What's the Tor for k[x,y]?

Question

Let $R=k[x,y]$ where $k$ is a field, $I=(x,y)$. Let $\alpha(r)=(yr,-xr), \beta(f,g)=fx+gy$. We know

$0\rightarrow R\xrightarrow{\alpha} R^2\xrightarrow{\beta} R\rightarrow k\rightarrow 0$

is a free resolution of $k$ as an $R$ module, where the map after $\beta$ is the canonical quotient map $R\rightarrow R/I=k$.

(a) Show that $Tor_2^R(k,k)=k$

(b) show that $Tor_1^R(I,k)=k$

For (a) I am stuck after applying tensor with $k$: $0\rightarrow k\otimes R\xrightarrow{1\otimes \alpha} k\otimes R^2\xrightarrow{1\otimes \beta} k\otimes R\rightarrow k\otimes k\rightarrow 0$

For (b) I am in a similar place. I guess I lack some basic machinery to solve for kernels & images of tensored homomorphisms. Any suggestion?

Answer 1

1. Since $0\to R \to R^2 \to R \to k$ is a free (hence projective) resolution of $k$, $Tor^i(k,k)$ is the $i$-th homology of the sequence $$0\to R\otimes k\to R^2\otimes k \to R\otimes k \to 0$$ That is :

$$Tor^0(k,k)= \frac{ker ( R\otimes k \to 0)}{Im ( R^2\otimes k\to R\otimes k)}= \frac{R\otimes k}{Im(\beta\otimes 1)}$$

$$Tor^1(k,k)= \frac{ker (R^2\otimes k\to R\otimes 1)}{Im ( R\otimes k \to R^2\otimes k)}= \frac{ker( \beta\otimes 1)}{ Im( \alpha\otimes 1)} $$

$$Tor^2(k,k)= \frac{ker (R\otimes k \to R^2\otimes k)}{Im (0\to R\otimes k)}= \frac{\ker( \alpha\otimes 1)}{Im (0)}=\ker( \alpha\otimes 1)$$

Now, what are $R\otimes k$, $R^2\otimes k$, $\alpha\otimes 1$ and $\beta\otimes 1$?

• For $P\in R$, observe that $P\otimes_R 1= 1\otimes_R \overline{P}$ where $\overline{P}$ is the image of $P$ in $k$. This is simply $P(0,0)$. Thus, $R\otimes k$ is just $k$ (both as a $R$ -module and a $k$-vector space).

• Similarly, for $(P,Q)\in R^2$, we have $(P,Q)\otimes_R 1 = 1\otimes_R(P(0,0),Q(0,0))$. So $R^2\otimes k \cong k^2$

• As for $\alpha\otimes 1$, it sends $P\otimes 1$ to $$(YP,-XP)\otimes_R 1= 1\otimes_R (0\times P(0,0),-0\times P(0,0))=(0,0)$$

So $\alpha\otimes 1$ is just the null application !

• Similarly, $\beta\otimes 1$ sends $(P,Q)\otimes_R 1$ to $(PX+QY)\otimes_R 1=1\otimes_R \overline{PX+QY}=0$

because $\overline{PX+QY} = P(0,0)\times 0+ Q(0,0)\times 0=0$.

Putting this all together :

$$Tor^0(k,k)= \frac{R\otimes k}{Im(0: k^2\to k)} \cong k$$

$$Tor^1(k,k) \cong \frac{ker(0: k^2 \to k)}{ Im(0: k\to k^2)} =k^2$$

$$Tor^2(k,k)\cong ker( 0: k\to k^2)=k$$ 2. The short exact sequence $$0\to I\to R \to k\to 0$$ gives rise to the long exact sequence $$...\to Tor_2(I,k)\to Tor_2(R,k)\to Tor_2(k,k)\to Tor_1(I,k)\to Tor_1(R,k)\to Tor_1(k,k)\to Tor_0(I,k)\to Tor_0(R,k)\to Tor_0(k,k)\to 0$$

Because $R$ is free, $Tor^i(R,k)=0$ for $i\ge 1$ :

$$ 0 \to Tor_2(k,k)\to Tor_1(I,k)\to 0\to Tor_1(k,k)\to I\otimes k\to R\otimes k\to k\otimes k\to 0$$

Therefore $$Tor_1(I,k) \cong Tor_2(k,k)=k$$