Understanding operators on a vector space from the POV of algebraic geometry

Given an operator $T$ on a finite dimensional vector space $V$,
$V$ is a $K[x]$-module, where $K$ acts as scalars and $x$ acts as $T$.
More intrinsically:
- $V$ is a module over $\End V$ (duh)
- the algebra generated by $T$ is a subalgebra $A=K[T]$
- so $V$ is an algebra over this subalgebra (formally, by restriction)

We can understand $T$ by studying $V$ as an $K[T]$-module.

The struct thm (fg mod over pid)
[V is fg b/c it's finite dimensional as $K$-module, much less $K[T]$!]
tells us everything.

We can read off even more by doing some algebraic geometry,
reading off eigenspaces and generalized eigenspaces intrinsically.
Dunno notation for generalized eigenspace; I'll use
$V_\lambda$ for eigenspace, and
$\tilde V_\lambda$ for generalized eigenspace.

Consider $V$ as a coherent sheaf over $\Spec A$
(! This is just geometric language for ``it's a module over $A$'')
Then
$V_\lambda = V_{(x-\lambda)}$

The generalized eigenspace is the completion at $\lambda$:
$\tilde V_\lambda = \hat V_{(x-\lambda)}$

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a eigenspace is the submodule (hence subspace, closed under $T$)
annihilated by $(T-\lambda)$
a generalized eigenspace is the submodule (hence subspace, closed under $T$)
annihilated by $(T-\lambda)^\infty$

...these are basically the reduced part at the point $\lambda$,
and the completion/germ at that point $\hat R_{(x-\lambda)}$

[all of this works for algebras, where $\lambda\from A \to K$
 is a $K$-algebra map, aka a closed point $\Spec K \to \Spec A$]

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Given a module over a ring, there is a Galois correspondence (I think)
given by $R \times M \to M$,
where given $X \subset M$, $\Ann X \subset R$ is its \Def{annihilator},
  [it's elements of ring that kill all elements of module]
  and is an ideal,
while given $S \subset R$, $\ker S \subset M$ is its (kernel? annihilator? name!)
  [it's elements of module that are killed by all elements of ring]
  and is a submodule.

EG, for abelian groups,
  the $2$-torsion is what's annihilated by $2$.

For Z/4 + Z/3
...it's 12-torsion (Annihilator = (12))
The 2-torsion is 2Z/4 + 0
the 4-torsion is  Z/4 + 0
the 3-torsion is  0 + Z/3
the 6-torsion is 2Z/4 + Z/3
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Recall that $\Spec$ is contravarient:
$R \to S$ means $\Spec S \to \Spec R$

Ideals in $R$ correspond to closed subschemes of $\Spec R$
$R \to R/I$: closed subset: quotient out by $I$ (quotient ring, factor ring)
  $V(I) = \Spec R/I < \Spec R$
$R \to R_I$: localize away from $I$! complement, open subset: invert $I$
$R \to R_{(I)}$: localize at $I$: invert everything outside of $I$
$R \to \widehat R/I$: the completion at $I$ (``R I hat''); $I$-adic topology
  (the germ at $I$; aka, $R/I^\infty$)

(There are some maps between them, no?
 $\widehat R/I \to R/I$ and $\to R/I^n$)

EG:
Z/2: the point 2
Z[1/2]: the complement of 2
Z_{(2)}: 
\widehat Z/2 = Z/2^\infty: the germ at 2

Given a function on $\Spec R$, we can restrict it to any of these;
that corresponds to:
$R/I$ restrict to $V(I)$
$R_I$ restrict to $D(I)$ (complement, distinguished open set)
...

Now we have a *kernel* for each of these restriction maps,
which corresponds to functions that *vanish* at...
$R/I$ vanish at $I$
  $R/I^k$ vanish at $I$ to order $k$
  $\widehat R/I$ vanish at $I$ to all orders
$R_I$ vanish away from $I$ (on an open set, so actually zero)

Note that these yield respectively:
- rings you map *to*
- ideals of functions that vanish (kernels)
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We can do the same thing for modules!

These yield respectively:
- modules you map to
- submodules
...and this finally gets us back to Annihilators and Torsion, maybe? (v.i.)
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Ok, actually, Torsion is *much subtler than I thought*.
Remember, $n$-torsion of an abelian group is a *derived functor*!
It's: $\Tor(\bZ/n\bZ,A) = \Tor_1(\bZ/n\bZ,A)$
...given by the free resolution:
$0 \to Z \to^n Z \to 0$
(or if you prefer: $0 \to Z \to^n Z \to Z/n \to 0$)

So I'm not even convinced that we get a map
$\Tor(R/I,M) \to M$
OIC:
$0 \to I \to R \to R/I \to 0$
yields
$\Tor(R,M) \to \Tor(R/I,M) \to I \otimes M \to R \otimes M \to R/I \otimes M$
and:
$0 \to \Tor(R/I,M) \to I \otimes M \to M \to M/I$
($R \otimes -$ is exact)

OIC:
the $I$-torsion of $M$ maps to $I \otimes M$
...which for a PID we can identify with $M$.
...but it seems that you can take what's annihilated by an ideal...

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Rule of thumb:
if you have a map that has a kernel and cokernel,
it just might be coming from a derived functor.
EG of Tor (of 0 -> Z -n-> Z -> Z/2 -> 0)
...and group (co)homology.
Hey, these are very similar! (both are Ext/Tor).
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OIC:
Given a commutative algebra $A$ over an algebraically closed field $K$
(e.g., $K[x_1,\dots,x_n]$),
the residue field at each point of $\Spec A$ is $K$,
and is naturally identified with $K$ via the inclusion $K \to A \to A/m$:
we have constant functions.
The inclusion of a closed point into $\Spec A$ yields a map $A \to \cO_* = K$,
hence the closed points of $\Spec A$ are exactly $A^* := \Hom_K(A,K)$.
(This is all very useful for linear algebra.)
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Remember, elements of $R$ are ``functions'' on $\Spec R$,
given by $r \mapsto \bar r \in R/I$;
properly, elements of $R$ are global sections of the sheaf of regular functions,
aka, are regular functions.
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An operator is regular
iff the spectrum (of $K[T]$) is a regular subvariety of $\Spec K[x]$,
or something:
regular variety = non-singular,
so no double points
  (diag(1,1) isn't regular)
...but we're allowed to have non-reduced points.
  (1 1)
  (0 1) is ok
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Given $R \to S$,
one gets $\Mod R \to \Mod S$
via $S \otimes_R -$
which is called ``extension of scalars''
EG, from $\bR$ to $\bC$

(This is equivalent to asking for an $R$-action on $S$, I think:
 just look at where you send $1$;
 might have problems with non-commutative?)
