7
$\begingroup$

I'm trying to understand the proof in Eisenbud's Commutative Algebra that, given a right exact sequence $$K \to N \to M \to 0$$ of $R$-modules, we have an exact sequence $$K \otimes \wedge N \to \wedge N \to \wedge M \to 0.$$

This is Proposition A2.2 on page 576. Here's the relevant part of the setup:

$R$ will denote a graded ring concentrated in degree 0, and $M$ will denote a $\mathbb{Z}$-graded module. The tensor algebra $T_R(M)$ becomes a $\mathbb{Z}$-graded module with $i$-th graded component $$T_R(M)_i = \bigoplus_{j_1+\cdots+j_n=i} M_{j_1} \otimes \cdots \otimes M_{j_n}.$$

[...]

(d) An exact sequence $K \to M \to N \to 0$ of $R$-modules gives rise to exact sequences $$T_R(N) \otimes K \otimes T_R(N) \to T_R(N) \to T(M) \to 0,$$

[...]

(I take it that the expression for $T_R(M)_i$ is meant to include, when $i = 0$, a term where $n = 0$, contributing a copy of $R$ via the vacuous tensor product to $T_R(M)_0$ -- otherwise this definition doesn't seem as though it could actually give a grading of $T_R(M)$.)

The proof supplied says that this last fact follows immediately from the basic properties of tensor products (namely right exactness, associativity, commutativity, base change, and distributivity). I'm not quite sure what's going on here, but I took this to mean that we've tensored by $T_R(N)$ on the left and right, getting an exact sequence

$$T_R(N) \otimes K \otimes T_R(N) \to T_R(N) \otimes N \otimes T_R(N) \to T_R(N) \otimes M \otimes (N) \to 0$$

and that we've somehow obtained identifications $T_R(N) \otimes N \otimes T_R(N) \cong N$ and $T_R(N) \otimes M \otimes T_R(N) \cong N$ of graded modules.

But this doesn't seem to be true: for instance, if $N = N_1$ is concentrated in degree 1, then we have

$$T_R(N)_0 = R$$

but

$$[T_R(N) \otimes N \otimes T_R(N)]_0 = 0$$

since $T_R(N)_k = 0$ for $k < 0$ and $N_k = 0$ for $N \leq 0$.

Or, if that's just some trick of grading, suppose that $N$ is an ideal of $R$ generated by a zerodivisor, in which case it appears that the entire module $T(N) \otimes N \otimes T(N)$ would be killed by a particular element of $R$, whereas the module $T(N)$ contains a copy of $R$ and thus would not have such a property.

Can someone help me out here? What's going on?

Alternatively, is there a more straightforward way to arrive at the exact sequence $K \otimes \wedge N \to \wedge N \to \wedge M \to 0$, which is all I really need at the moment?

$\endgroup$
8
  • $\begingroup$ is the original exact sequence an exact sequence of $\mathbb Z$-graded modules? $\endgroup$
    – Avitus
    Commented Nov 4, 2013 at 19:59
  • $\begingroup$ Interesting question. It doesn't explicitly say so there, but the goal is to get results about this graded symmetric algebra gadget which encompasses both the symmetric and exterior algebras, so it appears that we need the option of having these be graded. (Why do you ask? Just for clarification?) $\endgroup$ Commented Nov 4, 2013 at 20:17
  • $\begingroup$ In the grey box the ring $R$ is concentrated in degree 0 while the module $M$ is graded. The presence of a grading makes dealing with tensor algebras a bit more complicated, due to the presence of the weight (length of strings $M\otimes ...\otimes M$) and the grading itself. A small question: in the tensor algebra definition, is the tensor product over $R$, am I right? If yes, and $R$ is not necesseraly commutative, then the module $M$ is an $R$-$R$-bimodule, not simply a left (or right) $R$-module. $\endgroup$
    – Avitus
    Commented Nov 4, 2013 at 20:22
  • $\begingroup$ $R$ is assumed to be commutative throughout the entire book, and further to be Noetherian in this chapter. $\endgroup$ Commented Nov 4, 2013 at 20:30
  • $\begingroup$ I see, then notation is ok :). Let us suppose all modules are concentrated in degree $0$. Then $\wedge^0N=\wedge^0M=R$ by definition. What are the maps in the induced sequence $K\simeq K\otimes_R R\rightarrow R\rightarrow R\rightarrow 0$? Does the book describe these maps in some detail? $\endgroup$
    – Avitus
    Commented Nov 4, 2013 at 20:37

