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, 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), $x\otimes y\ne 0$. 
