# How does one show that this tensor product is not torsion-free?

I am having trouble showing that a particular tensor product is not torsion-free. Let $R = k[[x,y]]$, where $k$ is a field (this is the ring of formal power series in $x$ and $y$ with coefficients in $k$). Let $I$ be the non-principal ideal $(x,y)$; naturally, $I$ is an $R$-module. Then form the tensor product $I \otimes_{R} I$, which is also an $R$-module. How does one show that $I \otimes_{R} I$ is not a torsion-free $R$-module? In other words, how would one go about explicitly finding a non-zero element $x \in I \otimes_{R} I$ and a non-zero $p \in R$ such that $p \cdot x = 0$? Any hints are welcome. Thanks!

-

I think $x \otimes y - y \otimes x$ should do the trick. Observe that $$xy \cdot (x \otimes y - y \otimes x) = xy \otimes xy - xy \otimes xy$$and hope that the original element is nonzero (morally it isn't since to drag something under the tensor, you'd need that 1 is in the ideal)!

Formally, I think we have a bilinear map $I \otimes_R I \rightarrow k[[x,y]]/(x,y)^3$ sending $x \otimes y \mapsto x^2, x \otimes x \mapsto xy, y \otimes x \mapsto y^2, y \otimes y \mapsto xy$.

-
It's better to be uncooked than to be half-baked. – Rudy the Reindeer Dec 10 '12 at 9:53
I am still trying to verify the last part, but I will endorse your answer first. :) Thanks! – Haskell Curry Dec 10 '12 at 18:25
you're very welcome. glad to help! – uncookedfalcon Dec 10 '12 at 19:23