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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.

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I changed \text{ker} to \ker. This not only prevents italicization but also provides proper spacing before and after $\ker$ in expressions like $a\ker b$. –  Michael Hardy Mar 15 '13 at 18:18
    
I wasn't aware of the \ker command. I will use it from now on. Thank you. –  A.P. Mar 15 '13 at 18:27

2 Answers 2

up vote 4 down vote accepted

We shall prove that $$\ker(I\otimes_RI\to I^2)\simeq R/I.$$ In order to do this note first that $$I\cdot\ker(I\otimes_RI\to I^2)=0,$$ so $\ker(I\otimes_RI\to I^2)$ is an $R/I=k$-vectorspace.

Now we prove that this vector space has dimension $1$ by showing that $\ker(I\otimes_RI\to I^2)$ is generated by $X\otimes Y-Y\otimes X$. Let's consider an element of $\ker(I\otimes_RI\to I^2)$. One can write it as follows: $$a(X\otimes X)+b(X\otimes Y) +c(Y\otimes X)+d(Y\otimes Y)$$ with $aX^2+(b+c)XY+dY^2=0$. Then $X\mid d$ and $Y\mid a$, so $d=Xd_1$ and $a=Ya_1$. It follows that $a_1X+b+c+d_1Y=0$, and thus $c=-a_1X-d_1Y-b$. We get $$a(X\otimes X)+b(X\otimes Y) +c(Y\otimes X)+d(Y\otimes Y)=(b+a_1X+d_1Y)(X\otimes Y-Y\otimes X).$$

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Great calculation: bravo! –  Georges Elencwajg Mar 15 '13 at 21:05
    
Nice answer, thanks! –  A.P. Mar 15 '13 at 23:06

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?

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Yes, that's exactly what I just realised. Every manipulation I do must keep an element in $I\otimes_R I$ and $1\notin I$. –  A.P. Mar 15 '13 at 13:00
    
Congratulations A.P.! That's precisely the point. –  Georges Elencwajg Mar 15 '13 at 13:13
    
Dear @YACP, the sad truth is that I don't know a concise way to directly prove this equality for the kernel. (And, by the way, I have absolutely no desire that you delete your fine answer: I am upvoting it right now!) –  Georges Elencwajg Mar 15 '13 at 19:05

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