Symmetric powers of modules

Let

$0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$

be an exact sequence of $A-$modules, where $A$ is a commutative ring with one.

Then for each natural number $i$ one can look at the induced maps

$Sym^i(N) \rightarrow Sym^i(P)$

and

$Sym^i(M) \rightarrow Sym^i(N)$.

The first question is, if they are still surjective resp. injective, i.e. is $Sym^i$ an exact functor?

The second question is: what are the kernel resp. cokernel of these maps?

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Surjectivity: yes
The morphism $Sym^i(N) \rightarrow Sym^i(P)$ is surjective.
Its kernel is generated by products of the form $mn_2n_3...n_i$ with $m \in M$ and $n_k\in N$

Injectivity:no
The morphisms $Sym^i(N) \rightarrow Sym^i(P)$ needn't be injective just because $f:M\to N$ is injective.

Let me give an example where $f:M\to N$ is injective but $S^2(f):S^2(M) \to S^2(N)$ is not.
Take $A=\mathbb Z/(4) , N=A, M=2A$ and $f: M\to N$ is the inclusion .
The crucial point is that $S^2(N)=N\otimes_A N$ and $S^2(M)=M\otimes_A M$ because these modules are generated by a single element (Check it on the definition of symmetric product!).
So it suffices to show that $f^{\otimes 2} : M\otimes_A M \to N\otimes_A N:2a\otimes 2b\mapsto 2a\otimes 2b=0$ is not injective.
Since this map is zero it suffices to show that $M\otimes_A M\neq 0$.
But this is clear because $M\simeq A/(2)$ as $A$-modules and $A/(2)\otimes_AA/(2) \simeq A/(2)$ as $A$-modules.

Edit
Since Cyril asks, let me indicate the plan of the proof that the kernel of $Sym^i(N) \rightarrow Sym^i(P)$ is generated as an $A$-module by by the $mn_2n_3...n_i$ with $u \in M$ and $n_k\in N$.
It suffices to prove the analogous result that the kernel of $\otimes^iN \rightarrow \otimes^i P$ is generated by the $n_1\otimes n_2\otimes...\otimes n_i$with one of the $n_k\in M$ and then take suitable quotients.
The assertion on the kernel of the $i$-th tensor producte is based on the right-exactness of the tensor product.
Full details can be found in Bourbaki , Algebra, Chapter III, §6.2, Proposition 4, page 499.

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Georges, do you mean the kernel is simply generated as an $A-$module by products of the above form, i.e. every element in the kernel is just a sum of such products? How does one see this exactly? That's an important point for me. – Cyril Dec 10 '11 at 19:06
Dear @Cyril: yes, that is exactly what I mean. I'll add a few words of explanation in an edit. – Georges Elencwajg Dec 10 '11 at 23:00

Sym is not exact: just think about dimension (I will work over some field $k$). In your set-up, if M,N,P have dimensions m,n,p then n=m+p. The symmetric square of a vector space of dimension k has dimension k choose 2, and in general n choose 2 will be different from (m choose 2) plus (p choose 2). (They are always different unless m=p=1).

Sym does preserve surjections. I will write monomials in the symmetric power of P as $p_1 \vee \cdots \vee p_r$; such elements span the $r$th symmetric power. If $n_i$ is a preimage of $p_i$ under your surjection, $n_1\vee\cdots\vee n_r$ is a preimages of $p_1 \vee \cdots \vee p_r$ under the induced map on the symmetric power.

However Sym is not right-exact. Regard M as a subspace of N, consider an element of the symmetric square of the form $m \vee n$ where $m \in M, n \in N \setminus M$. This is in the kernel of the induced map $\operatorname{Sym}^2 (N) \to \operatorname{Sym}^2 (P)$ but not in the image of $\operatorname{Sym}^2(M) \to \operatorname{Sym}^2(N)$. Thus

$$\operatorname{Sym}^2(M) \to \operatorname{Sym}^2(N) \to \operatorname{Sym}^2(P) \to 0$$ is not exact. (You seem to have the wrong idea of what "exact" means: e.g. for a covariant functor $F$ to be right-exact it must send exact sequences $A\to B\to C \to 0$ to exact sequences $FA \to FB \to FC \to 0$ which is stronger than simply preserving surjections).

Sym is also not a linear functor, that is,

$$\operatorname{Sym}^2 : \hom(V,W) \to \hom(\operatorname{Sym}^2(V),\operatorname{Sym}^2(W))$$

is not a linear map. This causes technicalities when trying to define the derived functors, but that's not to say it is impossible. See http://arxiv.org/abs/0911.0638

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Thanks a lot, Georges and mt_, this helped much! Especially the hint at the derivation of $Sym$ is nice. Just one question which concerns another possible (in-)compatibility: in Algebraic geometry one often considers a vector bundle $\mathcal E$ and $Sym^i(\mathcal E)$ of it. This is again a vector bundle. Does then $Sym^i$ commute with taking the dual of the vector bundle? – Cyril Dec 10 '11 at 17:44
@Cyril: even for ordinary vector spaces it's false in positive characteristic (the dual of $\text{Sym}^i(V)$ is the divided power, which is not isomorphic to $\text{Sym}^i(V^{\ast})$ as a $\text{GL}(V)$-representation). – Qiaochu Yuan Dec 11 '11 at 2:36
@Qiaochu: does one know under which conditions the statement is true? Do you know a reference for these things? – Cyril Dec 11 '11 at 19:04
In Gelfand and Manin's Methods of homological algebra in the Reference Guide section they mention derived functors of non-additive functors and give references discussing symmetric/exterior powers. – m_t_ Dec 8 '14 at 18:21