Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
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

share|improve this answer
    
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. –  mt_ Dec 8 at 18:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.