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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?

Pierre-Yves Gaillard
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WLOG
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2 Answers2

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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 $xyt=0$.

Pierre-Yves Gaillard
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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)

Bombyx mori
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  • and for the torsion part ? – WLOG May 18 '14 at 07:06
  • @WOLG: This should be easy, jut multiply $xy\otimes xy$. – Bombyx mori May 18 '14 at 11:10
  • Is your map well-defined? Because $\phi(y \otimes xy) = f(y)g(xy)=1 \cdot 0 = 0 \neq \phi(y^2 \otimes x) = f(y^2)g(x)=1$. – Watson Oct 04 '16 at 20:37
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    @Watson: I think this is a good point. The issue with $f,g$ is that $f,g$ are not really $R$-homomorphisms from $I$ to $R$, namely we do not want to have $f(xy)=x(f(y))=x!=y(f(x))=0$. They are only $R$-homomorphisms from $I$ to $R/I$ as an $R$-module like what Pierre did. This is a subtle point I should have made clear. – Bombyx mori Oct 05 '16 at 18:32
  • Thank you for your reply. However, I still don't understand: even if we consider $\phi : I \times I \to R \to R/I$ (instead of just $\phi : I \times I \to R$), we still have $$\phi(y,xy) = 0 + I \neq 1 + I = \phi(y^2,x) \in R/I.$$ Maybe I don't understand what you mean by "extending $f,g$" to all elements in $I$. – Watson Oct 08 '16 at 10:08
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    @Watson: In this case we have $\phi(y,xy)=x\phi(y,y)=0$, as the image is in $R/I$. Similarly $\phi(y^2,x)=y\phi(y,x)=0$. – Bombyx mori Oct 08 '16 at 12:00
  • Does finding such a map show that $x \otimes y \neq y \otimes x$ because this would imply that $0 \neq 1 = \phi(x \otimes y) - \phi(y \otimes x) = \phi(x\otimes y - y \otimes x)$ where the last equality comes from the fact $\phi$ is a homomorphism, is this reasoning right? I am having troubles rigorously justifying that finding such a map implies $x \otimes y \neq y \otimes x$. Thanks for the clarificatioN! – user110320 Dec 18 '17 at 23:39