### Frobenius theorem on real division algebras

#### by Dan Shved

This post is a proof of the well known Frobenius theorem. It’s nothing groundbreaking, I’m just writing it down to organize my own thoughts.

**Definition.** A *real division algebra* is a set together with operations , and such that:

- is a vector space over .
- is a division ring.
- for every and .

Examples of real division algebras: itself, the algebra of complex numbers , the algebra of quaternions .

**Theorem** (Frobenius)**.** Every finite-dimensional real division algebra is isomorphic to one of , or .

Before we begin, some notation and terminology conventions:

- We will omit and everywhere and write simply and insdead of and .
- Brackets mean the linear span of over . means the dimension of over .
- “Homomorphism” always means “homomorphism of -algebras sending unity to unity”.
- In the same vein, a “subalgebra” of a real division algebra always means “a subset closed with respect to addition and both multiplications, and also containing the unity”.

*Proof of Frobenius theorem.* Let be a finite-dimensional real division algebra, and be its unity. Clearly, is a homomorphism, so is a subalgebra of , and it is isomorphic to . If , then it follows that . Therefore we can assume that and .

**Exercise 1.** Any finite field extension of is isomorphic (as a real division algebra) to either or . (Hint: any irreducible polynomial over has degree at most ).

**Exercise 2.** is central in . In other words, for every and we have .

**Exercise 3.** Let . Then is a subalgebra of . Moreover, .

By the last exercise we have an element such that , and is isomorphic to . If then . Let us assume then that and .

- Show that is a commutative ring.
- Show that is in fact a field, and therefore a finite field extension of .
- Conclude that such an cannot exist. (Hint: is algebraically closed and cannot have any nontrivial finite extensions).

If we take two elements and , we can form their product . Thus is a vector space over the field . Note that is *not* a -algebra: that would mean that , and there’s no reason why the second equality should hold.

Now, consider a map defined by .

**Exercise 5.** is an involution, i.e. .

**Exercise 6.** preserves multiplication, i.e. for any .

**Exercise 7.** is -linear, i.e. when and .

Here comes the main idea. From exercises 5 and 7 it follows that eigenvalues of can only be equal to or , and (as a vector space over ) is the direct sum of the corresponding eigenspaces: where

Clearly, . Exercise 4 shows that in fact .

Since , . Pick an arbitrary . From exercise 3 we have , where . Acting by on both sides and using exercises 6 and 7, we get:

By choice of we have , therefore

It follows that . Since , must be negative. Introduce new elements

**Exercise 8.** Elements and are linearly independent over .

**Exercise 9.** Establish all the equalities below that haven’t been established yet:

(Hint: remember that , which means that ).

It follows that is a subalgebra of , and . All that’s left to prove is that .

Take an arbitrary . Using exercise 6 we get

therefore . But then . So, and . It means that , which implies , as required.

THANK YOU!!!!

You are welcome. And thanks for the comment, I appreciate it.

Thank you very much for putting this together, it’s a great guide! 🙂

You are welcome. I’m glad that someone finds it useful.