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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.

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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

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

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  • $\begingroup$ We rather have $f(y,x)=-1$, no? And, which map $k[x,y]\to k$ did you mean? $\endgroup$ – Berci Oct 28 '13 at 1:06
  • $\begingroup$ @Berci Of course :) The map is just reduction mod $I$. $\endgroup$ – Alex Youcis Oct 28 '13 at 1:07
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    $\begingroup$ If anyone is wondering how I produced this, it is a common (classic) computation one performs in the basics of module theory. The usual goal is not to show that this element is non-zero, but the shifted problem of trying to show that $\Lambda^2(I)$ is non-zero. There are more sophisticated techniques to handle this that aren't so constructive. $\endgroup$ – Alex Youcis Oct 28 '13 at 1:11
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    $\begingroup$ Nice answer! But I think it would be better to write $$f(ax+by+\text{higher order},cx+dy+\text{higher order})=ad$$ to cover the characteristic 2 case. (And also to say explicitly that "the map $k[x,y]\to k$" is the evaluation at 0.) $\endgroup$ – Pierre-Yves Gaillard Dec 1 '18 at 12:16
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    $\begingroup$ Great comment, @Pierre-Yves! $\endgroup$ – Georges Elencwajg Mar 7 '20 at 21:20

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