Castelnuovo-Mumford regularity over different rings

Question

Let $S = k[x_1, \ldots, x_n, t]$ be the polynomial ring in $n+1$ variables over a field $k$ and let $R = k[x_1, \ldots, x_n]$. I have stumbled upon the following definition/result.

Let $\{f_1, \ldots, f_r \} \subset S$ be a set of homogeneous polynomials which have finitely many (projective) solutions over the algebraic closure of $k$, and let $J = \langle f_1, \ldots, f_r \rangle.$ Assume that $t \nmid 0$ mod $J$. The ideal \begin{equation} H = J + (t)/(t) \subset S/(t) \cong R \end{equation} is called an $\textbf{Artinian reduction}$ of $J$. In this case, we have $\mathrm{reg}_S(J) = \mathrm{reg}_{S/(t)}(H)$.

Now, the only result I know of that looks similar to this is the Independence Theorem of Local Cohomology (Thm. 13.1.6 in Brodmann & Sharps Local Cohomology)

Theorem: Let $S,R$ be graded rings , $I$ an ideal of $S$, $M$ a graded $R$-module and let $f: S \to R$ be a homogeneous homomorphism. Then $H_I^{i}(M) \cong H_{f(I)R}^{i}(M)$ for all $i \in \mathbb{Z}_{\geq 0}$

I tried to take $f:S \to S/(t)$ as the homomorphism setting $t = 0$ and applying the theorem above. But I don't see, how the conditions in the definition can be used to show the equality of regularity. What I was wondering, is, if the conditions in the definition of Artinian reduction are necessary for the result that $\mathrm{reg}_S(J) = \mathrm{reg}_{S/(t)}(H)$, or if this could be true for all modules $H$ of the form $H = J + (t)/(t) \subset S/(t)$?

Am I looking in the right direction with the independence theorem, or are there other results I'd have to consider?

Answer 1

Consider the exact sequences $$0\to J\to S\to S/J\to 0,$$ $$0\to\frac{J+(t)}{(t)}\to\frac{S}{(t)}\to \frac{S}{J+(t)}\to0,$$

and $$0\to \bar S(-1)\xrightarrow{t} \frac SJ\to\frac{S}{J+(t)}\to0,$$ where $\bar S=S/(J:t)$.

Since $t$ is a nonzerodivisor on $S/J$, $\bar S=S/(J:t)=S/J$. Furthermore, $S= k[x_1, \ldots, x_n,t]$, so every local cohomology of $S$ except the $n+1$-th one vanishes. Similarly, every local cohomology of $R=S/(t)$ except the $n$-th one vanishes. So for most values of $i$, we have the isomorphisms $$\newcommand{\m}{\mathfrak m}H^i_{\m}(S/J)\cong H^{i+1}_{\m}(J), \,H^i_{\m}(S/J+(t))\cong H^{i+1}_{\m}(J+(t)/(t)).$$ (I will use the same letter to denote the maximal homogeneous ideals. Thanks to the independence theorem, the distinction doesn't matter.) Moreover, from the third exact sequence, we get the corresponding long exact sequence of local cohomologies $$H^{i-1}_\m(S/J+(t))\to H^{i}_\m(S/J(-1))\to H^{i}_\m(S/J)\to H^{i}_\m(S/J+(t)).$$ Let $m\geq 0$ be an integer and assume that $J+(t)/(t)$ is $m$-regular. Then $H^{i}_{\m}(J+(t)/(t))_d=0$ for all $d\geq m-i+1$. So $$H^{i-1}_{\m}(S/J+(t))_d=0.$$ Since $d\geq m-(i+1)+1$, we also get $$H^i_{\m}(S/J+(t))_d\cong H^{i+1}_{\m}(J+(t)/(t))_d=0.$$ Then the long exact sequence yields an isomorphism $$H^{i}_\m(S/J(-1))_d=H^{i}_\m(S/J)_{d-1}\cong H^{i}_\m(S/J)_d.$$ But $H^{i}_\m(S/J)$ is Artinian, so $H^{i}_\m(S/J)_e=0$ for $e\gg 0$, so iterating the above isomorphism, we see that $H^{i}_\m(S/J)_{d-1}$ itself is $0$. This in turn implies that $H^{i+1}_{\m}(J)_{d-1}=0$ for all $d-1\geq m-i$.

To conclude that $J$ is $m$-regular, we still have to prove the cases that are left out. I hope the reader can take it from here (see here for some details). Finally, one has to prove the converse as well, i.e., $J$ is $m$-regular implies $J+(t)/(t)$ is $m$-regular. A similar argument should work.