Example of Gorenstein local ring of dimension 1

Question

The ring $k[[x,y]]/(xy)$ is Gorenstein. Why?

Answer 1

What is the definition of Gorenstein rings you want to use?

Using finiteness of self-injective dimension as the definition, you can do it as follows:

• Given any f.g. $R$-module $M$ over a Noetherian local ring $R$,

  $$\text{inj.dim}_R(M)\ =\ \text{sup}\{i\geq 0\ |\ \text{Ext}_R^i(k,M)\neq 0\}.$$

  In particular,

  $$\text{inj.dim}_R(R)\ =\ \text{sup}\{i\geq 0\ |\ \text{Ext}_R^i(k,R)\neq 0\}.$$

• For f.g. $R$-modules $M,N$ and any element $t\in{\mathfrak m}$ which is both $M$-regular and $R$-regular, and $tN=\{0\}$, we have

  $$\text{Ext}_R^{\ast}(N,M)\ \cong\ \text{Ext}_{R/tR}^{\ast-1}(N,M/tM).$$

Combining these two facts in case $N=k$ you have

$$\text{inj.dim}_R(M) = \text{inj.dim}_{R/tR}(M/tM)+1.$$

Using this with $R = M = {\mathbb k}[[x,y]]$ and $t := xy$, you get

$$\text{inj.dim}_{{\mathbb k}[[x,y]]/(xy)}({\mathbb k}[[x,y]]/(xy)) = \text{inj.dim}_{{\mathbb k}[[x,y]]}({\mathbb k}[[x,y]]) -1 \\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\ = \text{sup}\{i\geq 0\ |\ \text{Ext}^i_{{\mathbb k}[[x,y]]}(k,{\mathbb k}[[x,y]])\neq 0\} - 1.$$

and for the polynomial ring you can use the Koszul resolution of the ${\mathbb k} = {\mathbb k}[[x,y]]/(x,y)$ to check that $$\text{sup}\{i\geq 0\ |\ \text{Ext}^i_{{\mathbb k}[[x,y]]}(k,{\mathbb k}[[x,y]])\neq 0\}=2.$$

Note that we didn't use anything particular about the example ${\mathbb k}[[x,y]]/(xy)$ here: the arguments show in general that any power series ring is Gorenstein (even regular) and that the quotient of a Gorenstein ring by a regular sequence is again Gorenstein.