What is the kernel of the induced map on symmetric algebras?

Question

Let $$0\to M\to N\to P\to 0$$ be an exact sequence of $R$-modules over a commutative ring $R$.Then consider the induced map on the $i^{th}$ graded piece of the symmetric algebras: $$S^i(N)\to S^i(P).$$ What's the kernel?

It's clear that any product $mn_2\ldots n_i$ is in the kernel, where $m\in M$ and each $n_k\in N$. But why is that it?

In this question:

Symmetric powers of modules

They mention using right exactness. How specifically is this used? I couldn't find the referenced Bourbaki chapter.

Answer 1

I shall assume your exact sequence is in the category of modules over some commutative ring $A$. Assume further that the map $f : N \to P$ is a surjective $A$-module homomorphism. Then, the kernel of the induced map $\operatorname{Sym} f : \operatorname{Sym} N \to \operatorname{Sym} P$ is the ideal of $\operatorname{Sym} N$ generated by $\operatorname{Ker} f \subseteq N$. (For a proof, see Theorem 62 in my A few classical results on tensor, symmetric and exterior powers, but I agree -- this is probably somewhere in Bourbaki.) Note that the modules $\operatorname{Sym} N$ and $\operatorname{Sym} P$ are graded, and the induced map $\operatorname{Sym} f : \operatorname{Sym} N \to \operatorname{Sym} P$ is degree-preserving; thus, its kernel is a graded submodule of $\operatorname{Sym} N$. Hence, for each positive integer $i$, the kernel of the induced map $\operatorname{Sym}^i f : \operatorname{Sym}^i N \to \operatorname{Sym}^i P$ is the $i$-th graded component of the kernel of $\operatorname{Sym} f$. Our characterization of the latter kernel thus shows that the kernel of the induced map $\operatorname{Sym}^i f : \operatorname{Sym}^i N \to \operatorname{Sym}^i P$ is spanned by all products of the form $n_1 n_2 \cdots n_i$ with $n_1, n_2, \ldots, n_i \in N$ and $n_1 \in \operatorname{Ker} f$.

If $f$ is not surjective, then this no longer holds. For a simple counterexample, set $A = \mathbb{Z}$ and $N = \mathbb{Z}$ and $P = \mathbb{Z}/4$ and $f\left(a\right) = \overline{2a}$ (where $\overline{2a}$ means the canonical projection of $2a \in \mathbb{Z}$ onto the quotient $\mathbb{Z}/4$). Then it is easy to see that $\operatorname{Ker} f = 2 \mathbb{Z}$, but $\operatorname{Ker}\left(\operatorname{Sym}^2 f\right)$ is the whole $\operatorname{Sym}^2 \mathbb{Z}$.