Basic Issue With the Hom Functor on Commutative Rings

Question

In the category of $A$-modules, one has the following property: if $f:M \rightarrow M''$ is a map of $A$-modules, and the induced map $f^*:Hom(M'',N) \rightarrow Hom(M,N)$ is injective for all $N$, then $f$ itself must be surjective. The proof I know of this fact would be to take $N = coker(f)$ and note that if $q:M'' \rightarrow N$ is projection, we have $f^*q=0$ and therefore $q=0$.

I would like to know if the same result holds for commutative rings with 1. That is, if the pullback $f^*:Hom(A,C) \rightarrow Hom(B,C)$ is injective for all commutative rings $C$ with 1, then is it true that $f:B \rightarrow A$ is surjective, even now that our abelian category logic no longer applies?

The keyword is "epimorphism" vs. "surjective homomorphism". With these you will find many many answers to your question. — Martin Brandenburg, Oct 31 '13 at 09:43

score 2 · Accepted Answer · answered Oct 31 '13 at 09:34

2

No, it is not true. For example, take the map $\mathbb Z \hookrightarrow \mathbb Q$.

answered Oct 31 '13 at 09:34

Very nice, thank you. – Cass Oct 31 '13 at 09:56
Any localization does. You can also combine "orthogonal" epis, for example $\mathbb{Z} \to \mathbb{Z}/2 \times \mathbb{Z}[1/2]$. – Martin Brandenburg Oct 31 '13 at 11:16
Indeed. Martin's both comments are very relevant. As a very "basic" reference, you can try Chap. I of Jacobson's "Basic Algebra", Vol. 2. – Nov 03 '13 at 14:24

Basic Issue With the Hom Functor on Commutative Rings

1 Answers1