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Let $A$ be a noetherian local ring with maximal ideal $m$. Let $p$ be a prime ideal such that if $B=A/p$, then $\mathrm{dim}\;B=1$. Take $x\not\in p$, $x\in m$, and set $C=B/xB$. Then $C$ has finite length.

  1. Why does $C$ have finite length?

  2. By additivity of the length, if $B$ and $C$ have finite length then $l(C)=0$ and so $C=0$, am I right?

  3. Now take a finitely generated module $M$. Suppose to know $\mathrm{Ext}^{i+1}(C,M)\neq0$ and suppose $l(C)>1$. Then the exact sequence $0\rightarrow k\rightarrow C\rightarrow C^\prime\rightarrow0$ yields an exact sequence $\mathrm{Ext}^{i+1}(C^\prime,M)\rightarrow\mathrm{Ext}^{i+1}(C,M)\rightarrow\mathrm{Ext}^{i+1}(k,M)$. The notes where I'm studying now claim that this shows that there always exists a module $N$ with $1\leq l(N)< l(C)$ and such that $\mathrm{Ext}^{i+1}(N,M)\neq 0$. My questions now are: why such an $N$ exists and why this implies $\mathrm{Ext}^{i+1}(k,M)\neq0$?

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What do you mean "Let $p$ be another maximal ideal"? Local rings only have one maximal ideal! Maybe you meant just "another ideal"? – rschwieb Oct 5 '12 at 10:54
sorry I meant prime – Chris Oct 5 '12 at 11:13
here is a link, but the only version avaiable is in french:… – Chris Oct 5 '12 at 23:43
lemma 3 on page 12 – Chris Oct 5 '12 at 23:58
up vote 1 down vote accepted
  1. Prove that the ideal $p+xA$ is $m$-primary. (In general, a finitely generated module $C$ over a Noetherian ring $A$ has finite length if and only if $\mathrm{Supp}(C)\subseteq\mathrm{Max}(A)$. In your case $\mathrm{Supp}(C)=V(p+xA)=\{m\}$.)

  2. Yes, you are right, but $B$ cannot have finite length.

  3. If such an $N$ there exists, then you have two cases: $l(N)>1$ and keep going, or $l(N)=1$ and then $N\cong k$.

If $\mathrm{Ext}^{i+1}(N,M)=0$ for all $N$ with $1\le l(N)<l(C)$, then $\mathrm{Ext}^{i+1}(k,M)=0$ and $\mathrm{Ext}^{i+1}(C',M)=0$. This implies $\mathrm{Ext}^{i+1}(C,M)=0$, a contradiction.

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could you give me a reference where I can find a proof that every finitely generated module over a noetherian local ring has finite injective dimension? – Chris Oct 6 '12 at 0:35
I never said that finitely generated modules over noetherian rings have finite injective dimension (because this is false!). Instead, a characterization via $\mathrm{Ext}(k,-)$ functors of when the injective dimension of a finitely generated module over a noetherian local ring (with the residue field $k$) is finite can be found, for instance, in I. Kaplansky, Commutative Rings, Theorem 212. – user26857 Oct 6 '12 at 10:52

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