2 Answers 2

8
$\begingroup$

In fact this right exactness is so basic and formal that it holds in any linear cocomplete $\otimes$-category. But let me stick to modules here. It is enough to prove that $$K \otimes \Lambda^{p-1}(N) \to \Lambda^p(N) \to \Lambda^p(M) \to 0$$ is exact for every $p$. By the definition of a cokernel (i.e. its universal property), this means that for every "test" module $T$ the sequence $$0 \to \hom(\Lambda^p(M),T) \to \hom(\Lambda^p(N),T) \to \hom(K \otimes \Lambda^{p-1}(N),T)$$ is exact. Now this can be checked directly, using that $\hom(\Lambda^p(M),T)$ is (by definition) the set of alternating maps $M^p \to T$ (or alternating homomorphisms $M^{\otimes p} \to T$ if you don't want to go to sets here).

We have to show: Composing with $N^p \to M^p$ gives a bijection between alternating maps $M^p \to T$ and alternating maps $N^p \to T$ which vanish when you insert $K$ in the first variable. Let us denote the maps here by $i : K \to N$ and $f : N \to M$. If $\beta : N^p \to T$ is alternating with $\beta(i(K),\dotsc)=0$, then by antisymmetry $\beta(\dotsc,i(K),\dotsc)=0$. It follows that $\beta$ extends to a well-defined map $\gamma : M^p \to T$ such that $\gamma \circ f^p = \beta$. It is clear that it is alternating.

Once again the functorial approach reveals (a) the most efficient proof, and (b) what is really "behind" the statement.

$\endgroup$
2
$\begingroup$

A sketchy approach. Let

$$K\stackrel{\alpha}{\rightarrow} N\stackrel{\beta}{\rightarrow} M\rightarrow 0 $$

be an exact sequence of $R$-modules, i.e. $\operatorname{ker}\beta=\operatorname{im}\alpha$ and $\beta$ is surjective.

We want to prove that the induced exact sequence

$$K\otimes \wedge N \stackrel{\Phi}{\rightarrow} \wedge N\stackrel{\wedge(\beta)}{\rightarrow} \wedge M\rightarrow 0 $$

is exact, where the maps $\Phi$ and $\wedge(\beta)$ are defined as follows.

The morphism of $R$-modules $K\otimes \wedge N \stackrel{\Phi}{\rightarrow} \wedge N$ is given by the composition

$$K\otimes \wedge N \stackrel{\alpha\otimes 1_{\wedge N}}{\rightarrow} N\otimes\wedge N \stackrel{\rho}{\rightarrow} \wedge N, $$

where $\rho(n\otimes n_1\wedge\dots\wedge n_r):=n\wedge n_1\wedge\dots\wedge n_r$ is the restriction to the component $N\subset T_R(N):=\bigoplus_{n\geq 0}N^{\otimes n}$ of the left action

$$T_R(N)\otimes \wedge N\rightarrow \wedge N $$

of the tensor algebra (which is an associative algebra with unit) on the exterior algebra $\wedge N$. In summary

$$\Phi(k\otimes n_1\wedge\dots\wedge n_r)=\alpha(k)\wedge n_1\wedge\dots\wedge n_r. $$

The morphism $\wedge(\beta)$ is given by

$$\wedge(\beta)(n_1\wedge\dots\wedge n_r):=\beta(n_1)\wedge\dots\wedge\beta(n_r).$$

It follows from the above definitions that $\operatorname{im}\Phi\subset \operatorname{ker}\wedge(\beta)$, as $\beta\circ\alpha=0$. $\wedge(\beta)$ is clearly surjective; what is left to be proven is that $\operatorname{ker}\wedge(\beta)\subset \operatorname{im}\Phi$. This last fact is proven using Prop. 3 pag 509 in Bourbaki "Algebra I".

$\endgroup$
4
  • $\begingroup$ Thanks! Any clue about the approach presented in Eisenbud? I'd still like to understand that, so I may split that off into another question. $\endgroup$ Commented Nov 4, 2013 at 21:58
  • $\begingroup$ But you cannot prove the remaining inclusion with elements! You end the proof when it actually should start ... $\endgroup$ Commented Nov 5, 2013 at 11:13
  • $\begingroup$ I do not fully understand the comment about elements: prop.3 pg 509 and prop. 4 pag 499 in Bourbaki "Algebra I" describe the kernel of $\wedge\beta$ in terms of the kernel of $\beta$, which is equal to image of $\alpha$. $\endgroup$
    – Avitus
    Commented Nov 5, 2013 at 16:38
  • 1
    $\begingroup$ Actually, the result in Bourbaki shows every part of my question, though it's not phrased in terms of exact sequences. $\endgroup$ Commented Nov 6, 2013 at 4:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .