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Stream: deprecated: mathematics

Topic: fundamental theorem of noncommutative algebra

John Baez (Feb 23 2021 at 20:58):

I find "fundamental theorems" enjoyable, though I don't always understand why people consider some theorems to be "fundamental". I just learned about this one:

• Wikipedia, Fundamental theorem of noncommutative algebra.

John Baez (Feb 23 2021 at 21:00):

It says that if $V$ is a finite-dimensional complex vector space and $A$ is a proper subalgebra of the algebra $\mathrm{End}(V)$ of linear transformations of $V$, then there's a linear subspace $0 \subset L \subset V$ that's mapped into itself by every element of $A$.

John Baez (Feb 23 2021 at 21:00):

(Wikipedia manages to obfuscate the statement with lots of notation.)

John Baez (Feb 23 2021 at 21:03):

How do you prove this? I'd try to prove the contrapositive.

Does it only work over the complex numbers, or (some) other fields too?

Jens Hemelaer (Feb 24 2021 at 09:28):

I didn't hear about the name "Fundamental theorem of noncommutative algebra" before, but the name makes sense, since it is a generalization of the fundamental theorem of algebra in a non-obvious way.

The theorem follows from the following stronger result (which is more easily memorizable, for me): an algebra representation $\phi : A \to \mathrm{End}(V)$ is absolutely irreducible if and only if $\phi$ is surjective. Over an algebraically closed field, irreducible and absolutely irreducible are the same thing, so if $\phi$ is not surjective, then there is a proper subrepresentation, corresponding to a subspace $L \subset V$ that is preserved under the $A$-action.

The proof of this stronger result contains a proof of the "fundamental theorem of noncommutative algebra".

I think that the result holds if and only if the field is algebraically closed. Suppose that $K$ is not algebraically closed, then I believe that $L \subseteq \mathrm{End}(L)$ gives a counterexample, for any finite extension $L/K$.

John Baez (Feb 24 2021 at 16:21):

Thanks, this is great! What does "absolutely irreducible" mean? My guess: irreducible after passing to the algebraic closure. (Any field extension $L/K$ should let us extend a representation $\phi: A \to \mathrm{End}(V)$ of algebras over $K$ to one over $L$, $A \otimes_L K \to \mathrm{End}(V \otimes_L K)$.)

John Baez (Feb 24 2021 at 16:23):

Okay, that guess was right - I clicked on your link and that's what it says.

John Baez (Feb 24 2021 at 16:33):

The proof is so slick! I was vaguely imagining a proof not using any big theorems, and it seemed like it would be very hard. Artin-Wedderburn is so great... that's the theorem I might call the fundamental theorem of noncommutative algebra.

Fawzi Hreiki (Feb 24 2021 at 16:35):

Why would the Artin-Wedderburn theorem be the fundamental theorem?

John Baez (Feb 24 2021 at 16:42):

It's just my favorite theorem about noncommutative algebra. I don't really take this "fundamental theorem" concept seriously, but maybe I could argue that Artin-Wedderburn is the "fundamental theorem of semisimple algebras".

John Baez (Feb 24 2021 at 16:54):

It makes everything about semisimple algebras much easier.

Fawzi Hreiki (Feb 24 2021 at 16:55):

And representation theory of course.

John Baez (Feb 24 2021 at 16:58):

Right, when combined with this version of Maschke's theorem: for any finite group G and field k whose characteristic does not divide the order of G, the group algebra k[G] is semisimple. (Or various other theorems...)