Determining whether a certain element in a tensor product is zero

Question

Let $I = (x,y) \subset k[x,y]$, where $k$ is a field. Prove that

a) $x \otimes y - y \otimes x =0$ inside $k[x,y] \otimes_{k[x,y]} k[x,y]$.

b) $x \otimes y - y \otimes x \not= 0$ inside $I \otimes_{k[x,y]}I$.

a) Since $x, y, 1 \in \operatorname{k}[x,y]$, we have $x \otimes y = 1 \otimes xy = 1 \otimes yx = y \otimes x$.

b) We cannot say $x \otimes y = 1 \otimes xy$ since $1 \not\in I$. I tried to find a bilinear map $f$ from $I \times I$ to $\operatorname{k}[x,y] \otimes_{$\operatorname{k}[x,y]} \operatorname{k}[x,y]$ so that $f(x,y)-f(y,x)$ is not zero, but couldn't really think of any map like that...so I was wondering if anybody could give me a hint.

Answer 1

Hint: You can regard $k$ as a $k[x,y]$-module via the map $k[x,y]\to k$. Then, consider the map $I\times I\to k$ defined by

$$f(ax+by+\text{higher order},cx+dy+\text{higher order})=ad-bc$$

Check that $f$ really is a $k[x,y]$-bilinear map $I\times I\to k$. Note though that

$$f(x,y)=1\qquad f(y,x)=-1$$