ConceptComplete

Ext and Tor - Key Properties

Balance and long exact sequences are fundamental properties that make Ext and Tor computationally tractable.

Theorem5.5Balance of Ext

Ext can be computed using either a projective resolution of the first variable or an injective resolution of the second: ExtRn(M,N)Hn(HomR(P,N))Hn(HomR(M,I))\text{Ext}^n_R(M, N) \cong H^n(\text{Hom}_R(P_\bullet, N)) \cong H^n(\text{Hom}_R(M, I^\bullet))

where PMP_\bullet \to M is projective and 0NI0 \to N \to I^\bullet is injective.

Theorem5.6Balance of Tor

Tor is symmetric in its arguments: TornR(M,N)TornR(N,M)\text{Tor}_n^R(M, N) \cong \text{Tor}_n^R(N, M)

This follows from the symmetry of tensor product and the fact that we can resolve either variable.

Theorem5.7Long Exact Sequence for Ext (First Variable)

Given a short exact sequence 0ABC00 \to A \to B \to C \to 0, there is a long exact sequence: 0Hom(C,N)Hom(B,N)Hom(A,N)Ext1(C,N)Ext1(B,N)0 \to \text{Hom}(C, N) \to \text{Hom}(B, N) \to \text{Hom}(A, N) \to \text{Ext}^1(C, N) \to \text{Ext}^1(B, N) \to \cdots

Theorem5.8Long Exact Sequence for Ext (Second Variable)

Given a short exact sequence 0ABC00 \to A \to B \to C \to 0, there is a long exact sequence: 0Hom(M,A)Hom(M,B)Hom(M,C)Ext1(M,A)Ext1(M,B)0 \to \text{Hom}(M, A) \to \text{Hom}(M, B) \to \text{Hom}(M, C) \to \text{Ext}^1(M, A) \to \text{Ext}^1(M, B) \to \cdots

Theorem5.9Long Exact Sequence for Tor

Given a short exact sequence 0ABC00 \to A \to B \to C \to 0 where A,B,CA, B, C are flat or the sequence is split, there is a long exact sequence: Tor1(M,B)Tor1(M,C)MAMBMC0\cdots \to \text{Tor}_1(M, B) \to \text{Tor}_1(M, C) \to M \otimes A \to M \otimes B \to M \otimes C \to 0

ExampleComputing Tor Explicitly

For M=Z/2ZM = \mathbb{Z}/2\mathbb{Z} and N=Z/3ZN = \mathbb{Z}/3\mathbb{Z}, use the resolution: 0Z×2ZZ/2Z00 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0

Tensoring with Z/3Z\mathbb{Z}/3\mathbb{Z} gives: 0Z/3Z×2Z/3ZZ/2ZZ/3Z00 \to \mathbb{Z}/3\mathbb{Z} \xrightarrow{\times 2} \mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/3\mathbb{Z} \to 0

Since gcd(2,3)=1\gcd(2,3) = 1, multiplication by 2 is invertible on Z/3Z\mathbb{Z}/3\mathbb{Z}, so: Tor1Z(Z/2Z,Z/3Z)=0\text{Tor}_1^{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/3\mathbb{Z}) = 0

Definition5.10Ext and Yoneda Product

The Yoneda product gives Ext a ring structure: ExtRm(M,N)×ExtRn(N,P)ExtRm+n(M,P)\text{Ext}^m_R(M, N) \times \text{Ext}^n_R(N, P) \to \text{Ext}^{m+n}_R(M, P)

This makes nExtRn(M,M)\bigoplus_n \text{Ext}^n_R(M, M) into a graded ring.