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!


1 Answer 1


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 $x \otimes y \mapsto x^2, x \otimes x \mapsto xy, y \otimes x \mapsto y^2, y \otimes y \mapsto xy$.

  • $\begingroup$ It's better to be uncooked than to be half-baked. $\endgroup$ Dec 10, 2012 at 9:53
  • $\begingroup$ I am still trying to verify the last part, but I will endorse your answer first. :) Thanks! $\endgroup$ Dec 10, 2012 at 18:25
  • $\begingroup$ you're very welcome. glad to help! $\endgroup$ Dec 10, 2012 at 19:23

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