TITLE: Vanishing of Tor on $\mathbb{C}[x,y]$
QUESTION [5 upvotes]: Let $I\unlhd \mathbb{C}[x,y]$. Then is it possible that 
$$Tor_1^{\mathbb{C}[x,y]}\left(\frac{\mathbb{C}[x,y]}{(x)},\frac{\mathbb{C}[x,y]}{I}\right)=0$$
for all $I$? I'm really not too sure what to do here, I thought I'd just see if anyone had any intuition?

REPLY [3 votes]: There is some statement that I don't remember about only needing to check against the residue field. So (instead of hunting through Eisenbud) let's do that.
First, there is a free resolution of $C[x,y]  / (x,y)$:

Start with $C[x,y]^{\oplus 2} \to (x,y)$. Here the map is the direct sum of $m_x : \mathbb{C}[x,y] \to \mathbb{C} [x,y]$ (multiplication by $x$) and $m_y$. 
The kernel is the submodule (diagonal action of the coordinate ring) generated by $y \oplus (-x)$. This submodule is free, so this gives us a free resolution.

Next we tensor with $C[x,y]/(x)$. 
The result of thisis to get a chain complex $C[y] \to C[y]^2 \to C[y]$. The only place there can be homology is in the middle degree (you have to compute the maps to see this). When I computed the homology there, I got $C[y] / (y) = C$.

The conclusion seems to be that $Tor_1^{\mathbb{C}[x,y]}(C[x,y]/(x,y), C[x,y]/(y)) = \mathbb{C}$.

I hope I did the computation right. Maybe you can double check me.