I'm asked to prove that the ideal $I=(x,y)$ in $R=k[x,y]$ is not a flat R-module.
My approach was to use the exact sequence $$0\rightarrow I \to R \to R/I \to 0$$ to induce a non injective map $$I\otimes I\to R\otimes I$$ because the element $x\otimes y-y\otimes x$ is sent to $0$ by the inclusion.
But I need to show that such element is not zero in the domain. My intuitive approach is that such element is nonzero because in $I\otimes I$ I'm not allowed to move $x$ or $y$ "inside" the tensor because $1\notin I$. But I cannot formalise it properly. So how to prove that such tensor is not zero?
It might be easier to approach this problem the following way: We have the following free resolution of $I$:
$$0 \to R \to R \oplus R \to I \to 0,$$ where the first map is given by $1 \mapsto (-y,x)$ and the second map is given by $(1,0) \mapsto x, (0,1) \mapsto y$.
After tensoring with $R/I$ we get the following exact sequence:
$$0 \to Tor_1(I,R/I) \to R/I \to R/I \oplus R/I \to I/I^2 \to 0$$
The map $R/I \to R/I \oplus R/I$ is given by $1 \mapsto (-y,x)=0 \in R/I \oplus R/I$, hence this is the zero map. So we deduce $Tor_1(I,R/I) \cong R/I$, in particular $I$ is not flat.
Furthermore we deduce that the kernel of your map (since this kernel is precisely $Tor_1(I,R/I)$) is generated by an element, which is annihilated by $I$. This is just your element $x \otimes y - y \otimes x$.