Ext and Tor Functors — Mathematics Knowledge System

Overview

Ext and Tor are the derived functors of Hom and tensor product respectively. They measure the failure of these functors to be exact. Ext classifies module extensions and encodes obstruction theory; Tor detects torsion and measures how far a module is from being flat. Together they are the main computational tools of homological algebra.

Intuition

Ext$^1(M,N)$ classifies all short exact sequences $0 \to N \to E \to M \to 0$ up to equivalence. If Ext$^1 = 0$, every such sequence splits. Tor$_1(M,N) = 0$ means $M$ is flat — it tensors without introducing new relations. Higher Ext and Tor measure deeper obstruction.

Formal Definition

Definition

Given an $R$-module $M$, take a projective resolution $P_\bullet \to M$. Applying $\text{Hom}_R(-,N)$ and taking cohomology gives $\text{Ext}^n_R(M,N)$. Applying $-\otimes_R N$ and taking homology gives $\text{Tor}_n^R(M,N)$. Both are independent of the choice of resolution.

\text{Ext}^n_R(M,N) = H^n(\text{Hom}_R(P_\bullet, N))

Ext via projective resolution

\text{Tor}_n^R(M,N) = H_n(P_\bullet \otimes_R N)

Tor via projective resolution

0 \to \text{Hom}(C,N) \to \text{Hom}(B,N) \to \text{Hom}(A,N) \to \text{Ext}^1(C,N) \to \cdots

Long exact sequence for Ext

\cdots \to \text{Tor}_1(C,N) \to A \otimes N \to B \otimes N \to C \otimes N \to 0

Long exact sequence for Tor

Worked Examples

Use the projective resolution $0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n \to 0$.
$0 \to \mathbb{Z} \xrightarrow{\cdot n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$
Tensor with $\mathbb{Z}/m$: $\mathbb{Z} \otimes \mathbb{Z}/m \cong \mathbb{Z}/m$, and the map becomes multiplication by $n$ on $\mathbb{Z}/m$.
$0 \to \mathbb{Z}/m \xrightarrow{\cdot n} \mathbb{Z}/m$
$\text{Tor}_1 = \ker(\cdot n: \mathbb{Z}/m \to \mathbb{Z}/m) = \{a \in \mathbb{Z}/m : na \equiv 0 \pmod m\}$.
$\ker(\cdot n) \cong \mathbb{Z}/\gcd(m,n)$
Therefore:
$\text{Tor}_1^{\mathbb{Z}}(\mathbb{Z}/n,\, \mathbb{Z}/m) \cong \mathbb{Z}/\gcd(m,n)$

Answer: $\text{Tor}_1^{\mathbb{Z}}(\mathbb{Z}/n, \mathbb{Z}/m) \cong \mathbb{Z}/\gcd(n,m)$.

Practice Problems

Difficulty 7/10

What does $\text{Ext}^n_R(M, N) = 0$ for all $n \geq 1$ say about $M$?

Difficulty 8/10

Prove that $\text{Tor}_n^R(M,N) \cong \text{Tor}_n^R(N,M)$ (commutativity of Tor).

Difficulty 9/10

State the universal coefficient theorem in topology and identify the role of Tor and Ext.

Common Mistakes

Common Mistake

$\text{Ext}^n(M,N)$ depends on the choice of projective resolution.

Any two projective resolutions of $M$ are chain homotopy equivalent, so they give the same Ext groups. Ext is well-defined and functorial.

Common Mistake

$\text{Tor}_1(M,N) = 0$ means $M$ is projective.

$\text{Tor}_1(M,N) = 0$ for all $N$ means $M$ is flat, which is weaker than projective. Over a PID, flat implies projective implies free, but over general rings, flat $\subsetneq$ projective $\subsetneq$ free.

Quiz

Summary

Ext$^n_R(M,N)$ and Tor$_n^R(M,N)$ are the derived functors of Hom and tensor product.
They are computed using projective resolutions and measure the failure of these functors to be exact.
Ext$^1$ classifies extensions; Tor$_1 = 0$ characterizes flat modules.
Long exact sequences in Ext and Tor arise from short exact sequences of modules.
The universal coefficient theorem in algebraic topology expresses homology with coefficients via Tor and cohomology via Ext.

References

WebsiteWikipedia — Ext functor
WebsiteWikipedia — Tor functor