd socle degree $s$ has
trivial negative tangents if $r=4$ and $s\ge 15$, if $r=5$ and $s\ge 5$,
or if $r\ge6$ and $s\ge 3$ but $(r,s) \neq (7,3)$.  The case $s=3$ was
proved by Szafarczyk, as noted above.  So the authors used Macaulay 2 to
show that, in the 18 cases $r \ge 5$ and $r + s \le 16$ and $(s,r) =
(5,12),\ (11, 6)$ when $k = \bb Z/101$, a randomly generated Gorenstein
$C$ is I-compressed and has trivial negative tangents.  Hence by
\eqref{prLift}(5)--(6), the conjecture holds in these cases when $k =
\bb Z/101$ and also when $k = \bb Q$.

If $s$ is even and $C$ is Gorenstein and $I$-compressed, it's not hard
to see, by looking at $A/I^2$ and noting $(I^2)_{s+1} = 0$, that $\dim
T_{-1} = \dim A_{s+1}$, and so $C$ never has trivial negative tangents.
Nevertheless, if $s = 4$ and $r \ge 8$, or $s = 6$ and $r \ge 5$, or $s
= 8$ and $r \ge 5$, or $s \ge 10$ and $r \ge 4$, then $\mb F + r > rd$,
because $\mb F = \binom{r+s}{r} - d$ and $d = 2\binom{r+n-1}{r}+
\binom{r+n-1}{r-1}$; thus then a general $C$ isn't smoothable.
\end{example}

\begin{proposition}\label{prLift}
Let $A$ be a polynomial ring in $r$ variables over a Noetherian ring
$k$, and $K$ a field containing a residue field $R$ of $k$.  Set $\mb
h_m := \mb h^{\rm I}_m$ for all $m$.  Set $\bb P := \bb
F\Lambda_A^{\{\mb h_m\}}\x_k\bb A_k^r$ and $\bb H := \Hilb_{\bb
A_k^r/k}$.  Let $\theta\: \bb P \to \bb H$ be the translation map
\eqref{eqTr}.  Set $\mb F := \sum_p \mb t(p)\big(\sum_{q\le
p}\bigl(\binom{q+r-1}{r-1} - \mb h_{\bar s}(q)\bigr) \bigr)$.  Assume
$\mb t$ is permissible.

Fix $C \in \mb F\Lambda_{A_K}^{\{\mb h_m\}}$.  Let $T$ be the tangent
space at the $K$-point $C$ to $\bb H_K$.  Let $Z$ be the
scheme-theoretic image of $\theta$ in $\bb H_R$; let $Z\sm$ be the
smooth locus of $Z$, and $Z\0\sm$ the intersection of $Z\sm$ with the
smooth locus of $\bb H_R$.  Given a subscheme $G$ of a Noetherian scheme
$H$, call $G$ a {\rm component} if $Z = \Spec(\mc O_H/\mc Q)$
where $\mc Q$ is the primary ideal of a minimal prime in the primary
decomposition of $(0)$.

\(1) Then $Z$ is absolutely integral of dimension $\mb F+r$.  Moreover,
$Z$ is a component, necessarily elementary, iff $Z$ has a point at which
$\bb H_R$ is smooth of dimension $\mb F+r$.

\(2) If either $\dim_K(T/F^0T) \le r$ or $\dim_K(T) \le \mb F+r$, then
both are equalities.

\(3) Assume $C$ has trivial negative tangents.  Then $\dim_K(T/F^0T)\le

\(4) Assume $\dim_K(T/F^0T) \le r$.  Then $\bb H_R$ is smooth of dimension
$\mb F+r$ at $C$.

\(5) Assume either \(a) $C$ has trivial negative tangents or \(b)
$\car(K) = 0$ and $Z$ is a component of\/ $\bb H_R$.  Then $\theta$
restricts to an isomorphism $\theta^{-1}Z\sm \risom Z\sm$.  Moreover, a
$K$-point $C'$ of $Z$ has trivial negative tangents iff $C'$ lies in

\(6) Assume that $k$ is a Dedekind domain with fraction field $L$ and
that $Z$ is a component of\/ $\bb H_R$.  Then the scheme-theoretic image
$Y$ of\/ $\theta_L$ is a component of\/$\bb H_L$.
\end{proposition}

\begin{proof}
{\bf For (1),} note that $\bb F\Lambda_{A_R}^{\{\mb h_m\}}$ is smooth
and absolutely irreducible of dimension $\mb F$ owing to \eqref{coFdim}.
So $\bb P_R$ is smooth and absolutely irreducible of dimension $\mb
F+r$.  Thus $Z$ is absolutely integral.  But $\theta$ is injective on
field-valued points; hence $\theta$ cannot have a fiber of positive
dimension.  Thus $\dim Z =\mb F+r$.

Suppose $Z$ is a component.  Then necessarily it’s elementary, as the
$K'$-points of $\bb P$ for every algebraically closed field $K'\supset
k$ are the quotients $C'$ of $A_{K'}$ such that $\Spec(C')$ is supported
at a $K'$-point, so is absolutely irreducible.  Moreover, $Z$ contains a
nonempty open subscheme $U$ that's open in $\bb H_R$.  Since $Z$ is
integral of dimension $\mb F+r$, so is $U$.  Thus $U$ contains a point
at which $\bb H_R$ is smooth of dimension $\mb F+r$.

Conversely, suppose $Z$ contains a scheme point $z$ at which $\bb H_R$
is smooth of dimension $\mb F+r$.  Then $z$ lies in a unique integral
component $Z'$ of $\bb H_R$ of dimension $\mb F+r$.  But $Z$ is
integral; so $Z$ lies in some irreducible component $Z''$ of $\bb H$.
But $z\in Z$; so $Z''= Z'$.  But $\dim Z = \dim Z'$.  Thus $Z = Z'$.
Thus (1) holds.

{\bf For (2),} let $T_{\bb F}$ be the tangent space at the $K$-point
$C$ to $\bb F\Lambda_{A_K}^{\{\mb h_m\}}$.  First, let's prove $T_{\bb
F}= F^0T$.

Recall that $\bb H \Lambda_{A}^{\{\mb h_m\}}$ is an open subscheme of\/
$\bb H \Psi_{A}^{\mb h_{\bar s}}$ by \eqref{prSmOp}(1).  But $\bb F
\Psi_{A}^{\mb h_{\bar s}}$ is the preimage of\/ $\bb H
\Lambda_{A}^{\{\mb h_m\}}$ in $\bb F \Psi_{A}^{\mb h_{\bar s}}$ under
the retraction map by definition; see \eqref{sbMLev}.  Thus $T_{\bb F}$
is also the tangent space at $C$ to $\bb F \Psi_{A}^{\mb h_{\bar s}}$.
Note $T_{\bb F}\subset T$.

Let $I$ be the ideal of $C$.  Fix a standard basis $\{f_i\}$ of $I$.  A
$\tau\in T = \Hom_A(I,C)$ corresponds to a flat deformation of $C$ over
the dual numbers $K[e]$; say its ideal has basis $\{f_i+g_ie\}$ with
$g_i\in A$.  Then $\tau \in T_{\bb F}$ iff the initial terms of the
$f_i+g_ie$ define a flat deformation of $G_\bu C$.  The latter holds iff
the values $v(f_i)$ and $v(f_i+g_ie)$ are equal for all $i$ by Bruce
Bennett's \cite{Bennett}*{Thm.\,(2.13). p.\,29}; so iff $v(f_i) \le
v(g_i)$ for all $i$.  But $g_i$ reduces to $\tau(f_i)\in C$.  So if
$\tau \in T_{\bb F}$, then $\tau \in F^0T$.  Conversely, if $\tau \in
F^0T$, then replace $g_i$ with another lifting of $\tau(f_i)$ so that
$v(f_i) \le v(g_i)$.  Thus $T_{\bb F}= F^0T$.

Note that $\bb F\Lambda_{A}^{\{\mb h_m\}}$ is smooth and irreducible of
fiber dimension $\mb F$ by \eqref{coFdim}.  So $\dim F^0T = \mb F$.  So
$\dim_K(T/F^0T) + \mb F = \dim_K(T)$.  Thus if $\dim_K(T/F^0T) \le r$,
then $\dim_K(T) \le \mb F+r$, and if $\dim_K(T) = \mb F+r$, then
$\dim_K(T/F^0T) = r$.

Note $\dim Z =\mb F+r$ by (1).  Hence $\bb H_K$ is of dimension at least
$\mb F+r$ at the $K$-point $C$.  Thus $\dim_K(T) \ge \mb F+r$.  Hence,
if $\dim_K(T) \le \mb F+r$, then $\dim_K(T) = \mb F+r$, and so
$\dim_K(T/F^0T) = r$.  Thus (2) holds.

{\bf For (3),} consider the standard right-exact sequence
\begin{equation}\label{eqLift1}
\Der_K(A_K,C)\big/F^0\Der_K(A_K,C) \xto{\ d\ } T\big/F^0T
\to T^1(C)\big/F^0T^1(C)\to 0.
\end{equation}
So $C$ has trivial negative tangents, or $T^1(C)/F^0T^1(C) = 0$, iff
$d$ is surjective.  But $\Omega^1_{A/k} = A(-1)^{\oplus r}$.  So
$\Der_K(A_K,C) = C[1]^{\oplus r}$.  So the first term in \eqref{eqLift1}
is of $K$-dimension $r$.  Thus (3) holds.

{\bf For (4),} note $\dim_K T = \mb F+r$ by (2).  But (1) imples $\bb
H_K$ is of dimension at least $\mb F+r$ at $C$.  So $\bb H_K$ is smooth
of dimension $\mb F+r$ at $C$.  Hence, $\bb H_R$ is too by the Jacobian
criterion.  Thus (4) holds.

{\bf For (5),} note $Z$ and $\bb P_R$ are integral of dimension $\mb
F+r$ by (1) and its proof.  Recall $\theta$ is injective on field-valued
points; that is, $\theta$ is radicial, see \cite{EGAIihes}*{(3.5.4),
p.\,115}.  So $\theta_R$ is purely inseparable by
\cite{EGAIihes}*{(3.5.8), p.\,116}.  Thus (b) implies $\theta$ induces a
birational map $\theta'_R\: \bb P_R \to Z$.  Let's prove (a) too implies
$\theta'_R$ is birational.

