Finding $\text{Ext}_R^n(M,R)$ and $\text{Ext}_R^n(M,N)$ for a particular case Suppose we have the ring $R=\mathbb{Z}[x_1,x_2]$ and ideals $I=(2x_2)$, $J=(x_1x_2)$.
Now, consider $M:=R/I$ and $N:=R/J$.
I am trying to determine the following:
(1) $\text{Ext}_R^n(M,R)$ for all $n\ge 0$.
(2) $\text{Ext}_R^n(M,N)$ for all $n\ge 0$.
So, I tried the following for (1):
Firstly, I tried to find a projective resolution of $M$ and my guess is  $0\rightarrow R\xrightarrow{\cdot 2x_2}R\xrightarrow{\pi} M\rightarrow 0$, but I am not sure if I am in the right direction.
Any help is highly appreciated and many thanks for that.
 A: *

*Like you said, $0\to R \xrightarrow{\cdot 2x_2} R \to M\to 0$ is a free resolution $M$, therefore $\mathrm{Ext}^k(M,R)$ is the $k$-th homology group of the sequence $0 \to\mathrm{Hom}(R,R) \xrightarrow{\cdot 2x_2} \mathrm{Hom}(R,R)\to 0$. Hence :

$\mathrm{Ext}^0(M,R)=\frac{\ker(\cdot 2x_2:\mathrm{Hom}(R,R)\to \mathrm{Hom}(R,R))}{\mathrm{im}(0: 0\to \mathrm{Hom}(R,R))}=\frac{\ker(\cdot 2x_2 : R\to R)}{(0)}=0$
$\mathrm{Ext}^1(M,R)=\frac{\ker(\mathrm{Hom}(R,R)\to 0)}{\mathrm{im}(\cdot 2x_2: \mathrm{Hom}(R,R)\to \mathrm{Hom}(R,R))}=\frac{\ker( 0: R\to 0)}{\mathrm{im}(\cdot 2x_2: R\to R)}=R/(2x_2)=M$
and $\mathrm{Ext}^k(M,R) =0$ for $k\ge 2$.


*Similarly, we can use the sequence $0 \to \mathrm{Hom}(R,N) \xrightarrow{\cdot 2x_2} \mathrm{Hom}(R,N)\to 0$ to compute $\mathrm{Ext}^k(M,N)$. We find :

$\mathrm{Ext}^0(M,N)=\frac{\ker(\cdot 2x_2:\mathrm{Hom}(R,N)\to\mathrm{Hom}(R,N))}{\mathrm{im}(0: 0\to\mathrm{Hom}(R,R))}=\ker(\cdot 2x_2 : N\to N) = \mathbb{Z}[x_1]$ (seen as the $R$-module $\mathbb{Z}[x_1,x_2]/(x_2)$)
$\mathrm{Ext}^1(M,N)=\frac{\ker(\mathrm{Hom}(R,N)\to 0)}{\mathrm{im}(\cdot 2x_2: \mathrm{Hom}(R,N)\to\mathrm{Hom}(R,N))}=\frac{\ker( 0: N\to 0)}{\mathrm{im}(\cdot 2x_2: N\to N)}=N/(2x_2 N)$
and $\mathrm{Ext}^k(M,N)=0$ for $k\ge 2$.
