$A \otimes_k A \to \bigwedge^2 A$ homs Let $k$ be a unital commutative ring, and A be a $k$-module.
Is there a homomorphism $f: A \otimes_k A \to \bigwedge^2 A$ such that $f(a \otimes a) \neq 0$ for some $a \in A$? I can take a hint :)
Motivation:
In his textbook on Lie algebra, Serre defines a Lie algebra as a k-module with a homomorphism $A \otimes_k A \to A$ that factors through $\Lambda^2 A$. He then goes on to explain that it means that [x,x]=0 for all $x \in A$.
 A: What Serre means is the following.  You probably know that the tangent square $A \otimes_k A$ is universal for $k$-bilinear maps out of $A \times A$.  The exterior square $\Lambda^2 A$ is universal for alternating $k$-bilinear maps out of $A \times A$.  Any such map is first of all a $k$-bilinear map, so it factors through $A \otimes_k A$.  But the alternating property causes it to further factor through the natural quotient $A \otimes_k A \to \Lambda^2 A$.  Serre has in mind a specific map here.
A: This answer is no loger relevant, since the OP now asked (in a comment...) what he had meant to askm which happens not to be what he asked :)
There obviously exist examples of $k$s and $A$s such that such homomorphisms do exist (take, for example, $A$ free of rank at least $2$ over $k$...)
Now it is not true that such homomorphisms exist for all choices of $k$ and $A$ (take, for example, $A$ free of rank $1$...)
I am not sure neither of these facts is what you intended to ask, though.
Later: one thing you might have intended to ask if there is such a homomorphism depending naturally on $A$. The answer to that is also no, because of my second example above.
