Let $R$ be a commutative ring. The symmetric algebra functor $\operatorname{Sym}$ is defined from the category of $R$-modules to the category of (graded) commutative $R$-algebras.
It takes an $R$-module homomorphism $f:M\rightarrow N$ to an $R$-algebra homomorphism $\operatorname{Sym}(f):\operatorname{Sym}(M)\rightarrow \operatorname{Sym}(N)$.
Because $\operatorname{Sym}$ is left adjoint to the forgetful functor,
it is right exact.
Therefore, if $f$ is surjective, $\operatorname{Sym}(f)$ is also surjective.
My question is that if $f$ is injective, $\operatorname{Sym}(f)$ is also injective?
First I thought that it is false but I couldn't find any counterexamples.
Is it true if $R$ is sufficiently good (e.g. field, integral domain, etc.) or $M$, $N$ are particularly nice (e.g. free, projective, …) ?
Best Answer
For a simple counterexample, consider the homomorphism $\mathbb{Z}/(2)\stackrel{2}\to\mathbb{Z}/(4)$ of $\mathbb{Z}$-modules. If $x$ is the generator of $\mathbb{Z}/(2)$, then $x^2\neq 0$ in $\operatorname{Sym}(\mathbb{Z}/(2))$, but the induced map $\operatorname{Sym}(\mathbb{Z}/(2))\to \operatorname{Sym}(\mathbb{Z}/(4))$ sends $x^2$ to $4$ times a generator of $\operatorname{Sym}^2(\mathbb{Z}/(4))$, which is $0$.
(For a more dramatic example of the same idea, you could take the inclusion of any nontrivial cyclic subgroup into $\mathbb{Q}/\mathbb{Z}$, since $\mathbb{Q}/\mathbb{Z}\otimes\mathbb{Q}/\mathbb{Z}=0$ so all of its symmetric powers above the first are trivial.)
It is true over a field, since over a field any injective homomorphism has a left inverse and so will still have a left inverse after applying Sym.