# Why this element in this tensor product is not zero?

$R=k[[x,y]]/(xy)$, $k$ a field. This ring is local with maximal ideal $m=(x,y)R$. Then the book proves that $x\otimes y\in m\otimes m$ is not zero, but I don't understand what's going on, if the tensor product is $R$-linear, then $x\otimes y=1\otimes xy=1\otimes 0=0$, where is the mistake? And also the book proves that this element is torsion:

$(x+y)(x\otimes y)=(x+y)x\otimes y=(x+y)\otimes(xy)=(x+y)\otimes0=0$

why $(x+y)x\otimes y=(x+y)\otimes(xy)$?

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What book is "the book" ? – Georges Elencwajg Apr 18 '12 at 16:06

For your first question, $1$ does not lie in $m$, so $1 \otimes xy$ is not actually an element of $m \otimes m$.

$R$-linearity implies $a(b\otimes c)=(ab)\otimes c$, but (and this is important), if you have $(ab)\otimes c$, and $b$ does not lie in $m$, then "$ab$" is not actually a factorization of the element inside $m$, and if we try to factor out $a$, we get $a(b\otimes c)$, which is non-sensical since $b$ does not lie in $m$.

For your second question, $(x+y)x\otimes y=x((x+y)\otimes y)=(x+y)\otimes(xy)$. The reason we were allowed to take out the $x$ in this case was because $x+y$ lies in $m$, so all of the elements in the equations remained in $m \otimes m$.

Cheers,

Rofler

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For your first question, I suppose your tensor product is over $R$. It is enough to show that $x\otimes y$ is non zero in $(m/m^2)\otimes_R (m/m^2)$. As both sides in the latter tensor product are $k$-vector spaces, this tensor product identifies itself to the tensor product over $k$.

Now $v_1\otimes v_2\ne 0$ in a tensor product of vector spaces $V_1\otimes_k V_2$, if $v_1, v_2\ne 0$ (complete them to respective basis $\{e_i\}_i, \{f_j\}_j$ of $V_1, V_2$, and use the fact that the $e_i\otimes f_j$ form a basis of $V_1\otimes V_2$, or use the isomorphism $V_1\otimes V_2\to L(V_1^{\vee}, V_2)$, $v_1\otimes v_2\mapsto \{\phi \mapsto \phi(v_1)v_2\}$).

In your case, as the classes of $x,y$ in $m/m^2$ are non zero (they even form a basis), $x\otimes y\ne 0$.

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