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?