Assume (a).  Then the map $d$ in \eqref{eqLift1} is surjective.  Its
source is plainly equal to $\Der_K(A_K,K)$.  And under this
identification, $d$ becomes the tangent map of the restriction of
$\theta$ to the fiber over $C$ of the projection from $\bb P$ onto its
first factor.  The proof of (1) yields $T_{\bb F}= F^0T$.  Hence the
tangent map $d\theta'_R$ is surjective at $(C,0)$.  But its source is of
dimension $\mb F+r$, as $\bb P_R$ is smooth and irreducible of dimension
$\mb F+r$ by the proof of (1).  And $\dim_K(T) = \mb F+r$ by (2) and
(3).  Thus $d \theta_R$ is an isomorphism of $K$-vector spaces at

Hence $\theta_K$ is smooth at $(C,0)$ by d)\,\implies\,a) of
\cite{EGAIV}*{(17.11.1), p.\,82}, as the scheme points of $P_K$ and $\bb
H_K$ representing $(C,0)$ and $C$ have the same residue field, namely
$K$.  So $\theta_R$ is smooth at its $K$-point $(C,0)$.  But $\theta_R$
is purely inseparable, as noted above.  Thus $\theta'_R\: \bb P_R \to Z$
is birational.

Hence, $\theta^{-1}Z\sm \to Z\sm$ is bijective and birational.  But
$Z\sm$ is smooth, so normal.  Thus \cite{EGAIII0}*{(4.4.9), p.\,137}
implies $\theta^{-1}Z\sm \to Z\sm$ is an isomorphism.

Finally, replace $K$ by its algebraic closure.  Then $C'$ is supported
at a $K$-point of $\bb A_K^r$.  Change coordinates so that this point is
the origin.  First, assume $C'$ has trivial negative tangents. Then
$H_R$ is smooth of dimension $\mb F+r$ at $C'$ by (3) and (4) with $C'$
for $C$.  Hence, $Z$ is a component of $\bb H_R$ by (1), and it's the
only component containing $C'$.  Thus $C'$ lies in $Z\0\sm$.

Conversely, assume $C'$ lies in $Z\0\sm$.  Then $d\theta$ yields an
isomorphism from the tangent space of $\bb P_K$ at $(C',0)$ onto that,
$T'$ say, of $\bb H_K$ at $C'$, since $\theta^{-1}Z\sm \to Z\sm$ is an
isomorphism.  So let $T_{\go f}$ be the tangent space of the fiber
through $C'$ of the projection from $\bb P$ onto its first factor; then
$d\theta$ yields an isomorphism from $T_{\go f}$ onto $T'/F^0T'$, see
the reasoning above for $C$.  Furthermore, with $C'$ for $C$, the map
$d$ in \eqref{eqLift1} is therefore an isomorphism.  Thus
$T^1(C')\big/F^0T^1(C') = 0$; that is, $C'$ has trivial negative
tangents.  Thus (5) holds.

{\bf In (6),} $k$ is a domain; so $\bb F\Lambda_A^{\{\mb h_m\}}$ is
integral by \eqref{coFdim}.  So $\bb P$ is too. Thus $\Im(\theta)$ is
integral, so lies in an irreducible component $X$ of $\bb H$.

By hypothesis, $Z$ is a component of\/ $\bb H_R$.  Also, $Z$ is
integral by (1).  Hence there's a scheme point $x \in Z\sm$ that lies in
no other component of $\bb H_R$ than $Z$.  So $x \in Z\0\sm$.  Note
$\dim Z = \mb F + r$ by (1), and $x \in X$.  So $X_R$ is smooth of
dimension $\mb F+r$ at $x$.

Consider the local rings $\mc O_{X,x}$ and $\mc O_{X_{\rm red},x}$.
Note that $\mc O_{X,x}\ox_kR = \mc O_{X_R,x}$ and $\mc O_{X_{\rm
red},x}\ox_kR = \mc O_{(X_{\rm red})_R,x}$.  But $X_R$ is smooth at $x$;
hence, $\mc O_{X_R,x}$ is reduced, so equal to $\mc O_{(X_{\rm
red})_R,x}$.  Thus $\mc O_{X,x}\ox_kR = \mc O_{X_{\rm red},x}\ox_kR$.

Let $\go n$ be the nilradical of $\mc O_{X,x}$.  So the sequence $0 \to
\go n \to \mc O_{X,x} \to \mc O_{X_{\rm red},x} \to 0$ is exact.  So it
remains exact after tensoring over $k$ with $R$, as $X_{\rm red}$ is
$k$-flat since $k$ is a Dedekind domain and $X_L \neq \emptyset$.  So
$\go n\ox_kR = 0$.  So $\go n = 0$ by Nakayama's Lemma over $\mc
O_{X,x}$.  So $\mc O_{X,x} = \mc O_{X_{\rm red},x}$.  So $\mc O_{X,x}$
is $k$-flat.  But $X_R$ is $R$-smooth of dimension $\mb F+r$ at $x$.
Hence $X$ is $k$-smooth of fiber dimension $\mb F+r$ at $x$.  But $X$
has no embedded nilpotents by the definition of the scheme structure on
an irreducible component.  Thus $X$ is an $(\mb F+r+1)$-dimensional
integral component of\/ $\bb H$.

However, $L$ is the fraction field of $k$; hence, $X_L$ is an $(\mb
F+r)$-dimensional integral component of\/ $\bb H_L$.  Now, $Y$ is
integral of dimension $\mb F+r$ by (1) with $L$ for $R$.  But $Y \subset
X_L$.  So $Y = X_L$.  Thus (6) holds.
\end{proof}

\begin{remark}\label{rePGor}
Preserve the conditions of \eqref{prLift}.  Recall the first step in
the proof of (1): it shows $T_{\bb F}= T_{\mb h} = F^0T$ where $T_{\mb
h}$ is the tangent space at $C$ to $\bb F \Psi_{A}^{\mb h_{\bar s}}$; in
fact, it shows $T_{\mb h} = F^0T$ even when $C$ isn't I-compressed.
Notice that a small modification of the proof shows that, if $C$ is
homogeneous, then the tangent spaces at $C$ to $\bb H\Lambda_A^{\{\mb
h_m\}}$ and $\bb H \Psi_{A}^{\mb h_{\bar s}}$ are equal, and equal to
$T_0$; in fact, the latter two spaces are equal even when $C$ isn't

When $C$ is homogeneous, but not necessarily $I$-compressed, then the
equation $T_{\mb h} = F^0T$ was already proved in \cite{PGor}; see the
left-hand expression in the second display on p.\,615, noting the
definition of $Z_H(S)$ on p.\,614 and Remark~1.11(i) on p.\,617.
(Moreover, the right-hand expression is equal to the obstruction space
for $\bb F \Psi_{A}^{\mb h_{\bar s}}$ at $C$; this obstruction theory
agrees with Jelisiejew's \cite{JJ-JLMS2019}*{Thm.\,4.2, p.\,258}.)  The
proof of \cite{PGor}*{Thm.\,1.10, p.\,615} suggests using either
Bennett's \cite{Bennett}*{Thm.\,(2.13), p.\,29} or the lengthy
elementary argument on pp.\,616--617 in \cite{PGor} (to treat the
tangent space, but not the obstruction space); of course, Bennett's
theorem yields more results, as noted above.
\end{remark}

\section{Compressed Quotients of Gorenstein Artinian Rings}\label{GCQ}

\begin{setup}\label{se11}
Keep the setup of \eqref{se10}; in particular, keep the Noetherian base
scheme $S$ and the $\mc O_S$-algebra $\mc A$.  Also, fix a Noetherian
ring $k$, a $k$-flat finitely generated graded algebra $A :=
\bigoplus_{p\ge0}A_p$ with $A/F^1A = k$.

Call a field $K$ an {\it $S$-field} if $K$ contains a residue field of
\indt{Sfie@$S$-field}
$S$.  Similarly, call a field $K$ an {\it $k$-field} if $K$ is a
\indt{kfie@$k$-field}

Given a graded $A$-module $M = \bigoplus_p M_p$ and $g_i\in A_{d_i}$
with $d_i>0$ for $1\le i\le m$, note that the usual Koszul complex
$K_\bu(g_1,\dotsc,g_m;M)$ is naturally graded: its terms are sums of
right-shifts of M; and its differentials are maps of degree 0.  For
example, $K_\bu(g_1;M)$ is the complex $M(-d_1) \to M$ whose
differential is multiplication by $g_1$.  Furthermore,
\begin{equation*}\label{eqse11a}
K_\bu(g_1,\dotsc,g_m;M) = K_\bu(g_1;A) \ox_A \dotsb
\ox_A K_\bu(g_m;A) \ox_A M,
\end{equation*}
and the grading on the left-hand side is induced by that on the factors
on the right.

Fix an indeterminate $z$.  If $M_p$ is locally free of rank $\mb m(p)$
\indn{HmbrMz@$\mb H(M,z)$}
for all $p$, set
$$\ts\mb H(M,z) := \sum_p \mb m(p)z^p.$$

Given Laurent series $\mb A(z) = \sum a_p z^p$ and $\mb B(z) = \sum
b_p z^p$ and $\mb C(z) = \sum c_p z^p$ with real coefficients and $c_q =
0$ for $q < 0$, let $\mb A(z) = \mb B(z) \mod z^n$ mean $a_p = b_p$ for
$p < n$.  Then $\mb A(z)\mb C(z) = \mb B(z)\mb C(z) \mod z^n$.  Also,
let $\mb A(z) \le \mb B(z) \mod z^n$ mean $a_p \le b_p$ for $p < n$.  If
$c_q \ge 0$ for $q\ge 0$, then $0\le\big(\mb B(z) - \mb A(z)\big)\mb
C(z) \mod z^n$ and so $\mb A(z)\mb C(z) \le \mb B(z)\mb C(z) \mod z^n$.
Moreover, let $\mb A(z) \le \mb B(z)$ mean $a_p \le b_p$ for all $p$.
If $0 \le \mb C(z)$, then $\mb A(z)\mb C(z) \le \mb B(z)\mb C(z)$.
\end{setup}

\begin{lemma}\label{leCplx}
Let $M \xto\mu N \xto\nu P$ be a sequence of filtered $A$-modules.
Assume the filtrations are exhaustive, and there's $n$ with $G_pM \to
G_pN \to G_pP$ exact for all $p < n$ and with\/ $\nu\mu M \subset
F^{n+1}P$.  Set $D := \Im\nu$ and $F^pD := D \cap F^pP$ for all $p$.

\(1) Then $F^qM/F^pM \to F^qN/F^pN \to F^qP/F^pP$ is exact for all $q <
p \le n$.

\(2) Then $G_p D = \Im G_p\nu$ for all $p\le n$.

