## Atiyah-Macdonald: Modules and Exact Sequences June 23, 2010

Posted by Sarah in Uncategorized.
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A module is a generalization of the notion of ideals. Rings, ideals, quotient rings, and vector spaces are all modules. If A is a ring, then an A-module is an abelian group M on which A acts linearly:
$a(x+y) = ax + ay$
$(a+ b) x = a x + bx$
$(ab) x = a(bx)$
$1x = x$
Submodules are subgroups of M closed under multiplication by elements of A; the sum and intersection of submodules is again a submodule.

The annihilator of M is the set of all a in A such that aM = 0. A module is called faithful if its annihilator is zero.

Direct sum of two modules is just defined as an ordered pair (x, y), with componentwise addition and scalar multiplication. An A-module is called free if it’s isomorphic to the direct sum of modules isomorphic to A. (A finitely generated A-module is isomorphic to a quotient of $A^n$ for some n.)

Proposition.
Let M be a finitely generated A-module. Let a be an ideal of A and let $\phi$ be an A-module automorphism of M st $\phi(M) \subset aM$. Then $\phi$ satisfies an equation of the form

$\phi^n + a_1 \phi^{n+1} \dots + a_n = 0$.

Proof.
Let $x_1 \dots x_n$ be a set of generators of M. Then $\phi(x_i) = \sum a_{ij} x_j$ for some $a_{ij}$. That is,
$\sum (\delta_{ij} \phi - a_{ij}) x_j = 0$
Left-multiplying by the adjoint of the matrix $(\delta_{ij} \phi - a_{ij}$, we see that the determinant annihilates each $x_i$, hence is the zero endomorphism of M. Expanding out the determinant gives us an equation of the desired form.

Exact sequences. Suppose we have a sequence of homomorphisms between modules,
$M_{i-1} \to M_i \to M_{i+1} \dots$
This is exact at $m_i$ if $Im(f_i) = Ker(f_{i+1})$. The sequence is exact if it is exact at every module.
$0 \to M' \to M$ is exact iff f is injective;
$M \to M^* \to 0$ is exact iff g is surjective.
A short exact sequence:
$0 \to M' \to M \to M^* \to 0$.
Every long exact sequence can be split into these.
If the above is an exact sequence, then
$0 \to Hom(M^*, N) \to Hom(M, N) \to Hom(M', N)$ is also exact; note that the order reverses when we go to the dual space.

Next time I’ll introduce the boundary map and the Snake Lemma.