Property related to the Krull dimension of modules which form an exact sequence

Question

This is a claim from matsumura's commutative ring theory given in the proof of dimension theorem, and I confused about it for a while:

Given a finite generated module $M,M',M''$over noetherian ring $R$ suppose we have exact sequence $$0\longrightarrow M'\overset{i}{\longrightarrow} M\overset{j}{\longrightarrow} M''\longrightarrow 0$$ I know by the exactness of localization $\mathrm{Supp}(M)=\mathrm{Supp}(M')\cup\mathrm{Supp}(M'')$, but the author subsequently deduces from it that: $$\mathrm {dim }M = \max (\mathrm{dim} M’, \mathrm {dim }M'')$$ I know $\mathrm{dim}M=\mathrm{dim }\mathrm{Supp}(M)$ i.e the length of maximal strictly ordered chain in $\mathrm {Supp}(M)$ but I can figure out why the chain completely lies in supp(M') or supp(M''). Did I misunderstand something ?

Answer 1

Recall $\mathrm{Supp}(M)=\{\mathfrak{p} \text{ prime ideal in }A\mid M_{\mathfrak{p}}\neq 0\}$.

Now observe that $$\mathrm{Supp}(M)=\mathrm{Supp}(M')\cup \mathrm{Supp}(M'').$$

Let $\mathfrak{p}_0\subset\cdots\subset \mathfrak{p}_n$ be a maximal chain of prime ideals in $\mathrm{Supp}(M)$.

Assume $M'_{\mathfrak{p}_s}\neq 0$ for all $s=0,\dots, n$. Then $\mathfrak{p}_0\subset\cdots\subset \mathfrak{p}_n$ is a chain of prime ideals in $\mathrm{Supp}(M')$. So $n=\dim M\leq \dim M'$.

On the other hand, assume $M'_{\mathfrak{p}_s}= 0$ for some $s=0,\dots, n$. (Then $M''_{\mathfrak{p}_s}\neq 0$.)

Assume $s$ is the largest such. Now observe that $M'_{\mathfrak{p}_r}= 0$ for $r\leq s$, since $M'_{\mathfrak{p}_r}$ is a localization of $M'_{\mathfrak{p}_s}$. Hence $M''_{\mathfrak{p}_t}\neq 0$ for $t\leq s$.

So we get a chain $\mathfrak{p}_{0}\subset\cdots\subset \mathfrak{p}_s$ in $\mathrm{Supp}(M'')$ and $\mathfrak{p}_{s+1}\subset\cdots\subset \mathfrak{p}_n$ in $\mathrm{Supp}(M')$.

Now observe that if $M''_{\mathfrak{p}_{t}}=0$ for some $t>s$, then $M''_{\mathfrak{p}_{r}}=0$ for all $r\leq t$. But that is impossible, since $M''_{\mathfrak{p}_{s}}\neq 0$.

So we get that $M''_{\mathfrak{p}_{t}}\neq 0$ for all $t>s$. So the whole chain $\mathfrak{p}_{0}\subset\cdots\subset \mathfrak{p}_n$ is entirely in $\mathrm{Supp}(M'')$.

This implies $\dim M=\max (\dim M', \dim M'')$.

Answer 2

Let $N$ be a finitely generated $R$-module and $q\subseteq p$ two prime ideals with $q\in Supp(N)$. Since $q_p\in Supp(N_p)$ and $$(N_p)_{q_p}\simeq N_q\neq0$$ we must have necessarily $N_p\neq0$, that is, $p\in Supp(N)$.

From this and $Supp(M)=Supp(M')\cup Supp(M'')$ we conclude that any chain of prime ideals in $Supp(M)$ lies completely in $Supp(M')$ or $Supp(M'')$ and vice-versa.