\(3) Fix $p<n$. Assume $(M/F^{p+1}M)\ox_Ak \to (N/F^{p+1}N)\ox_Ak$
vanishes, and assume $((G_\bu N)\ox_Ak)_p \risom G_p(N\ox_Ak)$.  Then
$((G_\bu D)\ox_Ak)_p \risom G_p(D\ox_Ak)$.
\end{lemma}

\begin{proof}
{\bf In (1),} as $\nu\mu M \subset F^{n+1}P \subset F^pP$, the
sequence is a complex.  So it's exact as $G_rM \to G_rN \to G_rP$ is
exact for $q \le r < p$ and $L \mapsto G_\bu L$ is biexact by
\eqref{prBiex}.

{\bf For (2),} fix $p$.  Note $G_p D \subset G_p P$ as $F^{p+1} D := (
D) \cap F^{p+1}P$.  So we have to show $G_p N \to G_p D$ is surjective.
Thus given $z\in F^p D$, we have to find $y\in F^pN$ with $\nu y - z \in
F^{p+1}P$.  But $D := \Im\nu$; so $z = \nu y'$ with $y' \in F^qN$ for
some $q$.

As $z\in F^p D$, owing to (1) there's $x \in F^qM$ with $y := y'-\mu
x \in F^pN$.  Then $\nu y = \nu y' - \nu\mu x$.  But $\nu\mu x \in
F^{n+1}P \subset F^{p+1}P$.  Also $z = \nu y'$.  Thus $\nu y - z \in

{\bf For (3),} form this induced commutative square with surjective
right column:
\begin{equation*}\label{eqeqCplx}
\begin{CD}
((G_\bu N)\ox_Ak)_p @>\nu_t>> ((G_\bu D)\ox_Ak)_p \\
@VV\simeq V                          @VVV\\
G_p(N\ox_Ak)    @>\nu_b>>      G_p(D\ox_Ak)
\end{CD}
\end{equation*}
To prove (3), let's prove that $\nu_t$ is surjective and that $\nu_b$
is an isomorphism.

Given any graded $A$-module $L$, note that $(L\ox_Ak)_p =
L_p\big/((F^1A)L)_p$ and that $((F^1A)L)_p = \sum_{q>0}A_qL_{p-q}$.
Hence the hypotheses of (3) and its assertion don't involve $F^nM$ and
$F^nN$ and $F^nP$.  Thus replacing $M$ and $N$ and $P$ by $M/F^nM$ and
$N/F^nN$ and $P/F^nP$, we may assume $F^rM$ and $F^rN$ and $F^rP$ vanish
if $r\ge n$.

For all $r$, let's see $M/F^rM \to N/F^rN \to P/F^rP$ is now exact.
Indeed, given $y \in N$ with $\nu y \in F^rP$, say $y \in F^qN$.  Note
(1) holds now with $p$ replaced by any $r$.  Thus (1) yields $x \in
F^qM$ with $\mu x -y \in F^rN$, as desired.

For all $r$, the image of $N/F^rN$ in $P/F^rP$ is plainly $D/(D\cap
F^rP)$.  However, $F^rD := D \cap F^rP$.  Thus $M/F^rM \to N/F^rN \to
D/F^rD \to 0$ is exact.

For all $r$, form the following commutative diagram with exact rows and
\begin{equation*}\label{eqeqCplxa}
\begin{CD}
@.       @.  M/F^{r+1}M @>>> M/F^rM @>>> 0\\
@.      @.          @VVV           @VVV\\
0 @>>> G_rN @>>> N/F^{r+1}N @>>> N/F^rN @>>> 0\\
@.  @VV\gamma V  @VV\beta V    @VV\alpha V\\
0 @>>> G_rD @>>> D/F^{r+1}D @>>> D/F^rD @>>> 0
\end{CD}
\end{equation*}
As $\alpha$ and $\beta$ are surjective, a diagram chase shows $\gamma$
is too.  So $G_\bu N \to G_\bu D$ is surjective; so $(G_\bu N)\ox_Ak \to
(G_\bu D)\ox_Ak$ is too.  Thus $\nu_t$ is surjective, as desired.

For all $r$, note $(M/F^rM)\ox_Ak \to (N/F^rN)\ox_Ak \to (D/F^rD)\ox_Ak
\to 0$ is exact owing to the second paragraph above.  The first map
vanishes for $r := p+1$ by hypothesis.  So it vanishes for all $r \le
p+1$, as $(M/F^{p+1}M)\ox_Ak \to (M/F^rM)\ox_Ak$ is surjective.  Thus
$(N/F^rN)\ox_Ak \risom (D/F^rD)\ox_Ak$ for all $r \le p+1$.

But for any filtered $A$-module $L$, note $(L/F^rL)\ox_Ak =
(L\ox_Ak)\big/F^r(L\ox_Ak)$ as $F^r(L\ox_Ak)$ is the image of
$(F^rL)\ox_Ak$ in $L\ox_Ak$.  Hence, in the following commutative
diagram, the right two vertical maps are isomorphisms, as indicated:
\begin{equation*}\label{eqeqCplxb}
\begin{CD}
0 @>>> G_p(N\ox_Ak) @>>> (N\ox_Ak)\big/F^{p+1}(N\ox_Ak)
@>>> (N\ox_Ak)\big/F^p(N\ox_Ak) @>>> 0\\
@.        @V\nu_bVV              @VV\simeq V  @VV\simeq V\\
0 @>>> G_p(D\ox_Ak) @>>> (D\ox_Ak)\big/F^{p+1}(D\ox_Ak)
@>>> (D\ox_Ak)\big/F^p(D\ox_Ak) @>>> 0
\end{CD}
\end{equation*}
The horizontal sequences are exact. So $\nu_b$ is an isomorphism.  Thus
(3) holds.
\end{proof}

\begin{lemma}\label{leInj}
Fix $m$, and for $1\le i\le m$, fix nonzero $g_i\in A_{d_i}$ with
$d_i>0$.  Fix a $k$-flat, finitely generated, graded $A$-module $M =
\bigoplus_p M_p$.  Set $M^{(0)} := M$.  For $1\le q\le m$, set $M^{(q)}
:= M\big/\sum_{i=1}^qg_iM$, set $\mb P^{(q)}(z) := \prod_{i=1}^q
(1-z^{d_i})$, and for all $p$, define $\mu^{(q)}_p\: M^{(q-1)}_{p-d_q}
\to M^{(q-1)}_p$ by $\mu^{(q)}_p(x) := g_q x$.  Fix $n$.

\(1) Assume $k$ is a field.  Then $\mb H(M,z) \le \mb H(M^{(m)},z) \big/
\mb P^{(m)}(z)$.  Further, $\mu^{(q)}_p$ is injective for $1\le q\le m$
and all $p < n$ iff $\mb H(M,z) = \mb H(M^{(q)},z) \big/ \mb P^{(q)}(z)
\mod z^n$ for $1\le q\le m$, iff $\mb H(M,z) = \mb H(M^{(m)},z) \big/
\mb P^{(m)}(z) \mod z^n$.

\(2) Then $\mu^{(q)}_p$ is injective and $M^{(q)}_p$ is $k$-flat for
$1\le q\le m$ and all $p < n$ iff $\mb H(M,z) = \mb
H((M\ox_kK)^{(m)},z) \big/ \mb P^{(m)}(z) \mod z^n$ for every $k$-field

\(3) For all $r\ge 1$, the Koszul homology group $H_r(g_1,\dotsc,g_m;M)$
is $0$ in degree $p < n$, iff $H_1(g_1,\dotsc,g_m;M)$ is so, iff
$\mu^{(q)}_p$ is injective for $1\le q\le m$ and all $p < n$.
\end{lemma}

\begin{proof} {\bf For (1),} cf.\ \cite{RStan78}*{Cor.\,3.2, p.\,62}.
Given $p$, form this right exact sequence:
\begin{equation}\label{eqInj1}
M^{(m-1)}_{p-d_m} \xto{\mu^{(m)}_p} M^{(m-1)}_p \to M^{(m)}_p \to 0.
\end{equation}
Set $\mb m^{(j)}(p) := \dim M^{(j)}_p$.  Then \eqref{eqInj1} yields
$\mb m^{(m-1)}_p - \mb m^{(m-1)}_{p-d_m} \le \mb m^{(m)}_p$, with
equality iff $\mu^{(m)}_p$ is injective.  Thus
\begin{equation}\label{eqInj4}
\mb
H(M^{(m-1)},z)(1-z^{d_m}) \le \mb H(M^{(m)},z).
\end{equation}
Also, $\mb H(M^{(m-1)},z) (1-z^{d_m}) = \mb H(M^{(m)},z) \mod z^n$ iff
$\mu^{(m)}_p$ is injective for $p < n$.

Induction on $m$ gives $\mb H(M,z) \le \mb H(M^{(m-1)},z) \big/ \mb
P^{(m-1)}(z)$.  Now, multiply \eqref{eqInj4} by $1\big/\mb P^{(m)}(z)$.
The inequality is preserved as $1\big/\mb P^{(m)}(z)$ has nonnegative
coefficients.  But $\mb P^{(m)}(z) = \mb P^{(m-1)}(z)(1-z^{d_m})$.  Thus
\begin{equation}\label{eqInj3+}
\mb H(M,z) \le  \mb H(M^{(m-1)},z) \big/ \mb P^{(m-1)}(z)
\le \mb H(M^{(m)},z) \big/ \mb P^{(m)}(z)
\end{equation}

Further, suppose $\mu^{(q)}_p$ is injective for $1\le q\le m$ and all $p
< n$.  Then by induction, $\mb H(M,z) = \mb H(M^{(q)},z) \big/ \mb
P^{(q)}(z) \mod z^n$ for $1\le q < m$.  But $\mu^{(m)}_p$ is injective
for $p < n$.  So  $\mb H(M^{(m-1)},z) (1-z^{d_m}) = \mb H(M^{(m)},z)
\mod z^n$.  Thus also
$$\mb H(M,z) = \mb H(M^{(m-1)},z) \big/ \mb P^{(m-1)}(z)
= \mb H(M^{(m)},z) \big/ \mb P^{(m)}(z) \mod z^n. $$

Conversely, suppose $\mb H(M,z) = \mb H(M^{(m)},z) \big/ \mb P^{(m)}(z)
\mod z^n$.  Then in \eqref{eqInj3+}, the two inequalities become
equalities mod $z^n$.  So by the first and by induction,
$\smash{\mu^{(q)}_p}$ is injective for $1\le q < m$ and all $p < n$.
But as $\mb P^{(m)}(z)$ is an ordinary polynomial, the second equality
implies $\mb H(M^{(m-1)},z) (1-z^{d_m}) = \mb H(M^{(m)},z) \mod z^n$.
So as shown above, $\mu^{(m)}_p$ is injective for $p < n$.  Thus (1)

{\bf For (2),} first assume $\mu^{(q)}_p$ is injective and $M^{(q)}_p$ is
$k$-flat for $1\le q\le m$ and all $p < n$.  For any $k$-algebra $K$,
then $\mu^{(q)}_p\ox_kK$ is injective and $M^{(q)}_p\ox_kK$ is $K$-flat.
But $\mb H(M,z) = \mb H(M\ox_kK,z)$ and $M^{(q)}_p\ox_kK =
(M\ox_kK)^{(q)}$.  Thus (1) yields $\mb H(M,z) = \mb
H((M\ox_kK)^{(q)},z) \big/ \mb P^{(m)}(z) \mod z^n$ for every $k$-field

Conversely, assume $\mb H(M,z) = \mb H((M\ox_kK)^{(m)},z) \big/ \mb
P^{(m)}(z) \mod z^n$ for every $k$-field $K$.  Then (1) implies $\mb
H(M,z) = \mb H((M\ox_kK)^{(m-1)},z) \big/ \mb P^{(m-1)}(z) \mod z^n$ for
every $k$-field $K$.  So by induction on $m$, then $\mu^{(q)}_p$ is
injective and $M^{(q)}_p$ is $k$-flat for $1\le q < m$ and all $p < n$.
In particular, $M^{(m-1)}_p$ is flat.  Also, (1) implies
$\mu^{(m)}_p\ox_kK$ is injective.  So by the Local Criterion of
Flatness, $\mu^{(m)}_p$ is injective and $M^{(m)}_p$ is $k$-flat for all
$p < n$.  Thus (2) holds.

