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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. –  BBischof May 17 '11 at 5:04
    
Both maps are morphisms in $R$-mod. –  BBischof May 17 '11 at 5:13
    
Right again... sorry. –  BBischof May 17 '11 at 5:31
    
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 :-) –  Jack Schmidt May 17 '11 at 5:54
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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. –  Jack Schmidt May 17 '11 at 6:03

1 Answer 1

Here is one way to look at this problem. I will write $I$ for $\bar{R}$.

Firstly, $B/IB = (R/I) \otimes_R B$, that is, $B/IB$ is the image of $B$ under the induction functor (change of rings) $R$-mod $\to R/I$-mod. This functor will be written $\uparrow$. Frobenius reciprocity tells us

$$ \hom_{R/I}( B\uparrow, B'\uparrow) \cong \hom_R (B, B'\uparrow\downarrow) $$ where $\downarrow$ is restriction $R/I$-mod $\to R$-mod. We'll make this an identification, so we consider $\phi$ as an element of the right hand side. The question is then: "is $\phi$ in the image of $\hom_R(B,B') \to \hom_R(B,B'\uparrow \downarrow)$ ?" (the map on homs arises from $B' \to B\uparrow \downarrow$, the morphism given by the universal property of induction).

This stuff fits into the long exact sequence obtained by applying $\hom_R(B, -)$ to $IB' \to B' \to B'\uparrow\downarrow$. We get

$$0\to \hom_R(B, IB') \to \hom_R(B, B') \to \hom_R(B, B'\uparrow\downarrow) \stackrel{\omega}{\to} \operatorname{Ext}^1 _R(B, IB') \to \cdots $$

$\phi$ being in the image is equivalent to its being in the kernel of the connecting homomorphism $\omega$, which is multiplication by the short exact sequence above. So one answer to your question is: $\phi$ lifts if and only if it is killed by multiplication by the short exact sequence.

Really though, this is just a restatement of the problem. Clearly $B$ being projective is enough to guarantee the lift exists, as is any assumption that makes that ext group $\operatorname{Ext}^1 _R(B, IB')$ vanish. But for arbitrary $\phi$ I don't know general conditions other than the one above. It's sufficient for the induction functor to be full. $R/I$ being flat over $R$ (so induction is exact) certainly isn't enough.

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