From exercise 2.3, Atiyah Macdonald

(My concern is not about the exercise itself, but about two passages in the hint for the exercise.) Let $A$ be a local ring with maximal ideal $m$, and call $k$ its residue field $A/m$. Let $M,N$ be finitely generated $A$-modules. First it is said that $$k\otimes_A(M\otimes_A N)=(k\otimes_AM)\otimes _k (k\otimes_AN).$$ Where does this equality come from? Shouldn't I tensor with $k$ only one between $M$ and $N$ (obtaining isomorphic results)?

Then it is also said that if $V,W$ are $k$-vector spaces, then $V\otimes_k W=0$ only if either $V=0$ or $W=0$. Even for this result, I didn't see it among the proposition of the book, and I wouldn't know how to prove it using them. Thanks for any clarify

Solution 1:

To the first question: This follows from

$$(M \otimes_A k) \otimes_k (k \otimes_A N) \cong M \otimes_A (k \otimes_k k) \otimes_A N \cong M \otimes_A k \otimes_A N \cong k \otimes_A (M \otimes_A N).$$

To the second question: This can be done in several ways. The easiest is probably noting that $\dim(V \otimes_k W) = \dim(V)\cdot \dim(W)$ and thus $\dim(V \otimes_k W) = 0$ if and only if $\dim(V) = 0$ or $\dim(W) = 0$.

Another method would be to observe that $V \otimes_k W$ is generated by all the pure tensors $v \otimes w$, so if $V \otimes_k W = 0$, then $v \otimes w = 0$ for any $v \in V$ and $w \in W$. We show that if $w \neq 0$ and $v \otimes w = 0$, then $v = 0$ using the universal property of the tensor product. Suppose that $v \neq 0$ and choose $\varphi \in V^*$ such that $\varphi(v) = 1$. Since $w \neq 0$, we may also choose $\psi \in W^*$ such that $\psi(w) = 1$. Then the map

$$\beta \colon V \times W \to k, \quad (v',w') \mapsto \varphi(v')\psi(w')$$

is bilinear (and non-zero by construction). By the universal property of the thensor product there is a unique linear map $\tilde \beta \colon V \otimes_k W \to k$ such that $\tilde \beta(v' \otimes w') = \beta(v',w')$. But $\tilde \beta$ is the zero map and $\beta$ is non-zero, a contradiction. Hence $v = 0$.

Solution 2:

The trick to the first problem is to use associativity of the tensor product, together with the fact that $k\cong k\otimes_k k$.

$$\begin{split} k\otimes_A(M\otimes_A N) & \cong (k\otimes_A M)\otimes_A N \\ & \cong (M\otimes_A k)\otimes_A N \\ & \cong M\otimes_A (k)\otimes_A N \\ & \cong M\otimes_A (k\otimes_k k)\otimes_A N \\ & \cong (M\otimes_A k)\otimes_k (k\otimes_A N) \\ & \cong (k\otimes_A M)\otimes_k (k\otimes_A N). \end{split}$$

For the second part, that is going to depend what you know about tensor products of vector spaces. But one good approach is to use the hom-tensor adjunction. We have that if $X$ and $Y$ are non-zero vector spaces, then so is $\hom(X,Y)$, but if either is 0, then so is the hom. So it suffices for us to show that we can find a $Z$ such that $\hom(V\otimes W, Z)\neq 0$. By the adjunction, this is $\hom(V,\otimes \hom(W,Z))$, and as long as V, W, and Z are all non-zero, this will be non-zero too. If any one of the three are zero, then so will the hom. So we can transfer our understanding of the behavior of homs to give us understanding of the behavior of tensor products.