{\bf For (3),} cf.\ \cite{SerreAlLoc}*{Prop.\,3, p.\,IV-5}.  Form the
Koszul complexes $K_\bu := K_\bu(g_m;A)$ and $L_\bu :=
K_\bu(g_1,\dotsc,g_{m-1};M)$.  Then \cite{SerreAlLoc}*{Prop.\,1,
p.\,IV-2} yields, for all $r$, the following exact sequence of graded
modules and maps of degree 0: \begin{equation}\label{eqInj2}
0 \to H_0(K_\bu\ox_A H_r(L_\bu)) \to  H_r(K_\bu\ox_A L_\bu)
\to H_1(K_\bu\ox_A H_{r-1}(L_\bu)) \to 0.
\end{equation}

First assume $\mu^{(q)}_p$ is injective for $1\le q\le m$ and $p < n$.
Induct on $m$.  So assume $H_r(L_\bu)_p = 0$ for $r \ge 1$ and $p < n$.
So $H_0(K_\bu\ox_A H_r(L_\bu))_p = 0$, and if $r\ge 2$, then
$H_1(K_\bu\ox_A H_{r-1}(L_\bu))_p = 0$.  But $H_0(L_\bu)) = M^{(m-1)}$.
Furthermore, $K_\bu\ox_A M^{(m-1)} = K_\bu(g_m,M^{(m-1)})$.  So
$H_1(K_\bu\ox_A H_0(L_\bu))_p = 0$ as $\mu^{(m)}_p$ is injective.  Thus
\eqref{eqInj2} yields $H_r(g_1,\dotsc,g_m;M)_p = 0$, as desired.

Finally, fix $p < n$ and assume $H_1(g_1,\dotsc,g_m;M)_p = 0$.  Then
\eqref{eqInj2} yields $\Ker\mu^{(m)}_p = 0$ and $H_1(L_\bu)_p \big/
(g_m\cdot H_1(L_\bu)_{p-d_m}) = 0$.  As $d_m > 0$, the latter yields
$H_1(L_\bu)_p = 0$.  So $\Ker\mu^{(q)}_p=0$ for $1\le q< m$ by induction
on $m$. Thus (3) holds.
\end{proof}

\begin{proposition}\label{prOpens}
Fix $d_1,\dotsc,d_m >0$, and set $\mb P(z) := \prod_{i=1}^m(1-z^{d_i})$
and set $\bb P := \prod_{i=1}^m \bb P(\mc A^*_{d_i})$.  Fix an $S$-flat,
finitely generated, graded $\mc A$-module $\mc M = \bigoplus_p \mc M_p$.
Given an $S$-field $K$, set $A_K := \Gamma(\mc A_{\Spec K})$ and $M_K :=
\Gamma(\mc M_{\Spec K})$.  Given a $K$-point of $\bb P$, let
$g_1,\dotsc,g_m \in (A_K)_{d_i}$ correspond to it.  Set $M_K^{(m)} :=
M_K\big/\sum_{i=1}^mg_iM_K$.  Assume $S$ is irreducible.  Fix $n\ge 1$.

\(1) There's a nonempty open subset $U$ of $\bb P$ such that, given any
$S$-field $K$, a $K$-point of $\bb P$ lies in $U$ iff $\mb
H(M_K^{(m)},z) \le \mb H(M_{K^\di}^{(m)},z)\mod z^n$ for every
$K^\di$-point of $\bb P$ for every $S$-field $K^\di$.

\(2) Assume $\mb H(M_K,z)\mb P(z) = \mb H(M_K^{(m)},z) \mod z^n$ for
some $K$-point of $\bb P$ for some $S$-field $K$.  Then given any
$S$-field $K^\di$ and any $K^\di$-point of $\bb P$, the latter lies in
the open subset $U$ of \(1) iff\/ $\mb H(M_{K^\di},z)\mb P(z) = \mb
H(M_{K^\di}^{(m)},z) \mod z^n$.
\end{proposition}

\begin{proof}
First off, let's define a global version on $\bb P$ of the module
$M^{(q)}$ of \eqref{leInj}.  For $1\le i\le m$, set $\bb P_i := \bb
P(\mc A^*_{d_i})$, and denote the pullback to $\bb P$ of the
tautological injection $\mc O_{\bb P_i}(-1)\into (\mc A_{d_i})_{\bb
P_i}$ by $\mc L_i \into (\mc A_{d_i})_{\bb P}$.  Combine the latter with
the pullback of the multiplication map $\mc A\ox \mc M \to \mc M$ to get
a map $\mc L_i \ox_{\mc O_P}\mc M(-d_i)_{\bb P} \to \mc M_{\bb P}$.
Finally, set $\mc M^{(q)} := \cok\bigl(\bigoplus_{i=1}^q \mc L_i \ox_{\mc
O_P}\mc M(-d_i)_{\bb P} \to \mc M_{\bb P}\bigr)$.

Note that $S$ is irreducible, and that each $\smash{\mc A_{d_i}^*}$ is
locally free.  Thus $\bb P$ is irreducible.

{\bf For (1),} for $0\le q< n$, apply the theory of flattening
stratification to $\mc M^{(m)}_q$.  Let $U_q$ be the stratum of $\bb P$
on which the restriction of $\mc M^{(m)}_q$ has minimum rank, $r_q$ say.
Then $U_q$ is nonempty and open.  Set $U : = \bigcap_{q=0}^{n-1} U_q$.
Then $U$ is open.  Further, as $\bb P$ is irreducible, $U$ is nonempty
and irreducible.  Set $\mb R(z) := \sum_{q=0}^{n-1} r_qz^q$.  Then given
an $S$-field $K$ and a $K$-point of $\bb P$, the construction of $U$
yields $\mb R(z) \le \mb H(M_K^{(m)},z) \mod z^n$, with equality iff the
$K$-point lies in $U$.  Thus (1) holds.

{\bf For (2),} let's find an affine open subset $V^{(m)} \subset \bb P$
containing the given $K$-point such that, in the notation of
\eqref{leInj} with $M^{(q)} := \Gamma(V^{(q)},\mc M^{(q)})$, the map
$\mu^{(q)}_p$ is injective and the module $M^{(q)}_p$ is flat for $1\le
q\le m$ and $p < n$.  To find $V^{(m)}$, note $\mb H(M_K,z) = \mb
H(M_K^{(m-1)},z) \big/ \mb P^{(m-1)}(z) \mod z^n$ by \eqref{leInj}(1).
So induction on $m$ yields the analogous set $V^{(m-1)}$.  In
particular, $M^{(m-1)}_p$ is flat.  But $\mu^{(m)}_p\ox_kK$ is injective
by \eqref{leInj}(1).  So the Local Criterion of Flatness yields an
affine open subset $V^{(m)} \subset V^{(m-1)}$ containing the given
$K$-point such that the restriction of $\mu^{(m)}_p$ to $V^{(m)}$ is
injective and the restriction of $M^{(m)}_p$ is flat for all $p < n$, as

Both $V^{(m)}$ and $U$ are nonempty open subsets of $\bb P$, and $\bb P$
is irreducible.  So $V^{(m)} \cap U \neq \emptyset$. So $V^{(m)} \cap U$
contains a $K^\di$-point for some field $K^\di$.  Then \eqref{leInj}(2)
yields $\mb H(M_{K^\di},z) = \mb H(M_{K^\di}^{(m)},z) \big/ \mb
P^{(m)}(z) \mod z^n$. But $\mb P(z)$ is an ordinary polynomial.  So $\mb
H(M_{K^\di},z)\mb P(z) = \mb H(M_{K^\di}^{(m)},z) \mod z^n$.  But mod
$z^n$, for every $K^\di$-point of $U$, the value of $ \mb
H(M_{K^\di}^{(m)},z)$ is the same, and that value is strictly less than
the value of $ \mb H(M_{K^\di}^{(m)},z)$ for every $K^\di$-point of $\bb
P - U$.  Thus (2) holds.
\end{proof}

\begin{proposition}\label{prNonempty}
Assume $k$ a field; $A$ Gorenstein, Artinian of socle degree a,
generated by $A_1$.  Set $r := \mb a(1)$ and $\mb r(p) := \binom{p+
r-1}{r-1}$ and $v_0 := \inf\{\, v \mid \mb r(v) > \mb a(v)\, \}$.  Set
$\mb P(z) := \prod_j(1-z^j)^{\mb t(a-j)}$.  Define $\mb w^*(p)$ by
$\sum_p\mb w^*(p)z^p := \mb H(A^\dg,z) \mb P(z)$.  Fix $n \le v_0 - a$
with $\mb w^*(n-1)=0$.  Assume $\mb t^*(p)=0$ for $p\ge n$ and $p \le
-a$.  Set $\mb u(i) := \sum_{j > i}\mb t(j)$.  Assume $\mb u(1-n) \le
r$.  Set $\mb h(q) := \mb a(q)-\mb w(q)$ for $q > -n$ and\/ $\mb h(q) :=
\mb a(q)$ for $q \le -n$.  Then $\mb w^*(p) \ge 0$ for $p <n$, and $\mb
H\Psi_A^{\mb h, \mb t} \neq \emptyset$.
\end{proposition}

\begin{proof}
Since $A$ is Artinian, $A^\dg = A^*$.  Since $A$ is Gorenstein, $A^*$
is an invertible $A$-module, so free of rank 1.  Choose an $f\in
A^*_{-a}$ with $A^* = Af$.  Choose variables $X_1,\dotsc,X_r$ and set $R
:= k[X_1,\dotsc,X_r]$.  Fix a $k$-basis of $A_1$.  Mapping $X_i$ to its
$i$th element, view $A$ as a homogeneous quotient of $R$.  Thus $A^* =
Rf \subset R^\dg$.

