Commutative algebra

Yoshifumi Tsuchimoto

Let $\mathcal{C}$ be an abelian category. For any object $M$ of $\mathcal{C}$ , the extension group $\operatorname{Ext}^j_\mathcal{C}(M,N)$ is defined to be the derived functor of the ``hom'' functor

$$N\mapsto \operatorname{Hom}_\mathcal{C}(M,N).$$

We note that the $\operatorname{Hom}$ functor is a ``bifunctor''. We may thus consider the right derived functor of $\bullet \mapsto \operatorname{Hom}(\bullet,N)$ and that of $\bullet \mapsto \operatorname{Hom}(M,\bullet,N)$ . Fortunately, both coincide: The extension group $\operatorname{Ext}^\bullet_\mathcal{C}(M,N)$ may be calculated by using either an injective resolution of the second variable $N$ or a projective resoltuion of the first variable $M$ .

EXAMPLE 09.1   Let us compute the extension groups $\operatorname{Ext}^j_\mathbb{Z}(\mathbb{Z}/36\mathbb{Z}, \mathbb{Z}/108\mathbb{Z})$ .

1. We may compute them by using an injective resolution
   $$0 \to \mathbb{Z}/108\mathbb{Z}\to \mathbb{Q}/108\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}\to 0$$
   of $\mathbb{Z}/108\mathbb{Z}$ .
2. We may compute them by using a free resolution
   $$0 \leftarrow \mathbb{Z}/36\mathbb{Z}\leftarrow \mathbb{Z}\leftarrow 36 \mathbb{Z}\leftarrow 0$$
   of $\mathbb{Z}/36 \mathbb{Z}$ .

EXERCISE 09.1   Compute an extension group $\operatorname{Ext}^j(M,N)$ for modules $M,N$ of your choice. (Please choose a non-trivial example).

DEFINITION 09.2   Let $A$ be an associative unital (but not necessarily commutative) ring. Let $L$ be a right $A$ -module. Let $M$ be a left $A$ -module. For any ( $\mathbb{Z}$ -)module $N$ , an map

$$\varphi: L\times M \to N$$

is called an $A$ -balanced biadditive map if

1. $\varphi(x_1+x_2,y)=\varphi(x_1,y)+\varphi(x_2,y)$      $(\forall x_1,\forall x_2\in L, \forall y\in M)$ .
2. $\varphi(x,y_1+y_2)=\varphi(x,y_1)+\varphi(x,y_2)$      $(\forall x\in L, \forall y_1,\forall y_2\in M)$ .
3. $\varphi(x a, y)=\varphi(x, a y)$      $(\forall x\in L, \forall y\in M, \forall a\in A)$ .

PROPOSITION 09.3   Let $A$ be an associative unital (but not necessarily commutative) ring. Then for any right $A$ -module $L$ and for any left $A$ -module $M$ , there exists a ( $\mathbb{Z}$ -)module $X_{L,M}$ together with a $A$ -balanced map

$$\varphi_0: L\times M \to X_{L,M}$$

which is universal amoung $A$ -balanced maps.

DEFINITION 09.4   We employ the assumption of the proposition above. By a standard argument on universal objects, we see that such object is unique up to a unique isomorphism. We call it the tensor product of $L$ and $M$ and denote it by

$$L\otimes_A M.$$

LEMMA 09.5   Let $A$ be an associative unital ring. Then:

1. $A\otimes_A M \cong M$ .
2. $(L_1\oplus L_2) \otimes_A M \cong (L_1 \otimes M) \oplus (L_2 \otimes_A M ).$
3. For any $M$ , $L\mapsto L \otimes_A M$ is a right exact functor.
4. For any right ideal $J$ of $A$ and for any $A$ -module $M$ , we have
   $$(A/J) \otimes_A M \cong M/ J.M$$

In particular, if the ring $A$ is commutative, then for any ideals $I,J$ of $A$ , we have

$$(A/I) \otimes_A (A/J) \cong A/ (I+ J)$$

DEFINITION 09.6   For any left $A$ -module $M$ , the left derived functor $L_j F(M)$ of $F_M=\bullet \otimes_A M$ is called the Tor functor and denoted by $\operatorname{Tor}^A_j(\bullet,M)$ .

By definition, $\operatorname{Tor}^A_j(L,M)$ may be computed by using projective resolutions of $L$ .

EXERCISE 09.2   Compute $\operatorname{Tor}_j^\mathbb{Z}(\mathbb{Z}/n \mathbb{Z}, \mathbb{Z}/m\mathbb{Z})$ for $n,m\in \mathbb{Z}_{>0}$ .