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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.

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