# Equivalent properties on the vanishing of Bass numbers

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

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 $\mathrm{injdim}_{A/fA}M/fM<\infty$.
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