Knowlpedia

Ext and Tor as derived functors

Ext and Tor are the right/left derived functors of Hom and tensor, computed via injective/projective (or flat) resolutions.

Let RR be a commutative ring and M,NM,N be RR-modules. The functors

  • N↦Hom⁡R(M,N)N \mapsto \operatorname{Hom}_R(M,N) and M↦Hom⁡R(M,N)M\mapsto\operatorname{Hom}_R(M,N) are in their covariant and contravariant senses, respectively;
  • M↦M⊗RNM \mapsto M\otimes_R N is .

Their failure to be exact is measured by derived functors, giving rise to and .

Ext via resolutions

Fix MM. Choose an 0→N→I0→I1→⋯0\to N\to I^0\to I^1\to \cdots. Apply Hom⁡R(M,−)\operatorname{Hom}_R(M,-) degreewise to get a cochain complex Hom⁡R(M,I∙)\operatorname{Hom}_R(M,I^\bullet). Define

Ext⁡Rn(M,N):=Hn(Hom⁡R(M,I∙)).\operatorname{Ext}^n_R(M,N) := H^n(\operatorname{Hom}_R(M,I^\bullet)).

Equivalently (often more computationally), choose a ⋯→P1→P0→M→0\cdots \to P_1\to P_0\to M\to 0 and set

Ext⁡Rn(M,N):=Hn(Hom⁡R(P∙,N)).\operatorname{Ext}^n_R(M,N) := H^n(\operatorname{Hom}_R(P_\bullet,N)).

These agree and are well-defined up to canonical isomorphism because projective/injective resolutions exist (see and ).

Tor via resolutions

Fix NN. Choose a projective (or flat) resolution ⋯→P1→P0→M→0\cdots \to P_1\to P_0\to M\to 0. Apply −⊗RN-\otimes_R N to get a chain complex P∙⊗RNP_\bullet\otimes_R N. Define

Tor⁡nR(M,N):=Hn(P∙⊗RN).\operatorname{Tor}^R_n(M,N) := H_n(P_\bullet\otimes_R N).
Functorial consequences
Examples

Example 1 (Tor over Z\mathbb Z: Tor⁡1Z(Z/n,Z/m)\operatorname{Tor}_1^\mathbb Z(\mathbb Z/n,\mathbb Z/m))

Use the standard projective resolution

0→Z→⋅nZ→Z/n→0.0\to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \to \mathbb Z/n\to 0.

Tensor with Z/m\mathbb Z/m to get

0→Z/m→⋅nZ/m→(Z/n)⊗(Z/m)→0.0\to \mathbb Z/m \xrightarrow{\cdot n} \mathbb Z/m \to (\mathbb Z/n)\otimes(\mathbb Z/m)\to 0.

Then

Tor⁡1Z(Z/n,Z/m)=ker⁡(⋅n:Z/m→Z/m)≅Z/gcd⁡(n,m),\operatorname{Tor}_1^\mathbb Z(\mathbb Z/n,\mathbb Z/m)=\ker(\cdot n:\mathbb Z/m\to \mathbb Z/m)\cong \mathbb Z/\gcd(n,m),

and Tor⁡i=0\operatorname{Tor}_i=0 for i≥2i\ge 2 because the resolution has length 11.

Example 2 (Ext over Z\mathbb Z: Ext⁡Z1(Z/n,Z/m)\operatorname{Ext}^1_\mathbb Z(\mathbb Z/n,\mathbb Z/m))

Apply Hom⁡Z(−,Z/m)\operatorname{Hom}_\mathbb Z(-,\mathbb Z/m) to the same resolution:

0→Z→⋅nZ→Z/n→00\to \mathbb Z \xrightarrow{\cdot n} \mathbb Z \to \mathbb Z/n\to 0

to obtain

0→Hom⁡(Z,Z/m)→⋅nHom⁡(Z,Z/m)→Ext⁡1(Z/n,Z/m)→0.0\to \operatorname{Hom}(\mathbb Z,\mathbb Z/m)\xrightarrow{\cdot n}\operatorname{Hom}(\mathbb Z,\mathbb Z/m)\to \operatorname{Ext}^1(\mathbb Z/n,\mathbb Z/m)\to 0.

Since Hom⁡(Z,Z/m)≅Z/m\operatorname{Hom}(\mathbb Z,\mathbb Z/m)\cong \mathbb Z/m, we get

Ext⁡Z1(Z/n,Z/m)≅(Z/m)/n(Z/m)≅Z/gcd⁡(n,m),\operatorname{Ext}^1_\mathbb Z(\mathbb Z/n,\mathbb Z/m)\cong (\mathbb Z/m)/n(\mathbb Z/m)\cong \mathbb Z/\gcd(n,m),

and Ext⁡i=0\operatorname{Ext}^i=0 for i≥2i\ge 2.

Example 3 (a quick vanishing test: projective/flat inputs)

If MM is , then Ext⁡Rn(M,N)=0\operatorname{Ext}^n_R(M,N)=0 for all n≥1n\ge 1 and all NN. If NN is , then Tor⁡nR(M,N)=0\operatorname{Tor}^R_n(M,N)=0 for all n≥1n\ge 1 and all MM. Both statements follow because you can take a resolution of length 00 (exactness of the relevant functor), reflecting the general principle that derived functors measure failure of exactness (see ).