Form the map $\mu\: R(a) \to A^*$ by $\mu(x) := xf$. It factors as
$R(a)\onto A(a)\risom A^*$ where $R\onto A$ is the quotient map.  Fix $i
< v_0$.  Then $\dim_k R_i = \dim_k A_i$ by definition of $v_0$.  So
$R_i\risom A_i$.  Thus $\mu_i\: R(a)_{i-a} \risom (A^*)_{i-a}$.

For $1 \le p < a+n-1$, note $u(a-p-1)\le r$, and for $u(a-p) < j \le
u(a-p-1)$, set $g_j := X^p_j$.  Set $J:= \sum Rg_j$.  As the $g_j$ form
an $R$-regular sequence, \eqref{leInj}(1) yields $\mb H(R,z) = \mb
H(R/J,\,z) \big/ \prod_{j=1}^{a+n-2} (1-z^j)^{\mb t(a-j)}$.  Multiplying
both sides by $\mb P(z)$ gives $\mb H(R,z)\mb P(z) = \mb
H(R/J,\,z)(1-z^{a+n-1})^{\mb t(1-n)}$ as $\mb t^*(p)=0$ for $p \ge n$.
However, $\mb H(R,z) = \mb H(A^*(-a),z) \mod z^{v_0}$.  Thus
$$\ts \mb H(R/J,\,z) (1-z^{a+n-1})^{\mb t^*(n-1)}
= \sum_p\mb w^*(p-a)z^p  \mod  z^{v_0}.$$

Compare terms of degree $a+p$ with $p< n-1$.  Thus $\mb w^*(p) = \dim
(R/J)_{a+p} \ge 0$ as  $a+p < v_0$.

Similarly, $\dim (R/J)_{a+n-1} - \mb t^*(n-1) = \mb w^*(n-1) = 0$.
Thus there are $g_j$ in $R_{a+n-1}$, for $\mb u(1-n)< j \le \mb u(-n)$,
whose residues form a basis of $(R/J)_{a+n-1}$.

Set $D := \sum_{j=1}^{\mb u(-n)} Ag_jf\subset A^*$ and $C := D^*$.  Then
$J(a)_i\risom D_i$ for $i< n-1$.  But $\dim J(a)_i = \mb h^*(i)$ for
$i<n-1$.  Thus $\dim C_p = \mb h(p)$ for $p > 1-n$.  Now, by
construction, $D_{n-1} = (A^*)_{n-1}$.  But $A^* = Rf$; so $R_1\cdot
(A^*)_i = (A^*)_{i+1}$ for all $i$.  So $D_i = (A^*)_i$ for all $i\ge
n-1$.  Thus $\dim C_p = \mb h(p)$ for $p \le 1-n$.  Plainly the $g_jf$
form a minimal generating set of $D$.  Thus $D$ has local generator type
$\mb t^*$. Thus $C \in \mb H\Psi_{A}^{\mb h, \mb t}$.
\end{proof}

\begin{proposition}\label{prN28May22}
Assume $k$ is a field, and $A$ is generated by $A_1$.  Fix a finitely
generated, graded $A$-module $B$, and $C \in \mb H\Psi_B^{\mb h, \mb
t}$ for some $\mb h,\ \mb t$.  Fix $n$ with $\mb t^*(p)= 0$ for $p\ge
n$.  Fix $\wt C \in \mb H\Psi_C^{\wt{\mb h}, \wt{\mb t}}$ for some
$\wt{\mb h},\ \wt{\mb t}$.  Assume $(C^*)_p = (\wt C^*)_p$ for $p <
n-1$, and $A_1(C^*)_{n-1} \subset A_1(\wt C^*)_{n-1}$.  Set $d := \dim
(C^*/\wt C^*)_{n-1}$.

\(1) Then $\wt{\mb h}(q) = \mb h(q)$ for $q \neq 1-n$, and\/ $\wt{\mb
h}(1-n) = \mb h(1-n) - d$.

\(2) Then $\wt{\mb t}(q) = \mb t(q)$ for $q\neq 1-n$, and\/ $\wt{\mb
t}(1-n) = \mb t(1-n) - d$.

\(3) Set $\mb P(z) := \prod_j(1-z^j)^{\mb t(a-j)}$ and\/ $\wt{\mb P}(z)
:= \prod_j(1-z^j)^{\wt{\mb t}(a-j)}$.  Define $\mb w^*(p)$ by $\sum_p\mb
w^*(p)z^p := \mb H(\mc B^\dg,z) \mb P(z)$ and $\wt{\mb w}^*(p)$ by\/
$\sum_p\wt{\mb w}^*(p)z^p := \mb H(\mc B^\dg,z) \wt{\mb P}(z)$.  Set
$b:=\dim B_a$. Then $\wt{\mb w}^*(p) = \mb w^*(p)$ for $p < n-1$, and\/
$\wt{\mb w}^*(n-1) = \mb w^*(n-1) + bd$.
\end{proposition}

\begin{proof} {\bf For (1) and (2),} for $p < n-1$, notice $\wt{\mb
h}^*(p) = \mb h^*(p)$ as $(\wt C^*)_p = (C^*)_p$.

For the same reason, $((F^1A)\wt C^*)_p = ((F^1A)C^*)_p$ but for $p\le
n-1$.  However, $\wt{\mb t}^*(p) := \dim \bigl( ({\wt C}^*)_{p} \big/
((F^1A){\wt C}^*)_{p} \bigr)$ and $\mb t^*(p) := \dim \bigl( (C^*)_{p}
\big/ ((F^1A) C^*)_{p} \bigr)$ for all $p$.  So $\wt{\mb t}^*(p) = \mb
t^*(p)$ for $p < n-1$, and\/ $\wt{\mb t}^*(n-1) = \mb t^*(n-1) - d$ as
$d := \dim (C^*/\wt C^*)_{n-1}$.

Notice $\wt{\mb h}^*(n-1) = \mb h^*(n-1)-d$, also as $d := \dim (C^*/\wt

For $p\ge n$, note $\mb t^*(p) = 0$; so $((F^1A) C^*)_{p} = (C^*)_p$.
But $A_1$ generates $A$; hence, $((F^1A) C^*)_{p} = A_1(C^*)_{p-1}$.
Thus $A_1(C^*)_{p-1} = (C^*)_p$.

Recall $A_1(C^*)_{n-1} \subset A_1(\wt C^*)_{n-1}$.  So $(C^*)_n \subset
A_1(\wt C^*)_{n-1} \subset (\wt C^*)_n \subset (C^*)_n$.  So
\begin{equation}\label{eqN28}
A_1(\wt C^*)_{p-1} = (\wt C^*)_p = (C^*)_p.
\end{equation}
holds for $p = n$.  Suppose \eqref{eqN28} holds for some $p\ge n$.
Then $A_1(\wt C^*)_p = A_1(C^*)_p$.  But $A_1(C^*)_p = (C^*)_{p+1}$.  So
$A_1(\wt C^*)_p = (C^*)_{p+1}$.  But $A_1(\wt C^*)_p \subset (\wt
C^*)_{p+1}$ and $(\wt C^*)_{p+1} \subset (C^*)_{p+1}$.  Thus
\eqref{eqN28} holds for all $p\ge n$ by induction on $p$.  Thus $\wt{\mb
t}^*(p) = 0$ and $\wt{\mb h}^*(p) = \mb h^*(p)$ for all $p > n-1$.  Thus
(1) and (2) hold.

{\bf For (3),} note $\mb t(1-n) - \wt{\mb t}(1-n) = d$ by (2).  Hence
$\mb P(z) = \wt{\mb P}(z)(1-z^{a+n-1})^d$.  So $\sum\mb w^*(p)z^p =
\bigl(\sum\wt{\mb w}^*(p)z^p\bigr) (1-z^{a+n-1})^d$.  Compare terms of
degree $p$ for $p \le n-1$, and note $\mb w^*(-a) = \dim B^\dg_{-a} = b$.
Thus (3) holds.
\end{proof}

\begin{theorem}\label{thAffBdl}
Assume $\mc A$ is $\mc O_S$-Gorenstein and $\mc O_S$-Artinian of socle
degree $a$.  Set $\mb P(z) := \prod_j(1-z^j)^{\mb t(a-j)}$.  Define $\mb
w^*(p)$ by $\sum_p\mb w^*(p)z^p := \mb H(\mc A^\dg,z) \mb P(z)$.  Fix $n$
with $\mb w^*(p) \ge 0$ for $p <n$. Set $\mb h(q) := \mb a(q)-\mb w(q)$
for $q > -n$, and\/ $\mb h(q) := \mb a(q)$ for $q \le -n$. Assume $\mb
t^*(p)=0$ for $p\ge n$ and $p \le -a$.  Finally, assume $\bb
H\Psi_{\mc A}^{\mb h, \mb t} \neq \emptyset$.

\(1a) Set $\mb R := \sum_p \mb t(p)\bigl(\sum_{p'<p}\mb w(p')\bigr)$.
Then $\bb F\Psi_{\mc A}^{\mb h, \mb t}$ is an affine-space bundle
over\/ $\bb H\Psi_{\mc A}^{\mb h, \mb t}$ of fiber dimension $\mb R$.

\(1b) Set $\mb H := \sum_p \mb t(p)\mb w(p)$.  Then $\bb H\Psi_{\mc
A}^{\mb h, \mb t}$ is covered by nonempty open subschemes, with each
isomorphic to an open subscheme of the affine space over $S$ of fiber
dimension $\mb H$.

\(1c) Set $\mb F := \mb H + \mb R$.  Then $\bb F\Psi_{\mc A}^{\mb h,
\mb t}$ is covered by nonempty open subschemes, with each isomorphic to
an open of the affine space over $S$ of fiber dimension $\mb F$.

\(2) Assume $S$ irreducible.  Given $S^\di/S$ and $\mb h^\di$ and $\mc
C^\di \in \mb H\Psi_{\mc A_{S^\di}}^{\mb h^\di, \mb t}$, necessarily
$\mb h^\di(q) \le \mb h(q)$ for all $q$.

