Let \(R = \ZZ[x,y]\), and \(M = R/(x-y), N = R/(x,y)\). Construct free resolutions of \(M,N\) to compute:
\(\ext_R^*(M, M)\)
\(\ext_R^*(M, N)\)
\(\ext_R^*(N, M)\)
\(\ext_R^*(N, N)\)
Solution
Let \(R=\mathbb Z[x,y]\), \(M=R/(x-y)\), and \(N=R/(x,y)\).
1. A free resolution of \(M\) is\[ 0\to R\xrightarrow{x-y}R\to M\to0. \]
The element \(x-y\) is a non-zero-divisor in the domain \(R\), and its cokernel is \(M\).
2. A free resolution of \(N\) is the Koszul resolution\[ 0\to R\xrightarrow{\binom{-y}{x}}R^2\xrightarrow{(x\;y)}R\to N\to0. \]
The sequence \(x,y\) is regular in \(R\), so its Koszul complex is exact and resolves \(R/(x,y)\).
3. We have\[ \boxed{\operatorname{Ext}_R^0(M,M)\cong M,\quad \operatorname{Ext}_R^1(M,M)\cong M,\quad \operatorname{Ext}_R^i(M,M)=0\ (i\ge2).} \]
Applying \(\operatorname{Hom}_R(-,M)\) to 1 gives \[ 0\to M\xrightarrow{0}M\to0, \] because \(x-y=0\) in \(M\). Hence the degree-\(0\) and degree-\(1\) cohomology groups are both \(M\), and there are no higher terms.
4. We have\[ \boxed{\operatorname{Ext}_R^0(M,N)\cong N,\quad \operatorname{Ext}_R^1(M,N)\cong N,\quad \operatorname{Ext}_R^i(M,N)=0\ (i\ge2).} \]
1. Since \(x=y=0\) in \(N\), multiplication by \(x-y\) is zero, so the cochain complex is \(0\to N\xrightarrow0 N\to0\).
5. We have\[ \boxed{\operatorname{Ext}_R^0(N,M)=0,\quad \operatorname{Ext}_R^1(N,M)\cong\mathbb Z,\quad \operatorname{Ext}_R^2(N,M)\cong\mathbb Z.} \]
Identify \(M\cong\mathbb Z[t]\) via \(x=y=t\). Applying \(\operatorname{Hom}_R(-,M)\) to the Koszul resolution gives \[ 0\to M\xrightarrow{m\mapsto(tm,tm)}M^2\xrightarrow{(a,b)\mapsto t(a-b)}M\to0. \] The first map is injective, so \(H^0=0\). The kernel of the second map is the diagonal \(\{(a,a)\}\), and modulo the image \(tM\cdot(1,1)\) this gives \(M/tM\cong\mathbb Z\), hence \(H^1\cong\mathbb Z\). The second map has image \(tM\), so its cokernel is also \(M/tM\cong\mathbb Z\), giving \(H^2\cong\mathbb Z\).
6. Finally,\[ \boxed{\operatorname{Ext}_R^0(N,N)\cong N,\quad \operatorname{Ext}_R^1(N,N)\cong N^2,\quad \operatorname{Ext}_R^2(N,N)\cong N,} \]and higher Ext groups vanish.
After applying \(\operatorname{Hom}_R(-,N)\) to the Koszul resolution, both differentials are zero because \(x\) and \(y\) act trivially on \(N\). Hence the cohomology is simply \(N\), \(N^2\), \(N\) in degrees \(0,1,2\).