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I'm studying on this notes. I'm finding some difficulties on proposition 12 on page 15. Let me recall what we are trying to prove:

At first we are trying to prove that if inj dim$_{A_p}\;A_p<\infty$ then $\mu_i(p)=0$ if $i<ht(p)$ and 1 if $i=ht(p)$ ($\mu_i(p)$ is the Bass number $\mu_i(A,p)$). Here is how the proof begins: we argue by inductionon the Krull dimension of $A_p$. If it is 0 we are good. Suppose it's greater than 0, let $f$ be an $A_p$-regular element then $A/fA$ has finite injective dimension on itself. This is what I don't understand, why inj dim$_{A/fA}(A/fA)<\infty$? The notes claim that they are using the following property:

Let $M$ be a finitely generated module on a local ring $A$ and $0\rightarrow M\rightarrow E_0\rightarrow E_1\rightarrow\cdots$ ($d_0:E_0\rightarrow E_1$) a minimal injective resolution. If $f$ is $A$-regular and $M$-regular and if $D=d_0(E_0)$ then we have the following exact sequence:

$0\rightarrow Hom_A(A/fA,D)\rightarrow Hom_A(A/fA,E_1)\rightarrow\cdots$

that is a minimal injective resolution of the $A/fA$-module $Hom_A(A/fA,D)$ that is isomorphic to $M/fM$.

This was the implication i$\Rightarrow$ iv. Any help on this issue?

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It seems that you want to study the Gorentein rings. I suggest you to use the original Bass' paper which is very well written. – user26857 Oct 8 '12 at 10:16
up vote 1 down vote accepted

In fact, one uses a change of rings theorem for injective dimension. In the book of I. Kaplansky, Commutative Rings, this is Theorem 205 and it is called "Second theorem on injective change of rings". A slightly different proof of this theorem is given by the proof of Proposition 6 on page 9 of your notes and this says, in particular, the following: if $f$ is $A$-regular and $M$-regular, then $\mathrm{injdim}_AM<\infty$ implies $\mathrm{injdim}_{A/fA}M/fM<\infty$.

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So to prove that inj dim$_A/fA\;A/fA<\infty$ I need to show that $\mathrm{inj\;dim}_A\;A\infty$. I don't see why that is true. If we knew that $dim\;A<\infty$ then it's ok. But I don't see it. It doesn't suppose $A$ local, we have only $A$ noetherian. How can I bypass this issue? – Chris Oct 8 '12 at 10:42
You can reduce all to the local case, because the Bass numbers localize. Thus you have to prove the following: If $(A,m)$ is a local noetherian ring with $\mathrm{injdim}A<\infty$, then $\mu_i(m)=0$ for $i<\dim A$ and $\mu_i(m)=1$ for $i=\dim A$. Now I can see no issue if you want to apply the change of rings theorem for injective dimension as it is stated in Proposition 6 on page 9 of your notes. – user26857 Oct 8 '12 at 18:49
Ok if we reduce to the local case we can prove this proposition. But in those notes they don't make this reduction. So is it true that the injective dimension of $A/fA$ is finite over itself (even if $A$ is not local) or we must reduce to the local case? – Chris Oct 8 '12 at 23:54
In my opinion you must reduce the proof to the local case. It is the most natural thing to do and Bass does it. I also think that in your notes the removing of $\mathfrak{p}$ from $A_{\mathfrak{p}}$ at a moment is a typo. – user26857 Oct 9 '12 at 0:26

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