I thought that $$k[x,y]/(x,y) \otimes_{k[x,y]} (x,y) = 0,$$ as tensor product is bi-linear and, for example, $[f] \otimes x = [xf] \otimes 1 = 0 \otimes 1 = 0 $. However my colleague claims that this is not true.

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    $\begingroup$ I don't see why $1$ is an element of $(x, y).$ If $1$ is not an element of $(x, y),$ then why is $[xf]\otimes1$ a valid expression? $\endgroup$
    – Will R
    Mar 23, 2020 at 10:48

1 Answer 1


No this tensor product is not $0$. In general, if $I$ is an ideal in $R$ and $M$ an $R$-module, then $R/I \otimes_R M \cong M/IM$. Your example takes $R=k[x,y]$, $I=(x,y)$, and $M=I$, so $k[x,y]/(x,y) \otimes_{k[x,y]} (x,y) \cong (x,y)/(x,y)^2 \ne 0$.

As for your argument, you do not have $[f] \otimes x=[xf] \otimes 1$; indeed, the last expression makes no sense since $1 \notin (x,y)$.


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