(This writeup needs *work!*)
The structure theorem for finitely generated modules over a principal ideal domain
This is useful for abelian groups, but also for understanding Jordan
form algebraically (at least Hoffman-Kunze think so)
$\bZ, K[x]$ both have homological dimension 1 (they're PIDs and not
fields);
transforms $V^m \to W^n$ of vector spaces all are conjugate to
``1s on diagonal, then 0s, and 0s elsewhere''
(some call it the ``Fundamental theorem of linear algebra'')
transforms $M^m \to N^n$ of modules over a PID all are conjugate to
``$a_1 \divides a_2 \divides \dots \divides a_k$ on diagonal,
then 0s, and 0s elsewhere''
This form is called ``Smith Normal Form'',
and the entries are called the invariant factors of the transform.
[dumb proof: entries of matrix generate an ideal,
which has a generator (unique up to units);
put that generator in the upper left, kill top row and left column,
and repeat, noting that this generator divides all entries of the
submatrix; (H-K, p.258)]
there's a more natural proof in terms of finding good lattices
(given by Eric Somers in 250a, say.)
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See (my) discussion on wikipedia;
there are two natural decompositions,
the invariant factor decomposition
and the primary decomposition.
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The structure theorem is *weird* from an algebraic geometry perspective,
because it fundementally involves *torsion*;
if you want to understand it algebro-geometrically,
you have to look at Tor = Tor_1 !
Roughly,
the primary decomposition slices vertically,
while the invariant factor decomposition slices horizontally.
You can somewhat recover the primary decomposition by
completing the module at each prime ideal.
(For abelian groups, this is exactly tensoring with the p-adics.)
This works for torsion modules, but not if you have free components:
the primary factors are naturally the $p^\infty$-torsion
[they are naturally a submodule]
(union of all the $p^n$ torsion factors),
and that's *weird* to visualize.
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When understanding linear operators from this POV,
keep the abelian group version in mind -- it often helps.
BTW, primary decomposition is the generalized eigenspace one.
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application to linear operators:
an operator on a space is a K[x]-module
it is *not* necessarily cyclic!
it really is just "different components"
and "Chinese remainder theorem to combine 'em"
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if looking at everything from K[x]-module POV,
it's related to points on a line:
the spectrum of the operator is an (unreduced) subvariety of C
...and then things get very, very interesting for 2 operators:
if commuting, get a K[x,y]-module
(that's finite-dimensional as a K-module (dimension d),
hence a zero-dimensional subvariety of C^2,
(which are the simultaneous eigenvalues)
and x,y satisfy lots of relations)
oh!!!!!!!!!!!
this is what's behind the "operators satisfy polynomial"
if they're non-commuting, you get into non-commutative algebraic
geometry, which is totally insane
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BTW, write everything in terms of
"simple/semisimple/irreducible/completely reducible"
simple = irreducible
semisimple = completely reducible = sum of simple
[contrast "simple" with "complex" = 1 part, many parts]
vs.
"indecomposable"
Z/p is simple
Z and Z/q = Z/p^r are indecomposable
have a *composition series*,
whose terms are simple modules
NB: if have composition terms Z/p, Z/q (for p != q primes),
then actually are Z/p + Z/q (sorta chinese remainder);
it's just *primary* ness that's a problem
Z/p^2
and
K[x]/(x-a)^2
(=
[a 1]
[ a]
)
has an invariant subspace/submodule,
but is not decomposable
i.e., they are not irreducible but are indecomposable
Urk: what is the Jacobson radical of Z and of K[x]?
(since these are not semisimple rings,
have non-trivial Jacobson radicals)
[completely reducible iff every submodule has a complement
"the complementary property"]