# Spectrum of a Ring and Spectrum of a Linear Operator

A quick post before bed, an impressionist stroke on some nice things lurking in linear algebra. I love polynomials. They are the ultimate tools that make me feel like I’m touching something, calculating at the level of a polynomial is a good clean feeling. I want to show you that it is nice to think of a vector space over $$F$$ as a $$F[x]$$-module (thanks Emmy Noether). Thanks to Semon Rezchikov for helping me get over a few bumps in grasping some of the following.

Let $$A$$ be a finite type $$\mathbb{C}$$-algebra, then $$A = \mathbb{C}[x_1, \ldots, x_n]/(f_1, \ldots , f_k)$$, then Spec $$A$$ is the variety cut out of $$\mathbb{C}^n$$ by $$f_1, \ldots, f_k$$.

We can specify representation of (C[x]) on a vector space (V) by specifying where to send (x). That is, by specifying a linear operator (A).

We can alternatively phrase this. Given (A \in \text{End}_F(V)), we can treat (V) as an (F[x]) module by $$f(x)\cdot v =: f(A)v$$

Alright, now, take the image of (C[x]) in (\text{End}(V)), let’s denote this (C[A]). This is a subring of (\text{End}(V)), it’s commutative.

Here’s the cool part:

$$\text{Spec }C[A] = \text{Spec }A$$

But where is this identification coming from?

Well, $$C[A] = C[x]/(\text{ stuff that acts by 0 on V })$$

Let’s give “stuff that acts by 0 on V” a closer look as the subring of (F[x]) it is, let’s frak it up:

 $$\mathfrak{P}_A = { f \in F[x] f(A)v = 0, \forall v \in V }$$ This is an ideal of (F[x]), and since (F[x]) is a PID, any ideal is generated by one element. The generator of (\mathfrak{P}_A) is the minimal polynomial of (A) (i.e., characteristic polynomial with multiplicity 1).

So! Spec (C[A]/\mathfrak{P}_A) is the variety cut out of (\mathbb{C}) by all (g \in \mathfrak{P}_A \subset F[x]).

 Nice, now, the operator spectrum (\text{Spec }A) is defined as ({x \in C A-xI \text{ is not invertible }}), which is like eigenvalues, since (A-xI) is not invertible (\Leftrightarrow) (\det(A-xI) = 0). Not remembering the multiplicity, if the eigenvalues of (A) are (a, a, b), then (\text{Spec }A = {a,b}). Also, we’re assuming the eigenvalues have finite multiplicity, (otherwise nilpotent matrices might run us amuck)

Start with (\text{Spec }A). This is a set, it’s the eigenvalues* of A, look at the polynomial (p) whose roots are (\text{Spec } A). Look at the (F[x])-submodule generated by this (p_A), this is exactly (\mathfrak{P}_A).

*finite multiplicity.

That’s pretty nice!

It reminds me of something. The original reason for studying group determinants (Dedkind’s idea) probably come from Galois theory, where matrices were what Galois used to depict his “groups” as permutation groups of the roots. Dedekind looked at the determinent of such a “group matrix.”

Frobenius factored this “group determinent” over the complex numbers, and found two things:

1) the number of irreducible factors equals the number of conjugacy classes of G

2) each irreducible factor is homogeneous of some degree, and then each such irreducible factor appears to the power of its own degree.

Written on February 14, 2016