Eisenbud's proof of right-exactness of the exterior algebra

Question

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?

Answer 1

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.

Answer 2

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".