\(3) Given $p, q > 0$, set $r := a - (p + q)$. Assume $\mb a(r) > \mb
h(r)$. Assume $\mb t(a-p) \ge 2$ if $p = q$, or else $\mb t(a-p) \ge 1$
and\/ $\mb t(a-q) \ge 1$ if $p \neq q$. Then $\mb h(r) < \mb h^{\rm
I}_{\bar s}(r)$.
\end{theorem}

\begin{proof}
For general use below, set $m := \sum_p \mb t(p)$.

{\bf For (1a),} as for \eqref{thFHdim}, it suffices to apply
\eqref{prAffBdl}.  To do so, we must see that the universal sheaf, $\mc
C$ say, on $\bb H\Psi_{\mc A}^{\mb h, \mb t}$ is very effective.  So
given a Noetherian affine $T/\bb H\Psi_{\mc A}^{\mb h, \mb t}$ and a
potential lifting $\mc F$ of $\mc C_T$, we must show that the subsheaves
$\mc C_T^*$ and $G_\bu \mc F^*$ of $\mc A^\dg_T$ are equal and that the
surjection $\gamma\:(G_\bu \mc F^*)\ox_{\mc A_T}\mc O_T \onto G_\bu (\mc
F^*\ox_{\mc A_T}\mc O_T)$ is an isomorphism.  But these issues are
local.  Thus we may assume $T$ is local.

As $\mc A$ is $\mc O_S$-Artinian, $\mc A^\dg = \mc A^*$.  But $\mc A$ is
$\mc O_S$-Gorenstein, and $T$ is local.  So $\mc A^\dg_T$ is cyclic.
Now, $\mb t^*$ is the local generator type of $\mc C^*$.  Hence for
$1\le i\le m$, there's $g_i\in \Gamma((\mc A_{d_i})_T)$ with $d_i>0$,
with $\#\{\, i \mid d_i = a-q\,\} = \mb t(q)$ for all $q$, and with $\mc
C_T^* = \sum g_i\mc A^\dg_T$.  Note $\mb P(z) = \prod_i(1-z^{d_i})$.

Set $\mc M^{(m)} := \mc A^\dg_T/\mc C_T^*$.  Then $\mc M^{(m)}_{p}$ is
locally free of rank $\mb w^*(p)$ for $p < n$.  So given any $T$-field
$K$, then $\mb H(\mc M^{(m)}\ox K,z) = \mb H(\mc A^\dg,z) \mb P(z) \mod
z^n$.  Thus \eqref{leInj}(2),\,(3) yield $H_1(g_1,\dotsc,g_m; \mc
A^\dg_T)_p = 0$ for all $p<n$.

As $\mc F$ is a potential lifting of $\mc C_T$, there's $f_i \in
F^{d_i}\mc A_T$ with $g_i$ as its initial form and with $\mc F = \sum
f_iA^\dg$.  Form the Koszul complex $K_\bu := K_\bu(f_1,\dotsc,f_m; \mc
A^\dg_T)$.  Note $G_\bu K_\bu = K_\bu(g_1,\dotsc,g_m; \mc A^\dg_T)$.  So
\eqref{leCplx} applies to the complex $K_2\to K_1\to K_0$.

Hence \eqref{leCplx}(2) yields $G_p(\mc F^*) = (\mc C_T^*)_p$ for $p \le
n$.  But $G_p(\mc F^*) = (\mc C_T^*)_p$ for $p \ge n$ as $(\mc
C_T^*)_p \subset G_p(\mc F^*) \subset (\mc A_T^\dg)_p$ and $\mb h^*(p)
:= \mb a^*(p)$.  Thus $\mc F$ is a lifting of $\mc C_T$.

Fix $p\ge n$.  Then $(\mc C^*\ox\mc O_T)_p = 0$ as $\mb t^*(p) = 0$.
But $\mc C^*\ox\mc O_T =: \mc C^*_T = G_\bu\mc F^*$, and $\gamma_p$ is
surjective.  Thus $\gamma_p$ is an isomorphism.

Finally, fix $p< n$, and let's apply \eqref{leCplx}(3).  Note $K_2\ox\mc
O_T\to K_1\ox \mc O_T$ vanishes as $d_i > 0$ for all $i$.  But $K_1 =
\mc A^{\dg\oplus m}_T$.  Hence $G_\bu (K_1\ox \mc O_T) = K_1\ox \mc O_T$
and $G_\bu K_1 = K_1$.  Hence $(G_\bu K_1)\ox \mc O_T = G_\bu (K_1\ox
\mc O_T)$.  Thus \eqref{leCplx}(3) implies $\gamma_p$ is an isomorphism.

Thus $\mc C$ is very effective.  Thus (1a) holds.

{\bf For (1b),} form the functor $\mb U_{\mb t}$ whose $T$-points are
the homogeneous quotients $\mc E = \mc A_T^\dg / \mc D$ with $\mc E_p$
locally free of rank $\mb w^*(p)$ for $p < n$ and with $\mc D$ of local
generator type $\mb t^*$.  Let's use the hypotheses $\mb w^*(p) \ge 0$
for $p <n$ and $\mb t^*(p)=0$ for $p\ge n$ and $p \le -a$, but not $\bb
H\Psi_{\mc A}^{\mb h, \mb t} \neq \emptyset$, to represent $\mb
U_{\mb t}$ via recursion on $s - \bar s$ where $s := \sup \{\, q\mid \mb
t(q) \neq 0\,\}$ and where $\bar s := \inf \{\, q\mid \mb t(q) \neq
0\,\}$.  Set $\bar t := \mb t(\bar s)$.

If $\bar s = s$.  set $U' := S$ and $\wt{\mc E'} = \mc A^\dg$; so set
$\wt{\mc D}' := 0$, set $\mb w'^* := \mb a^*$, and set $\mb H' := 0$.

If $\bar s < s$, set $\mb t'(\bar s) := 0$ and $\mb t'(q) := \mb t(q)$
for $q \neq \bar s$, and define $w'^*$ and $\mb H'$ to be the objects
corresponding to $w^*$ and $\mb H$.  Trivially, $\mb t'^*(p)=0$ for
$p\ge n$ and $p \le -a$.  Further, $\sum_p\mb w'^*(p)z^p =
\bigl(\sum_p\mb w^*(p)z^p\bigr)(1-z^{a - \bar s})^{-\bar t}$; hence,
$\mb w'^*(p) \ge \mb w^*(p) \ge 0$ for $p < n$.  So assume that $\mb
U_{\mb t'}$ is representable by an $S$-scheme $U'$ and a universal
quotient $\wt{\mc E}' = \mc A_{U'}^\dg \big / \wt{\mc D}'$ with $\wt{\mc
E}'_p$ locally free of rank $\mb w'^*(p)$ for $p < n$, with $\wt{\mc
D}'$ of local generator type $\mb t'^*$, and with $U'$ covered by
nonempty open subschemes of the affine space over $S$ of fiber dimension
$\mb H'$.

For any $s$, note $\bigl(\sum_p\mb w'^*(p)z^p\bigr)(1-z^{a - \bar
s})^{\bar t} = \sum_p\mb w^*(p)z^p$ and $\mb w'^*(-a) = 1$; therefore,
comparing terms of degree $-\bar s$ yields $\mb w'^*(-\bar s) - \bar t =
\mb w^*(-\bar s) \ge 0$. Set $\bb G := \Grass_{\mb w(\bar s)}(\wt{\mc
E}'_{-\bar s})$, the Grassmann\-ian of rank-$\mb w(\bar s)$ locally free
quotients.  Note that $\bb G$ is covered by nonempty open subschemes of
the affine space over $U'$ of fiber dimension $(\mb w'^*(-\bar s)-\mb
w(\bar s))\mb w(\bar s)$, or $\bar t \mb w(\bar s)$.  Thus $\bb G$ is
covered by nonempty open subschemes of the affine space over $S$ of
fiber dimension $\mb H' + \bar t \mb w(\bar s)$ or $\mb H$.

Let's adapt the argument surrounding \eqref{eqLev1}.  Let $\mc
E^\di_{-\bar s}$ be the universal quotient of $(\wt{\mc E}'_{-\bar
s})_\bb G$.  Recall that $\wt{\mc E}' = \mc A_{U'}^\dg \big / \wt{\mc
D}'$.  So say $\mc E^\di_{-\bar s} = (\mc A^\dg_\bb G)_{-\bar s} \big/
\mc D^\di_{-\bar s}$.  Then set $\mc D^\di := \mc A_\bb G \mc
D^\di_{-\bar s} \subset \mc A^\dg_\bb G$ and $\mc E^\di := \mc A^\dg_\bb
G \big/ \mc D^\di$.  For $p < n$, let $U_p \subset \bb G$ be the
subscheme such that a map $T\to \bb G$ factors through $U_p$ iff $(\mc
E^\di_p)_T$ is locally free of rank $\mb w^*(p)$.  (Of course, $U_{-\bar
s} = \bb G$.)  Set $U := \bigcap_{p < n}U_p$, set $\wt{\mc E} := \mc
E^\di\mid U$, and set $\wt{\mc D} := \mc D^\di\mid U$.  Let's prove that
the pair $(U,\, \wt{\mc E})$ represents $\mb U_{\mb t}$.

First, note that, by construction, $\wt{\mc E} = \mc A_{U}^\dg \big/
\wt{\mc D}$ is a homogeneous quotient with $\wt{\mc E}_p$ locally free
of rank $\mb w^*(p)$ for $p < n$.  Furthermore, $\wt{\mc D}$ is of local
generator type $\mb t^*$ since $\wt{\mc D}'$ is of local generator type
$\mb t'^*$ and since $\wt{\mc D} / \wt{\mc D}'_U$ is generated by the
pullback of the universal subsheaf on $\bb G$, which is locally free of
rank $\bar t$.

Next, given any $S$-scheme $T$ and any homogeneous quotient $\mc E = \mc
A_T^\dg / \mc D$ with $\mc E_p$ locally free of rank $\mb w^*(p)$ for $p
< n$ and with $\mc D$ of local generator type $\mb t^*$, we have to show
that there's a unique $S$-map $T\to U$ with $\mc E = \wt{\mc E}_T$ as
quotients of $\mc A^\dg_T$.

If $\bar s = s$, recall $U' := S$ and $\wt{\mc E'} = \mc A^\dg$, and set
$\mc E' := \mc A^\dg_T$.  Then trivially there's a unique $S$-map $T\to
U'$, and $\mc E' = \wt{\mc E'}_T$.

