# Regularity of generic initial ideals: The only associated prime of $I$ containing $x_r$ is the maximal ideal.

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?

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

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.
• FWIW, I think Eisenbud is choosing $x_r$ generically so that $\mathfrak m$ is the only associated prime of $I$ containing $x_r$.