$x \otimes y - y \otimes x \neq 0$ in $I \otimes_{R} I$

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?

Solution 1:

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 $xyt=0$.

Solution 2:

Let $\phi: I\times I\rightarrow R/I$ to be the extension of product of $f:I\rightarrow R/I$ and $g: I\rightarrow R/I$. Here $f,g$ are given by extending $$ f(x)\rightarrow 0, f(y)\rightarrow 1, f(1)\rightarrow 1; g(x)\rightarrow 1, g(y)\rightarrow 0, g(1)\rightarrow 1 $$ to all elements in $I$. Then we have $$ \phi(x\otimes y)=f(x)g(y)=0, \phi(y\otimes x)=1 $$ So $x\otimes y\not=y\otimes x$.

(Thanks Watson for the update)