2024-08-29

Theorem (Jacobson Density)

Let be a field. Let be a -algebra. Let be a simple finite-dimensional -module. Let . Then the canonical homomorphism

is surjective.

In fact, by Schur’s Lemma, will turn out to be a division algebra. In many cases, will just be . In this case, is all linear operators, i.e., all matrices where . The theorem tells us that all of these matrices lie in the image of .

In general, is constructed to commute with , and this is the smallest subspace of that can map into.

Example of Morita Theory

Let be a field. Consider making into a -bimodule. Then is an equivalence of categories

An equivalence in the other direction is the right adjoint functor (because the left adjoint is tensor).

Last time

We had a finite Galois extension and .

Theorem (Category of vector spaces with semilinear Galois actions)

There is an equivalence of categories

Note that we can consider to be a . Here, a semilinear -action refers to

for , and where .

Proof

Step 1

We construct a -algebra such that the right hand side is . In the category of -vector spaces with semilinear -actions, there are two things that are acting: -action on the -vector spaces, and the -action floating around. So we should have and . We therefore declare to be generated by the symbols:

  • for each ,
  • for each .

We impose the relations:

  • relations in ,
  • relations in ; i.e.,
  • an algebra homomorphism and a monoid homomorphism (we can’t quite yet say group homomorphism because we don’t know yet that , but it will be a group homomorphism),
  • the interesting relation: (from the semilinear axiom).

By fiat, is the same as .

Step 2

Now let (because is Galois). Then

To see this, the semilinear relation allows us to separate the s to the left and the s to the right. Now if is a -basis of and , then is a spanning set.

Now consider the fact that is an -module.

  • acts by multiplication by ; and
  • acts by .

Last time, we showed that . Furthermore, is simple. In fact, is a simple -module and . In other words, since , every -module is a -vector space, and has no proper -subspaces.

Step 3

We now apply Theorem (Jacobson Density) with , :

On the left, we have . On the right, we have . By comparing dimensions, we must have an isomorphism:

In the context of Morita Theory, we have

And so we are done.

The conclusion

#TODO2
We have , i.e., -vector spaces that become isomorphic to after tensoring with . There is an injective map

We had a construction that from a cocycle, we obtain a semilinear action. However, the above theorem implies that every cocycle must come from a vector space (since semilinear -actions are the same as vector spaces over ):

This is called the effectivity of descent problem. We therefore have an isomorphism

Every vector space has a basis, so there is only one -dimensional vector space up to isomorphism. So the LHS of is a one point set .

Remark

Here, a form is what we called previously a twist.

Theorem (Hilbert 90)

More structure

For example, we can consider a vector space with a symmetric bilinear form. Over , we can talk about positive definite/negative definite/indefinite symmetric bilinear forms. Tensoring with , these all become the same.

Let be a vector space over . Let , i.e., is sent to the element . Let

Suppose is a cocycle. From what we concluded above, the construction wants us to do the following:

  1. Start with .
  2. Tensor up to .
  3. Use to twist the -action, so we get a new action of on .
  4. Take invariants under the twisted action.

Claim

under the twisted action.

Proof

We check if :

Conclusion

The above along with Theorem (Category of vector spaces with semilinear Galois actions) () allows us to conclude that

More generally

More generally, we can consider over a finite-dimensional vector space, and a tensor, i.e.,

Let

The same argument gives that

Example: algebras

Let be a finite-dimensional -algebra. The only piece of data is a multiplication map (because everything else is a property satisfied by this map)

Hence, we get that

Example: matrices

Let .

Theorem (Automorphisms of matrix algebra)

Every algebra automorphism of is conjugation by an invertible matrix, i.e.,

Conclusion

So we have

#TODO2

Example

We can take and , and when , we have . In this case, .