I was going through Eisenbud's proof of a result of Bayer and Stillman concerning the regularity of an ideal and its generic initial ideal, and I'm having trouble understanding his proof. The theorem I present below is a simplified version (Eisenbud proves the theorem for graded free modules). $\newcommand{\reg}{\operatorname{reg}} \newcommand{\gin}{\operatorname{Gin}} \newcommand{\In}{\operatorname{In}} \newcommand{\m}{\mathfrak{m}}$

Theorem. Let $S=k[x_1,\ldots,x_r]$ be a polynomial ring over an infinite field $k$. Let $I$ be a homogeneous ideal of $S$ and $>$ be the reverse lexicographic order. Then $$\reg S/I=\reg S/\gin(I).$$

Proof. Rechoosing the variables $x_1,\ldots,x_r$ generically, we may assume in particular that $\In(I) = \gin(I)$. Since the only associated prime of $I$ containing $x_r$ is the maximal ideal, we see that $(I:x_r)$ is of finite length....

It looks like Eisenbud is assuming $x_r$ is a zerodivisor on $S/I$. I think this is done via the generic change of coordinates, but I'm not sure. Anyway, even by assuming $x_r$ is a zerodivisor on $S/I$, I don't see the argument.

We know that the elements $x_r,\ldots,x_s$ form a regular sequence on $S/I$ iff $x_r,\ldots,x_s$ form a regular sequence on $S/\In(I)$. In particular, $x_r$ is a zerodivisor on $S/I$ iff $x_r$ is a zerodivisor on $S/\In(I)$. Moreover, since $\In(I)$ is Borel fixed, the associated primes of $\In(I)$ are all of the form $(x_1,\ldots,x_j)$ for $1\leq j\leq r$. Using these observations, we see that if $x_r$ is a zerodivisor on $S/I$ then $\m$ is an associated prime of $\In(I)$. Does this imply the above claim?

  • $\begingroup$ if you don't mind me asking, what page of Eisenbud is this proof on? (+1) $\endgroup$ Commented Jan 12, 2021 at 5:09
  • 1
    $\begingroup$ @AtticusStonestrom This is Corollary 20.21, page number 509. $\endgroup$
    – cqfd
    Commented Jan 12, 2021 at 7:18

1 Answer 1


I still don't see Eisenbud's argument, but the ultimate purpose of the argument is to prove that $(I:x_r)/I$ is of finite length. I present a different proof of this claim below.

From the above discussion, we see that $\newcommand{\reg}{\operatorname{reg}} \newcommand{\gin}{\operatorname{Gin}}\newcommand{\In}{\operatorname{In}}\newcommand{\m}{\mathfrak{m}}\newcommand{\p}{\mathfrak{p}}\m$ is the only associated prime of $\In(I)=J$ containing $x_r$. Let $\p\neq \m$ be a prime ideal. Since the associated primes of $J_\p$ correspond to the associated primes of $J$ contained in $\p$ (reason: the set of associated primes commutes with localisation), we see that $x_r$ is a nonzerodivisor on $M_\p=(S/J)_\p$ (the set of all zerodivisors is equal to the union of all associated primes). In other words, $$\ker(M_\p\xrightarrow{x_r}M_\p)=((J:x_r)/J)_\p=0$$ for all $\p\neq \m$. This means that the annihilator of $(J:x_r)/J$ is not contained in any prime $\p\neq \m$. Hence the radical of the annhilator of $(J:x_r)/J$ is precisely $\m$ (radical of $\mathfrak a=\cap_{\mathfrak a\subset \p}\p$). Now it is clear that $(J:x_r)/J$ is a module over $S/\m^n$ for some $n$ and hence has finite length.

Since $(J:x_r)/J$ is a finite length module, we have $(J:x_r)_d=J_d$ for $d\gg0$. Since we are using the reverse lexicographic order, we have $(J:x_r)_d=J_d$ if and only if $(I:x_r)_d=I_d$ (see Lemma 2.2 here). Hence we see that $(I:x_r)/I$ has finite length.

  • $\begingroup$ FWIW, I think Eisenbud is choosing $x_r$ generically so that $\mathfrak m$ is the only associated prime of $I$ containing $x_r$. $\endgroup$
    – cqfd
    Commented Jan 13, 2021 at 7:05

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