If $\bar s < s$, set $\mc D' := \mc A_T\big( \bigoplus_{j>\bar s}\mc
D_{-j}\big)$ and $\mc E' : = \mc A_T^\dg / \mc D'$.  Let's see that $\mc
E'_p$ is locally free of rank $\mb w'^*(p)$ for $p < n$ and that $\mc
D'$ is of local generator type $\mb t'^*$.  As these issues are local,
we may temporarily assume that $T$ is local.  Set $m' := \sum_{q>\bar
s}\mb t(q)$ and $m := \sum_{q}\mb t(q)$.  Proceed somewhat as in the
proof of (1a).

As $\mc A$ is $\mc O_S$-Artinian, $\mc A^\dg = \mc A^*$.  But $\mc A$ is
$\mc O_S$-Gorenstein, and $T$ is local.  So $\mc A^\dg_T = \mc A_Tf$ for
some $f\in \Gamma((\mc A^\dg_{-a})_T)$.  Now, $\mb t^*$ is the local
generator type of $\mc D$.  Hence for $1\le i\le m$, there's $g_i\in
\Gamma((\mc A_{d_i})_T)$ with $d_i>0$, with $\mc D = \sum g_i\mc
A^\dg_T$, and with $\#\{\, i \mid d_i = a-q\,\} = \mb t(q)$ for all $q$.
Rearrange the $g_i$ so that $a-d_i > \bar s$ iff $i \le m'$.  Then
$g_1f,\dotsc,g_{m'}f$ form a minimal generating set of $\mc D'$.  Thus
$\mc D'$ is of local generator type $\mb t'^*$.  Now, set $M :=
\Gamma(\mc E)$ and $M' := \Gamma(\mc E')$.  Given any $T$-field $K$,
note $\mb H(M\ox K,z) = \sum_p \mb w^*(p)z^p \mod z^n$.  So
\eqref{leInj}(1) yields $\mb H(M'\ox K,z) = \sum_p \mb w'^*(p)z^p \mod
z^n$.  Hence $M'_p$ is flat over $\Gamma(\mc O_T)$ for $p < n$ by
\eqref{leInj}(2).  Thus $\mc E'_p$ is locally free of rank $\mb w'^*(p)$
for $p < n$, as desired.

By recursion, for our original $T$, there's a unique $S$-map $T\to U'$
with $\mc E' = \wt{\mc E_s}_T$ as quotients of $\mc A^\dg_T$.  But $\mc
E_{-\bar s}$ is a rank-$\mb w(\bar s)$ locally free quotient of $\mc
E'_{-\bar s}$.  Hence there's a unique $U'$-map $T\to \bb G$ with $\mc
E_{-\bar s} = (\wt{\mc E}_{-\bar s})_T$ as quotients of $(\wt{\mc
E}'_{-\bar s})_T$.  So $\mc D_{-\bar s} = (\wt{\mc D}_{-\bar s})_T$.  So
$\mc D = \wt{\mc D}_T$.  Thus $\mc E = \wt{\mc E}_T$ as quotients of
$\mc A^\dg_T$.  Thus the pair $(U,\, \wt{\mc E})$ represents $\mb U_{\mb
t}$, as desired.

Finally, set $R_p := \Supp \wt{\mc E}_p \subset U$. Set $\wh U := U -
\bigl(\bigcap_{p\ge n}R_p\bigr)$ and $\wh{\mc E} := \wt{\mc E} \mid \wh
U$.  Then a map $T\to U$ factors through $\wh U$ iff $\wt{\mc E}_p = 0$
for $p \ge n$ by \eqref{leLev}(1).  But $(U,\, \wt{\mc E})$ represents
$\mb U_{\mb t}$.  So $(\wh U,\, \wh{\mc E})$ represents the functor
whose $T$-points are the $\mc E = \mc A_T^\dg / \mc D$ with $\mc E_p$
locally free of rank $\mb w^*(p)$ for $p < n$, with $\mc E_p=0$ for $p
\ge n$. and with $\mc D$ of local generator type $\mb t^*$.
Equivalently, set $\wh{\mc D} := \wt{\mc D} \mid \wh U$; then $(\wh U,\,
\wh{\mc D})$ represents the functor whose $T$-points are the flat $\mc D
\subset \mc A^\dg_T$ with Hilbert function $\mb h^*$, with local
generator type $\mb t^*$, and with $\mc A^\dg_T/\mc D$ flat.  Or, by
\eqref{thGMD3}, Macaulay Duality, $(\wh U,\, \wh{\mc D}^*)$ represents
the functor whose $T$-points are the flat quotients $\mc C$ of $\mc A_T$
with Hilbert function $\mb h$ and $T$-socle type $\mb t$.  Thus $\wh U =
\bb H\Psi_{\mc A}^{\mb h, \mb t}$.

Note $\wh U$ is an open subscheme of $U$, so one of $\bb G$.  So $\wh U$
is covered by open subschemes of the affine space over $S$ of fiber
dimension $\mb H$.  The latter open subschemes may be taken to be
nonempty as $\bb H\Psi_{\mc A}^{\mb h, \mb t} \neq \emptyset$.  Thus
(1b) holds.

{\bf Plainly, (1c)} follows immediately from (1a) and (1b).

{\bf For (2),} note $\mb P(z) = \prod(1-z^{d_i})$.  Now, the Hilbert
function of $\mc C_T$ is $\mb h$.  So $\dim((\mc A^\dg_{T}/\mc C_T^*)_p
= \mb a^*(p) - \mb h^*(p)$.  But $\mb h(q) := \mb a(q)-\mb w(q)$ for $q
> -n$.  Furthermore, $\sum_p\mb w^*(p)z^p := \mb H(\mc A^\dg,z) \mb
P(z)$.  Thus $\mb H((\mc A^\dg_{T}/\mc C_T^*),\, z) = \mb H(\mc
A_T^\dg,z) \mb P(z)\mod z^n$.

Let $K^\di$ be an $S^\di$-field, $T^\di := \Spec K^\di$.  But $\mb t^*$
is the local generator type of $\mc C^{\di*}$ and $\mc A_{T^\di}^\dg$ is
cyclic.  So for $1\le i\le m$, there's $g^\di_i\in \Gamma((\mc
A_{d_i})_{T^\di})$ with $\mc C_{T^\di}^{\di*} = \sum g^\di_i\mc

Set $\bb P := \prod_{i=1}^m \bb P(\mc A^*_{d_i})$.  Consider the
$T$-point of $\bb P$ defined by the $g_i$.  By \eqref{prOpens}(2), it
lies in the open subset $U$ of $\bb P$ provided by \eqref{prOpens}(1).
Consider the $T^\di$-point of $\bb P$ defined by the $g^\di_i$.  Note
$\mb H(\mc A^\dg,z) \mb P(z) \le \mb H( (\mc A^\dg_{T^\di}/\mc
C_{T^\di}^{\di}*),\, z) \mod z^n$ by \eqref{prOpens}(1).  So $\mb w^*(p) \le
\mb a^*(p) - \mb h^{\di*}(p)$ for $p < n$.  So $\mb h^\di(q) \le \mb
h(q)$ for $q > -n$.  But $h^\di(q) \le \mb a(q)$ for all $q$, and $\mb
h(q) = \mb a(q)$ for $q \le -n$.  Thus (2) holds.

{\bf For (3),} set $\mc N := \bigoplus_{i=1}^m \mc A_T(a-d_i)$, and use
the $g_i$ to define a surjection $\mc N \onto \mc C^*$.  Reorder the
$g_i$ so $d_1 = p$ and $d_2 = q$.  Then $(g_2,-g_2,0,\dotsc,0)\neq0$ in
$\Gamma(\mc N_{-r})$, but it maps to 0 in $\Gamma(\mc C^*_{-r})$.  Now,
$\dim \Gamma(\mc N_{-r}) = \mb g_{\bar s}(r)$ by \eqref{eqsbImax}.  So
$\mb g_{\bar s}(r) > \mb h(r)$.  But $\mb h^{\rm I}_{\bar s}(r) :=
\min\{\,\mb g_{\bar s}(r),\, \mb a(r)\,\}$ by \eqref{eqsbImaxh}, and
$\mb a(r) > \mb h(r)$.  Thus (3) holds.
\end{proof}

\begin{example}\label{exMetal}
Let's see what the theory in this section says about Migliore,
Mir\'{o}-Roig, and Nagel's Examples 2.14--2.16 on pp.\,343--345 in
\cite{MMN} of an algebra $A$ and a level $\mb t$ (that is, $ \bar s =
s$) with $\mb{H\Psi}_A^{\mb h^{\rm I}_{\bar s},\mb t} = \emptyset$.
Their A is an unspecified ``general'' complete intersection of forms of
the same degree;. They prove $A$ has a homogeneous quotient $C$ whose
Hilbert function $\mb h$ is maximal for $\mb t$, but $\mb h \neq \mb
h^{\rm I}_{\bar s}$.

By contrast, below, we construct specific such $A$ and $C$.  Now, take
$\mc M$ to be a polynomial ring over a field $k$ in \eqref{prOpens}.
Its (2) yields a scheme $U$ whose $K$-points, for any $k$-field $K$,
represent (with repetition) exactly the quotients $A^\di$ of $M_K$ with
the right Hilbert function; the sequence that defines an $A^\di$ is
regular, as $A^\di$ is Artinian.  The proof of \eqref{prOpens} provides
a flat $\mc O_U$-algebra $\mc A$ whose fibers are the $A^\di$.

Consider \eqref{thAffBdl}.  Its $\mb h$ turns out to be the Hilbert
function of our $C$; therefore, $C \in \mb H\Psi_{\mc A}^{\mb h, \mb
t}$.  So \eqref{thAffBdl}(1b) implies $\bb H\Psi_{\mc A}^{\mb h, \mb
t}$ is $U$-flat, where $U$ and $\mc A$ were introduced above.  So the
image of $\bb H\Psi_{\mc A}^{\mb h, \mb t}$ in $U$ is a nonempty open
subscheme, whose points represent (with repetition) exactly the desired
$A^\di$.  Moreover, \eqref{thAffBdl}(1b) describes the fiber over a
given $A^\di$; it's the scheme $\bb H \Psi_{A^\di}^{\mb h,\mb t}$
parameterizing all the desired quotients of $A^\di$.

There's more.  First, \eqref{thAffBdl}(1a) describes the retraction map
to $\bb H \Psi_A^{\mb h,\mb t}$ of the scheme $\bb F \Psi_A^{\mb h,\mb
t}$ of arbitrary filtered quotients of $A$ with Hilbert function $\mb h$
and socle type $\mb t$.  Second, \eqref{thAffBdl}(2) asserts $\mb h$ is
maximal among the Hilbert functions of homogeneous quotients of $A$ of
$k$-socle type $\mb t$.  Third, \eqref{thAffBdl}(3) yields $\mb h \neq
\mb h^{\rm I}_{\bar s}$.

