Let $A$ be a finite dimensional algebra over field K. $M,N$ are $A$-modules such that $0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$ and $0 \rightarrow Q_1 \rightarrow Q_0 \rightarrow N \rightarrow 0$ are minimal projective resolutions of them, respectively. Let $f: M \rightarrow N$ be an morphism. Then by the projectivity of $P_1,P_0$, we have $f_1,f_0$ such that the following diagram commutes: enter image description here

Also we know that $f_1,f_0$ may not be unique. I want to know under the condition that $Hom_A(M,Q_0)=0$, can we get $f_0,f_1$ uniquely determined by $f$?


The condition $Hom_A(M,Q_0)=0$ is not sufficient to ensure that $f_0$ and $f_1$ are uniquely determined by $f$.

To see a counter-example, let $Q=1\to 2\to 3$ be a quiver, $A = KQ$ be its path algebra, and let $M$ be the simple (left) $A$-module at vertex $2$ and $N$ be the simple module at vertex $1$. The projective resolutions of $M$ and $N$ look like this: $$ 0\to P(3) \to P(2) \to M\to 0 \quad \textrm{and} \quad 0\to P(2) \to P(1) \to N\to 0. $$ Now, if we take $f:M\to N$ to be the zero morphism, we see that letting $f_1$ be the inclusion of $P(3)$ into $P(2)$ and $f_0$ be the inclusion of $P(2)$ into $P(1)$ yields a commutative diagram as in your question. Since putting $f_0=f_1=0$ also yields such a commutative diagram, we get that $f_1$ and $f_0$ are not uniquely determined by $f=0$, even though $Hom_A(M, P(1)) = 0$.

However, a necessary and sufficient condition for $f_0$ and $f_1$ to be uniquely determined by $f$ would be that $Hom_A(P_0, Q_1)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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