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 .
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 2. is central in . In other words, for every and we have .
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 .
Clearly, . Exercise 4 shows that in fact .
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.