Extension of projective module - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:58:42Z http://mathoverflow.net/feeds/question/101294 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101294/extension-of-projective-module Extension of projective module vova 2012-07-04T08:31:30Z 2012-07-04T11:01:16Z <p>Hi all!</p> <p>I am interested in the following question in homological algebra.</p> <p>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 $B$ </p> <p>$0\rightarrow M\rightarrow N\rightarrow M\rightarrow0$</p> <p>What is the obstruction for $N$ to be a projective $B$-module?</p> <p>In other words, which elements from $\text{Ext}^{1}_{B}(M,M)$ correspond to projective modules?</p> http://mathoverflow.net/questions/101294/extension-of-projective-module/101300#101300 Answer by Anton Fonarev for Extension of projective module Anton Fonarev 2012-07-04T10:10:28Z 2012-07-04T10:58:46Z <p>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 $pd_B(M)&lt;\infty$. Then $pd_B(N) = pd_B(M)$.</p> <p>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.</p>