what is the projective dimension of $ (x,y)\mathbb{C}[x,y]_{(x,y)}$?

Question

For the local ring $R = \mathbb{C}[x,y]_{(x,y)}$ and its maximal ideal $M = (x,y)\mathbb{C}[x,y]_{(x,y)}$.

What is the projective dimension $\operatorname{pd}_R(M)$ of M?

My thought:

I tried to construct a minimal free resolution of $M$, $\cdots \rightarrow L_2 \rightarrow L_1 \rightarrow L_0 \rightarrow M \rightarrow 0 $.

Define $L_0 := R \oplus R$ since $ M = Rx + Ry$.

However, I couldn't prove that $L_0 \otimes_R k \cong M \otimes_R k$ , where $k = R/M$.

My questions are:

1. Is it correct to set $L_0 := R \oplus R$. If this is correct, then

2. How to prove $L_0 \otimes_R k \cong M \otimes_R k$.

Answer 1

Let $\varphi \colon L_{0} \to M$ denote the surjection sending $(r, s)$ to $rx +sy$. Note that $\varphi$ descends to a $k$-linear surjection $\varphi \otimes_{R} \mathrm{Id}_{k} \colon L_{0} \otimes_{R} k \to M \otimes_{R} k$. Since $L_{0} \otimes_{R} k$ is a $k$-vector space of dimension $2$, it follows that $\varphi \otimes_{R} \mathrm{Id}_{k}$ is an isomorphism of $k$-vector spaces if $M \otimes_{R} k$ has $k$-dimension $2$. On the other hand, $M \otimes_{R} k \cong M/M^{2}$ as $k$-modules, and it is not too hard to show that $M/M^{2}$ has $k$-basis given by the residue classes of $x, y$. This answers your question (2).

You can complete your complex by observing that the morphism $\psi \colon R \to L_{0}$ sending $r \in R$ to $(-ry, rx)$ is injective with image precisely $\ker(\varphi)$. Indeed, the containment $\mathrm{Im}(\psi) \subset \ker(\varphi)$ is obvious. On the other hand, if $rx + sy = 0$, then $rx = -sy$. Since $R$ is a UFD and $x, y$ are coprime irreducibles, one obtains $r = yu, s = xv$ for some $u, v \in R$. The equation $rx + sy = 0$ then reads $xyu + xyv = 0$, so $v = -u$.

This is a projective resolution of $M$ of minimal length. This follows because if there were a shorter resolution, then $M$ would be projective. A finitely generated projective module over a local ring is free of finite rank. But $M$ is an ideal of $R$, and so can only be free if it has rank $1$. This cannot be the case by the argument above: an $R$-linear surjection $R \to M$ would descend to a $k$-linear surjection $R \otimes_{R} k \to M/M^{2}$, which is impossible, since $M/M^{2}$ has dimension $2$ as a $k$-vector space.