Krull dimension and localization of a module

Question

Let $(R,\mathfrak m)$ be a $d$-dimensional noetherian local ring, and $M$ an $R$-module. If $\mathfrak p$ is a prime ideal of $R$ with height $d-1$, then $\dim M_{\mathfrak p}=\dim M-1$?

Answer 1

We have $\operatorname{Supp}(M_{\mathfrak p})=\{\mathfrak qR_{\mathfrak p}:\mathfrak q\subseteq \mathfrak p, \mathfrak q\in\operatorname{Supp}(M)\}$. If $\mathfrak p\notin\operatorname{Supp}(M)$, then $M_{\mathfrak p}=0$, so there is no reason to have $\dim M_{\mathfrak p}=\dim M-1$. If $\mathfrak p\in\operatorname{Supp}(M)$, then $\dim M_{\mathfrak p}\le\dim R_{\mathfrak p}=d-1$. Once again this can be different from $\dim M-1$.

Edit. Let me add a counterexample when $M$ is finitely generated. Take $R=k[X,Y,Z]_{(X,Y,Z)}$, $Q=(X)$, $P=(Y,Z)$, $I=P\cap Q$ and $M=R/I$. Obviously $P\in\operatorname{Supp}(M)$ and we have $\dim R=3$, $\operatorname{height}P=2$, $\dim M_P=\operatorname{height}P/I=0$ and $\dim M=\dim R/I=2$.

Answer 2

If $M$ is finitely generated, $\dim M_p = \dim (R_p/\text{ann}_{R_p}(M_p)) = \dim (R/\text{ann}(M))_p$, and as pointed out already, this need not be equal to $\dim (R/\text{ann}(M)) - 1$, e.g. if $\text{ann}(M) \not \subseteq p$.

EDIT: If $p \in \text{Supp}(M)$ though, and $\text{ann}(M)$ is prime, the equality does in fact hold (assuming $M$ is finite). I will add a proof of this later: perhaps the OP can try to think of one in the meantime.

Proof: First, the inequality $\dim(R/\text{ann}(M))_p \le \dim(R/\text{ann}(M))-1$ holds simply because $p$ is not equal to the maximal ideal $m$. It thus suffices to show $\dim(R/\text{ann}(M))_p \ge \dim(R/\text{ann}(M))-1$. Let $I = \text{ann}(M)$. By assumption $I \subseteq p$, so passing to $R/I$, we may assume $I = 0$, and we wish to show $\dim R_p \ge \dim R - 1$.

Now, the assumption $\text{ht }p = 1$ implies that $\dim R/p = 1$, so $R/p$ is a Noetherian local domain of dimension $1$. Pick a system of parameters (in this case of length 1) $\overline{x} \in R/p$, satisfying $\sqrt{(\overline{x})} = \overline{m}$. Then $m^n \subseteq (p, x)$ for some $n \in \mathbb{N}$. In particular, $m^{nk} \subseteq p^k + (x)$ for every $k \ge 1$. It suffices to show that there exists a $k$ with $p^k$ contained in an ideal generated by $\dim R_p$ elements, for then these elements, together with $x$, will form a system of parameters for $R$, so $\dim R \le \dim R_p + 1$.

In general, there need not be such a $k$, but there are sufficient conditions for existence: for example if $p$ is a complete intersection (note: this is really $p/I$, not the original $p$). For the condition mentioned earlier, if $I$ is prime, so $R/I$ is a domain. In this case, $R \hookrightarrow R_p$, so lifting a system of parameters for $R_p$ (necessarily consisting of $\dim R_p$ elements) back to $R$, we get an ideal of $\dim R_p$ elements which contains a power of $p$, as desired.

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In conclusion, we always have the inequality $\dim M_p \le \dim M - 1$, and as the example by user121097 shows, equality need not hold in general, but it does hold in the following cases:

i) $p$ is a complete intersection modulo $\text{ann}(M)$
ii) $\text{ann}(M)$ is a prime ideal