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Why are the only (associative) division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Here a division algebra is an associative algebra where every nonzero number is invertible (like a field, but without assuming commutativity of multiplication).

This is an old result proved by Frobenius, but I can't remember how the argument goes. Anyone have a quick proof?

Noah Snyder
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  • I am pretty sure it follows from this: http://books.google.com/books?id=ix4P1e6AkeIC&pg=PA218&lpg=PA218&dq=conway+division+algebra&source=bl&ots=0PbTsdZeTt&sig=hxmtgpV_SxHUR3KKZYOXNNOw5dA&hl=en&ei=EQpJTJC2H8LknAeApKTjDQ&sa=X&oi=book_result&ct=result&resnum=2&ved=0CBoQ6AEwAQ#v=onepage&q&f=false If I remember correctly from Functional Analysis last year... – BBischof Jul 23 '10 at 03:19
  • Is it that those are the only division algebras or if you have a division algebra it is isomorphic to the ones you listed? – Jonathan Fischoff Jul 23 '10 at 15:07
  • @Jonathan I don't know if I understand the difference of the two things you wrote. – BBischof Jul 30 '10 at 07:42
  • @BBischof I was confused by two things, I know you can construct something like complex numbers from a real clifford algebra http://en.wikipedia.org/wiki/Geometric_algebra#Complex_numbers, and I was just wrong about something related to spinors. I don't always use the word isometric correctly so maybe I confused you there, oh well. – Jonathan Fischoff Aug 04 '10 at 01:58
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    This is not true, the field of rational functions is a counterexample. The Frobenius theorem says that a _finite-dimensional_ associative division algebra over the reals is the reals, the complex numbers, or the quaternions (up to isomorphism). – aaa Sep 12 '11 at 11:23
  • You see, there are these Hopf fibrations...https://en.wikipedia.org/wiki/Hopf_invariant – Justin Young Oct 20 '20 at 15:40

4 Answers4

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Essentially one first proves that any real division algebra $D$ is a Clifford algebra (i.e. it's generated by elements of some inner product vector space I subject to relations $v^2=\langle v, v\rangle$): first one splits $D$ as $\mathbb R\oplus D_0$ where $D_0$ is the space of elements with $Tr=0$ and then one observes that minimal polynomial of a traceless element has the form $x^2-a=0$ (it's quadratic because it's irreducible and the coefficient of $x$ is zero because it is the trace). Now it remains to find out which Clifford algebras are division algebras which is pretty straightforward (well, and it follows from the classification of Clifford algebras).

This proof is written in Wikipedia.

Grigory M
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    There is a very [nice and simple proof](http://math.stackexchange.com/a/865139/34365) given by Jyrki Lahtonen [here](http://math.stackexchange.com/a/865139/34365). – Assad Ebrahim Jul 01 '15 at 20:30
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    @Assad that proof is indeed simple — but gives a proof only of a much weaker statement, AFAICS (only about dim=3) – Grigory M Jul 01 '15 at 22:47
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There is a simple proof of Frobenius's theorem in Lam's book on noncommutative rings, pp. 208--209. He attributes the argument to Palais.

One should consider this theorem to be two theorems: (1) $\mathbb C$ is the only $\mathbb C$-central division algebra and (2) $\mathbb R$ and $\mathbb H$ are the only $\mathbb R$-central division algebras. The reason there are so few choices is that $\mathbb C$ is alg. closed and $\mathbb R$ is nearly so. Division algebras with center equal to a particular field can be created using cyclic Galois extensions* and since $\mathbb Q$ has such extensions of arb. high degree there are $\mathbb Q$-central division alg. of arb. high dimension.

*There are further technical conditions to be satisfied on the cyclic extension in order for the construction of a division algebra to work, e.g., a finite field has a cyclic extension of each degree but there are no central div. alg. of $\dim > 1$ over a finite field. The relevant technical conditions are satisfied when the base field is the rational numbers.

KCd
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There is another proof using the theory of central simple algebras and quaternion algebras.

Denote by $D$ such a division algebra. Note that a division algebra is always simple. Moreover, since the center of a division algebra is a field, $D$ is central simple over either $\mathbb{C}$ or $\mathbb{R}$.

As there is no nontrivial division algebra over an algebraically closed field, $D$ must be $\mathbb{C}$ if $Z(D)=\mathbb{C}$.

Otherwise, $D$ is central simple over $\mathbb{R}$, then $\mathbb{C}$ splits $D$. Hence we obtain the following: $$ \sqrt{\mathbf{dim}_{\mathbb{R}}(D)} = \mathbf{ind}_{\mathbb{R}}(D)|[\mathbb{C}:\mathbb{R}]=2 $$ $D$ is nontrivial, hence the dimension of $D$ is 4. Then $D$ must be $\mathbb{H}$ by Wedderburn's theorem.

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I will write an answer which uses the Brauer group.

As $\mathbb C$ is the algebraic closure of $ \mathbb R$, $Br(\mathbb R)=Br(\mathbb C / \mathbb R)$. Also we know $Br(\mathbb C / \mathbb R) \cong H^2(Gal(\mathbb C / \mathbb R), \mathbb C ^{\times} )\cong \hat H^0(Gal(\mathbb C / \mathbb R),\mathbb C ) $. Here $\hat H $ stands for the Tate cohomology group and the second isomorphism is true because $Gal(\mathbb C / \mathbb R)$ is cyclic (it is $\mathbb Z /2 \mathbb Z $).

So it is clear that the only central division algebras over $\mathbb R $ are $\mathbb{R}$ itself and $ \mathbb H $.

If $D$ is a division algebra over $ \mathbb R$ whose center is not $\mathbb{R}$, then as it is a field algebraic over $\mathbb R$), $ \mathbb C =Z(D) $. So $D $ is a division algebra over $\mathbb C $ which implies $\mathbb C = D $.

Quantum Chill
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mathemather
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