TITLE: $x \otimes y - y \otimes x \neq 0$ in $I \otimes_{R} I$
QUESTION [11 upvotes]: Let $R = k[x,y]$ , $I = (x,y)$ , $k$ is a field.
I want to prove that :
1) $x \otimes y - y \otimes x \neq 0 $ in  $I \otimes_{R} I$
2) $x \otimes y - y \otimes x $ is a torsion element
My thoughts: to prove that $x \otimes y - y \otimes x \neq 0 $ in  $I \otimes_{R} I$ probably I should find a bilinear map $$\phi : I \times I \to R$$ such that $\phi(x,y) \neq \phi (y,x)$ , but which one?

REPLY [11 votes]: Assume that $k$ is a commutative ring, put $t:=x\otimes y-y\otimes x$, and identify $k$ to $R/I$.
1) Using the $R$-bilinear map 
$$
I\times I\to k,\qquad(f,g)\mapsto
\frac{\partial f}{\partial x}(0,0)\ 
\frac{\partial g}{\partial y}(0,0),
$$ 
it is easy to see that $t$ is nonzero.
2) We have $xyt=0$.