# Extension of projective module

Hi all!

I am interested in the following question in homological algebra.

Let we have two noncommutative rings with homomorphism $\phi:B\rightarrow A$ and $M$ be a projective $A$-module. Consider the following extension of $M$ over $B$

$0\rightarrow M\rightarrow N\rightarrow M\rightarrow0$

What is the obstruction for $N$ to be a projective $B$-module?

In other words, which elements from $\text{Ext}^{1}_{B}(M,M)$ correspond to projective modules?

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It's somehow weird to have a question starting with "two noncommutative rings" tagged ac.commutative-algebra. Could someone change it to homological? –  Anton Fonarev Jul 4 '12 at 10:17
I'd say there are too few hypotheses here to get an answer. –  Fernando Muro Jul 4 '12 at 11:00

Given a short exact sequence $0\to F_1 \to F \to F_2\to 0$, one has $pd(F)\leq \max \left( pd(F_1),pd(F_2) \right)$ with equality except when $pd(F_2)=pd(F_1)+1$. Suppose that $pd_B(M)<\infty$. Then $pd_B(N) = pd_B(M)$.
Now, $N$ is projective if and only if $pd_B(N)=0$. Therefore, one has to ask for $pd_B(M)=0$, which is the same as to say that $M$ is projective as a $B$-module.
It is also possible that $\mathrm{pd}_B(N)=0\ne\infty=\mathrm{pd}_B(M)$. –  user2035 Jul 4 '12 at 10:28
Counterexample: the ring $\mathbb{Z}/4$ and the extension $0\rightarrow \mathbb{Z}/2\rightarrow\mathbb{Z}/4\rightarrow\mathbb{Z}/2\rightarrow$. –  Fernando Muro Jul 4 '12 at 10:59
@Fernando, this obviously happens as $pd_{\mathbf{Z}/4}(\mathbf{Z}/2)=\infty$. I've already put this assumption that I had in mind. –  Anton Fonarev Jul 4 '12 at 11:05