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Let $(R,m)$ be a Noetherian local ring and let $M$ be a finitely generated $R$-module of dimension $d$. The Krull dimension of $M$ is defined to be the Krull dimension of $R/\operatorname{ann}(M)$. Suppose for simplicity that $M$ is faithful, i.e. $\operatorname{ann}(M)=0$. By the fundamental theorem of dimension theory, we know that $d$ is the smallest number of elements $x_1,\dots,x_d$ in $m$ such that the module $M/(x_1M+\cdots+x_d M)$ has finite length. By using a system of parameters, it is immediate to see that for any $x \in m$ we must have $\dim M/xM \ge \dim M -1$. This implies that $\dim \frac{R}{\operatorname{ann}(M/xM)}$ can drop at most by $1$, which in turn implies that the ideal $\operatorname{ann}(M/xM)$ has height at most $1$.

Question: Is there a direct way of seeing that $\operatorname{ht}(\operatorname{ann}(M/xM)) \le 1$?

At first (and second) sight this seems remarkable to me, since I seem unable to tell anything about the structure of $\operatorname{ann}(M/xM)=(xM:_R M)$ by direct inspection.

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    $\begingroup$ If $M=R$ then $\text{ann}(M/xM)=(x)$, so the question is about the height of principal ideal, which is bound above by $1$ by Krull's principal ideal theorem (e.g. en.wikipedia.org/wiki/Krull%27s_principal_ideal_theorem) (which as far as I remember also enters the proof of equivalence of the several definitions of dimension). $\endgroup$ – Hanno Jan 8 '15 at 7:37
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$\operatorname{Supp}M/xM=V(\operatorname{Ann}M/xM)$;
$\operatorname{Supp}M/xM=\operatorname{Supp}(R/xR\otimes_R M)=V(x)\cap\operatorname{Supp}M=V(xR+\operatorname{Ann}M)$;
$V(\operatorname{Ann}M/xM)=V(xR+\operatorname{Ann}M)$.

Let $\bar R=R/\operatorname{Ann}M$, and $\bar x$ the residue class of $x$. Then $\bar R/\bar x\bar R=R/(xR+\operatorname{Ann}M)$, so $\operatorname{ht}(\operatorname{Ann}M/xM)=\min_{P\in V(\operatorname{Ann}M/xM)}\operatorname{ht}P=\min_{\bar P\in V(\bar x\bar R)}\operatorname{ht}\bar P=\operatorname{ht}(\bar x\bar R)\le 1$.

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    $\begingroup$ For short, one can transfer the dimension properties from $M/xM$ to $\bar R/\bar x\bar R$. $\endgroup$ – user26857 Jan 8 '15 at 10:26

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