For Migliore et al.'s 2.14--2.16, respectively take $r := 3,\,4,\,3$ and
$e := 2,\,2,\,3$ in the following setup.  Let $k$ be any field, $R :=
k[X_1,\dotsc,X_r]$ the polynomial ring with $r \ge 2$.  Fix $e \ge 2$.
Set $f := X_1^{-e}\dotsb X_r^{-e}$ and $I := (0:_Rf)$; set $A := R/I$.
Plainly, $X_1^{i_i}\dotsb X_r^{i_r} \in I$ iff $i_j \ge e+1$ for some
$j$; hence, $I = (X_1^{e+1},\dots,X_r^{e+1})$.  Thus $A$ is a complete
intersection; its socle degree $a$ is $er$, and its initial degree $v_0$
is $e+1$.  Set $n := v_0-a = e+1-a$.

Fix $t\ge1$.  Take $\mb t(p) := t$ for $p = a-1$ and $\mb t(p) := 0$ for
$p\neq a-1$.  Migliore et al.'s examples have $t=2$; so $t = r-1$ in
2.14 and 2.16.  Let's do the general case: $t = r-1$.  Use the notation
of \eqref{prNonempty} and its proof.  So $\mb P(z) = (1-z)^t$, and $J:=
\sum_{j=1}^t RX_j$, and $D = Jf\subset A^*$.  But $1-a < n-1$ as $1 <
e$; so $\mb t(1-n) = 0$.  Plainly $R/J = k[X_r]$.  So $R/(I+J) =
k[X_r]/(X_r^{e+1})$.  However, $R/I = A^*(-a)$; hence, $R/(I+J) =
(A^*/D)(-a)$.  Thus $(A^*/D)_p = 0$ for $p\ge e+1-a =: n$.  Now, $\mb
H(R/J,\,z) = \sum_p\mb w^*(p-a)z^p \mod z^{v_0}$ by an argument in the
proof of \eqref{prNonempty}.  But $J_p = (I+J)_p$ for $p < v_0$.  So
$\dim (A^*/D)_p = \mb w^*(p)$ for $p < n$.  Thus the Hilbert function of
$D$ is $\mb h^*$, so that of $C := D^*$ is $\mb h$, as desired.

Migliore et al.'s 2.15 is a little different, requiring more attention
be paid to $J$ and to $\car k$.  Of course, $J:= Rg_1+Rg_2$ where
$g_1,\,g_2$ are linear forms, as $t=2$.  However, if $g_j := X_j$, then
$X_3X_4^2,\,X_3^2X_4 \notin (I+J)$; so $D_n \neq A^*_n$, where $n = -5$.

Fix $b \in k$, take $g_1 := X_1+X_2+X_3$ and $g_2:= X_2+b
X_3+X_4$, and define a map $u\: R_2\oplus R_2 \to A_{-5}^\dg$ by $u(x,y)
:= (xg_1+yg_2)f$.  It's straightforward to write out a 16 by 20 matrix
representing $u$; use as basis of $R_2$ its monomials, and as basis of
$A_{-5}^\dg$ the products $X_1^{i_1}\dotsb X_4^{i_4}f$ with $0\le
i_1,\dotsc,i_4 < 3$ and $\sum i_j = 3$.  Order the two bases
lexicographically.  Then Macaulay~2 (with {\tt ZZ[b]} as ring)
finds the first nonvanishing maximal minor to be $9b(1-b)$.
Thus, if $\car k\neq 3$ and $b \neq 0,\,1$, then $u$ is surjective,
and so $C \in \mb H\Psi_{ A}^{\mb h, \mb t}$.

The case $k = \Z/2$ is special.  Of course, there's no $b\in k$ with $b
\neq 0,\,1$; however, there is such a $b$ in any proper extension field.
So the $k$-scheme $\bb H \Psi_A^{\mb h,\mb t}$ is nonempty.
(Moreover, it has no $k$-points; indeed, Macaulay~2 was used to check
each of the six essentially different choices of the pair $(g_1,g_2)$.)
Thus \eqref{thAffBdl} still applies.

Assume $\car k = 3$.  Say $g_i = \sum_{j=1}^4 a_{ij}X_j$ for $i=1,\ 2$.
Set $\Delta := a_{11}a_{22} - a_{12}a_{21}$.  Note $g_i^3 = \sum
a_{ij}^3X_j^3$; so $\Delta^3 X_1^3,\ \Delta^3 X_2^3 \in
\bigl(g_1,g_2,X_3^3,X_4^3\bigr)$.  Assume $\Delta \neq 0$.  Then,
therefore, $I+J = \bigl(g_1,g_2,X_3^3,X_4^3\bigr)$.  Hence $R/(I+J)$ is
of socle degree $4$.  Hence $\dim (A^*/D)_p \neq 0$ for $p = n = -5$ and
$p = n+1 = -4$.  Denote the Hilbert function of $D$ by $\mb h_\Delta^*$.
Thus $\mb h_\Delta^*(p) \neq \mb h^*(p)$ for $p = -5,\,-4$.  (Moreover,
it's not hard to see using \eqref{prOpens} that $\mb h_\Delta^*$ is
maximal for $A$ and $\mb t$.) Thus $\bb H \Psi_A^{\mb h,\mb t} =
\emptyset$, although as seen above, $\bb H \Psi_A^{\mb h,\mb t} \neq
\emptyset$ when $\car k \neq 3$.  (In Ex.\,2.12 on p.\,340, Migliore et
al.\ work a similar example in characteristic 2 with $r := 3$ and $e :=
3$ and $t:=1$.)
\end{example}

\begin{remark}\label{reFrbg}
If true, Fr\"oberg's conjecture provides another major case where
\indt{Fr\"oberg's conjecture}
\eqref{thAffBdl} applies; namely, $A$ is an Artinian general complete
intersection over an infinite field $k$.  Fr\"oberg's conjecture is
suitably formulated in Migliore, Mir\'{o}-Roig, and Nagel's Conjecture
2.7 on p.\,339 in \cite{MMN} and in Nenashev's Conjecture 1 on p.\,273
in \cite{Nen}, with $A$ not necessarily Artinian.  The conjecture has
been proved in three instances: (1)~when $A$ has two or three
generators, (2)~when $\car k = 0$ and $A$ is a complete intersection, or
almost so, and (3)~wnen $k$ is an algebraically closed of characteristic
0 and $A$ is a polynomial ring modulo forms of the same degree.  The
proofs are generally attributed to Stanley, Fr\"oberg, Anick,
Hocster--Laksov, Aubry, and Nenashev; see \cite{Nen}*{p.\,275}.

Our treatment is geared to the setup of \eqref{thAffBdl}.  However, in
essence, it's the same as Migliore, Mir\'{o}-Roig, and Nagel's treatment
of Conjecture 2.9 and Remark 2.10 on p.\,340, although they don't assume
$A$ is Artinian.  The two treatments differ in one notable way: we use
Macaulay Duality, whereas they use linkage in Lemma 2.4 on p.\,338.
Nevertheless, this difference is superficial owing to \eqref{coLink}(2).

So say $A = R/I$ with $R := k[X_1,\dotsc,X_r]$ the polynomial ring,
$\deg X_i =1$, and $I$ is generated by $r$ general forms of degrees
$d_1,\dotsc,d_r$.  Note $\mb H(R,z) = 1/(1-z)^r$.  Set $\mb Q(z) :=
\prod_{i=1}^r (1-z^{d_i})$.  Thus \eqref{leInj}(1) yields $\mb H(A,z) =
\mb Q(z)\big/(1-z)^r$.

Set $a := \sum (d_i-1)$; so $a$ is the socle degree of $A$.  Choose an
$f\in A^\dg_{-a}\subset R^\dg_{-a}$ with $A^\dg = Af$.  Then $A(a)\risom
A^\dg$ by $g\mapsto gf$.  Thus $\mb H(A^\dg,z) = \mb H(A,z)z^{-a}$.

Set $\mb P(z) := \prod_j(1-z^j)^{\mb t(a-j)}$.  Define $\mb w^*(p)$ by
$\sum_p\mb w^*(p)z^p := \mb H(A^\dg,z) \mb P(z)$.  Set $n := \sup\{\,
m\mid \mb w^*(p) \ge 0 \text{ for }p < m\,\}$. Assume $\mb t^*(p)=0$ for
$p\ge n$ and $p \le -a$.

For all $i$, fix general forms $g_{ij}\in R_i$ for $j=1, \dotsc, \mb
t(a-i)$.  Set $J:= \sum_{i,j}Rg_{ij}$.  Then Fr\"oberg's conjecture
asserts $\mb H(R/(I+J),z) = \mb P(z)\mb Q(z)\big/(1-z)^r \mod z^{a+n}$
and $R_p = (I+J)_p$ for $p = a+n$.  Note $\mb P(z)\mb Q(z)\big/(1-z)^r =
\sum_p\mb w^*(p)z^{p+a}$.

Set $D := JA^\dg$.  Then $(R/(I+J))(a) \risom A^\dg/D$.  So Fr\"oberg's
conjecture yields $\dim (A^\dg/D)_p = \mb w^*(p)$ for $p < n$ and
$(A^\dg/D)_p = 0$ for $p= n$; so $(A^\dg/D)_p = 0$ for $p \ge n$, as
$R_1\cdot A^\dg_p = R_1\cdot R_{p+a}f = A^\dg_{1+p}$.  Moreover, the
conjecture implies that, if one of the $g_{ij}$ is omitted, then at
least one value of $\dim (A^\dg/D)_p$ changes, since $\mb t^*(p)=0$ for
$p\ge n$; hence, the products $g_{ij}f$ form a minimal generating set of
$D$, and so the (local) generator type of $D$ is $\mb t^*$.

Finally, set $C := D^*$.  Set $a(q) := \dim A_q$, and set $\mb h(q) :=
\mb a(q)-\mb w(q)$ for $q > -n$, and\/ $\mb h(q) := \mb a(q)$ for $q \le
-n$.  Then $\dim C_q = \mb h(q)$.  Therefore, $C\in \bb H\Psi_A^{\mb
h, \mb t}$.  Thus \eqref{thAffBdl} applies; so it describes the geometry
of $\bb H \Psi_A^{\mb h,\mb t}$ and  $\bb F \Psi_A^{\mb h,\mb t}
\onto \bb H \Psi_A^{\mb h,\mb t}$, and it asserts $\mb h$ is maximal,
but often $\mb h \neq \mb h^{\rm I}_{\bar s}$.
\end{remark}

\section*{References}

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\newpage

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\indexprologue[\vspace{-12pt}]{}
\printindex[notation]

\end{document}