Show that a tensor product is zero

Question

How can I show that $(k[x,y]/(x-y))\otimes_{k[x,y]}(k[x,y]/(x-1,y-2))$ is zero?

I tried by starting with some arbitrary element $a\otimes b$ to show it must be zero, but feel pretty lost on how to proceed.

Answer 1

This tensor product is annihilated by $(x-y)+(x-1,y-2)=(x-y,x-1,y-2)$. Can you show this ideal is trivial? (It contains $1$.)

Answer 2

First, here is some intuition: the quotient $k[x,y]/(x-y)$ amounts to insisting that $x = y$, while $k[x,y]/(x-1,y-2)$ means we set $x = 1$ and $y = 2$. This system of equations is inconsistent, so we expect the resulting tensor product to be $0$.

Or if you prefer geometry: $k[x,y]/(x-y)$ is the coordinate ring of the line $y = x$ in the plane, while $k[x,y]/(x-1,y-2)$ is the coordinate ring of the point $(1,2)$. Since $(1,2)$ doesn't lie on the line $y = x$ (i.e., the intersection of these two varieties is empty), then the tensor product should be $0$.

All right, now for a proof. I claim that $$ (k[x,y]/(x-y))\otimes_{k[x,y]}(k[x,y]/(x-1,y-2)) \cong \frac{k[x,y]}{(x-y, x-1, y-2)} \, . $$ Once we prove this we are done, since you've already shown that $1 \in (x-y, x-1, y-2)$, so $(x-y, x-1, y-2) = k[x,y]$. The claim follows from a more general fact.

Lemma: Let $R$ be a commutative ring, $I \trianglelefteq R$ an ideal and $M$ an $R$-module. Then $$ (R/I) \otimes_R M \cong M/IM $$ as $R$-modules.

(This is proved in example ($8$) of $\S10.4$ of Dummit and Foote (p. $370$). There is also a proof of a more particular case here.) Applying the lemma, we have \begin{align*} \frac{k[x,y]}{(x-y)}\otimes_{k[x,y]}\frac{k[x,y]}{(x-1,y-2)} & \cong \frac{k[x,y]/(x-1,y-2))}{(x-y)(k[x,y]/(x-1,y-2))}\\ &= \frac{k[x,y]/(x-1,y-2))}{(x-y, x-1, y-2)/(x-1,y-2))} \cong \frac{k[x,y]}{(x-y, x-1, y-2)} \end{align*} by the Third Isomorphism Theorem.