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

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
