Mathematics Lecture Notes

Theorem5.14Universal Coefficient Theorem for Cohomology

For a chain complex $C_\bullet$ of free $R$ -modules and any $R$ -module $G$ , there is a natural short exact sequence: $0 \to \text{Ext}^1_R(H_{n-1}(C_\bullet), G) \to H^n(\text{Hom}_R(C_\bullet, G)) \to \text{Hom}_R(H_n(C_\bullet), G) \to 0$

This sequence splits (non-naturally).

Proof

Apply the functor $\text{Hom}_R(-, G)$ to the exact sequences: $0 \to Z_n \to C_n \to B_{n-1} \to 0$ $0 \to B_n \to Z_n \to H_n \to 0$

Since $C_n$ is free (hence projective), the first sequence gives: $0 \to \text{Hom}(B_{n-1}, G) \to \text{Hom}(C_n, G) \to \text{Hom}(Z_n, G) \to \text{Ext}^1(B_{n-1}, G) \to 0$

Combining with cohomology formulas yields the result.

■

Theorem5.15Künneth Formula for Homology

For chain complexes $C_\bullet, D_\bullet$ of free $R$ -modules, there is a natural short exact sequence: $0 \to \bigoplus_{p+q=n} H_p(C_\bullet) \otimes_R H_q(D_\bullet) \to H_n(C_\bullet \otimes_R D_\bullet) \to \bigoplus_{p+q=n-1} \text{Tor}_1^R(H_p(C_\bullet), H_q(D_\bullet)) \to 0$

This splits (non-naturally).

Remark

The Künneth formula is essential for computing homology of product spaces. It shows that homology of a product is determined by homologies of the factors, up to Tor correction terms.

Theorem5.16Ext and Projective Dimension

For an $R$ -module $M$ : $\text{pd}_R(M) = \sup\{n : \text{Ext}^n_R(M, N) \neq 0 \text{ for some } N\}$

Thus projective dimension can be detected via Ext vanishing.

Theorem5.17Tor and Flat Dimension

For an $R$ -module $M$ : $\text{fd}_R(M) = \sup\{n : \text{Tor}_n^R(M, N) \neq 0 \text{ for some } N\}$

where $\text{fd}_R(M)$ is the flat dimension of $M$ .

ExampleComputing with Long Exact Sequences

To compute $\text{Ext}^2_{\mathbb{Z}}(\mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/2\mathbb{Z})$ , use: $0 \to \mathbb{Z} \xrightarrow{\times 4} \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to 0$

The long exact sequence gives: $\text{Ext}^1(\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}) \to \text{Ext}^2(\mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}) \to \text{Ext}^2(\mathbb{Z}, \mathbb{Z}/2\mathbb{Z})$

Since $\mathbb{Z}$ is projective, both outer terms vanish, so $\text{Ext}^2(\mathbb{Z}/4\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}) = 0$ .

Math Notes

Ext and Tor - Main Theorem