(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
Best Answer
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 $v = 0$.