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Today I tried to check this, but couldn't see how to do it. I think it is probably a standard result, but a brief check of Atiyah-Macdonald didn't yield anything, and I don't know what to google for. A reference is also appreciate.

Consider a associative, unital, $K$-algebra $R$ and a nontrivial ideal $\bar{R}$. Now consider two $R$ modules $B$ and $B'$. Also consider the quotients $\bar{B}=B/\bar{R}B$ and $\bar{B'}=B'/\bar{R}B'$.

If we have a map $\varphi:\bar{B}\to\bar{B}'$, is there some condition that we can lift to a map of $B\to B'$?

If the quotient map on $B'$ splits, we have this lift, but this is not iff.

Thanks in advance!

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    @Jack Thanks, you are right, $\bar{R}$ is not unital. I am not sure what structure you are asking about though.2011-05-17
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    Both maps are morphisms in $R$-mod.2011-05-17
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    Right again... sorry.2011-05-17
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    Problem statement looks good. My ideas to solve it don't seem to pan out. Quick reference check didn't find much. I nixxed all my bad ideas :-)2011-05-17
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    There are finite-dimensional algebras (commutative I think) which have simple modules X,Y with Ext(X,X)=Ext(X,Y)=Ext(Y,X)=K. Let R-bar be the Jacobson radical, B=X.X and B'=X.Y. Then Hom(B-bar,B'-bar)=K, but Hom(B,B')=0. R is the Brauer tree algebra of (2)-*-*, but I've misplaced my nice description of such an algebra. At any rate, this shows that R-bar being nilpotent is not sufficient.2011-05-17

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