TITLE: How to see that $\ker\left((X,Y)\otimes_R(X,Y)\to(X,Y)^2\right)\simeq k$ in $R=k[X,Y]$?
QUESTION [6 upvotes]: Let $k$ be a field, $R=k[X,Y]$ and $I=(X,Y)$, so that $R/I\simeq k$. I proved, using a projective resolution of $k$, that $\text{Tor}^R_2(k,k)= k$. I also proved that in general
$$
\text{Tor}^A_2(A/I,A/J)=\ker(I\otimes_AJ\to IJ)
$$
where $I,J$ are ideals in a commutative ring $A$. When trying to see this by computing $\ker(I\otimes_AJ\to IJ)$ for $A=R$ and $I=J=(X,Y)$, though, I get to an impasse.
Since $X,Y$ generate $I$, then $\{X\otimes X,X\otimes Y,Y\otimes X,Y\otimes Y\}$ is a system of generators of $I\otimes_R I$. In particular $\alpha=X\otimes Y - Y\otimes X \in \ker(I\otimes_RI\to I^2)$ since $\alpha \mapsto XY-YX=0$. But then $\forall f\in R$ we have $I\otimes_R I \ni f\alpha \mapsto f\cdot(XY-YX)=0$, so $R\alpha\subseteq\ker(I\otimes_RI\to I^2)$.
What am I missing here?
Edit: @GeorgesElencwajg cleared my doubt, though I still don't see how to prove that $\ker(I\otimes_RI\to I^2) = k$ without using homological algebra.

REPLY [5 votes]: You are missing nothing: there is actually no contradiction .
What  might confuse you is that $R\alpha =\ker(I\otimes_AJ\to IJ)$ seems huge, of the order of magnitude of $R$, but that is an illusion.
  For example $X^2\otimes _R Y-X\otimes _R XY$ is certainly in $\ker(I\otimes_AJ\to IJ)$, but  is actually zero since $X^2\otimes _R Y=X\otimes _R XY$: the element $X$  may jump  from the left of $\otimes _R$ to its right by $R$-bilinearity of the tensor product.
And this illustrates why artificial expressions like $X\otimes _R Y-X\otimes _R Y+X^2\otimes _R Y-X\otimes _R XY $ won't actually enlarge $\ker(I\otimes_AJ\to IJ)$.   
A word of warning however : do not jump too far!
For example you might try to force $X$ and $Y$ to jump back and forth like this:   $$X\otimes _R Y=1\otimes _R XY=Y\otimes _R X \quad \text {(NO!)}$$ and conclude that $$X\otimes_R Y-Y\otimes _R X=0  \quad \text {(NO!)}$$ but this is false. Do you see why?