Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm stuck on the following and could use a hint.

Let $f:P\longrightarrow M$ be a map of finite dimensional modules over a finite dimensional algebra $A$ (over probably an algebraically closed field $K$), with $P$ projective. $f$ induces a map $\operatorname{top}f:\operatorname{top}P\longrightarrow \operatorname{top}M$ where $\operatorname{top}M = M/\operatorname{rad}M$ and $\operatorname{rad}M$ is the Jacobson radical.

Show that $\operatorname{top}f$ an isomorphism implies $f:P\longrightarrow M$ is a projective cover.

Don't really know where to start.

share|cite|improve this question
up vote 1 down vote accepted

Let $\pi_P:P\to P/\mbox{rad}P$ and $\pi_M:M\to M/\mbox{rad}M$ be the natural projections. The induced homomorphism $\mbox{top}f$ is such that $\pi_M\circ f=\mbox{top}f\circ\pi_P$. If $y\in M$ then, being $\mbox{top}f$ surjective, $y+\mbox{rad}M=\mbox{top}f(x+\mbox{rad}P)=f(x)+\mbox{rad}M$, for some $x\in P$. This proves that $M=\mbox{rad}M+\mbox{Im}f$. Since $M$ is finite dimensional, $\mbox{rad}M$ is superfluous in $M$, therefore $M=\mbox{Im}f$, i.e. $f$ is an epimorphism.

On the other hand, if $f(x)=0$ then $\mbox{top}f(x+\mbox{rad}P)=f(x)+\mbox{rad}M=0$, so being $\mbox{top}f$ a monomorphism we have $x\in\mbox{rad}P$. Therefore $\mbox{Ker}f\leq\mbox{rad}P$. Again, since $P$ is finite dimensional, $\mbox{rad}f$ is superfluous, so $\mbox{Ker}f$ is also superfluous.

We conclude that $f:P\to M$ is a projective cover.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.