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I'm reading Derbyshire's Unknown Quantity.

It's an interesting exercise to enumerate and classify all possible algebras. Your results will depend on what you are willing to allow. The narrowest case is that of commutative, associative, finite-dimensional algebras over (that is, having their scalars taken from) the field of real numbers $\mathbb{R}$ and with no divisors of zero. There are just two such algebras: $\mathbb{R}$ and $\mathbb{C}$, a thing proved by Weierstrass in 1864.

What is this proof? I've googled Weierstrass algebra proof but found mainly the Stone-Weierstrass Theorem, which I'm not sure if this is the proof.

Red Banana
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    This is more commonly called Frobenius's theorem, although the usual statement drops the commutativity requirement. –  Jul 24 '13 at 00:14
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    http://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras) –  Jul 24 '13 at 00:14
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    Related http://math.stackexchange.com/questions/529/why-are-the-only-division-algebras-over-the-real-numbers-the-real-numbers-the-c – ADR Jul 24 '13 at 00:17
  • @SteveD But what's the proof given by Weierstrass in 1864? Is it the same thing? – Red Banana Jul 24 '13 at 01:46
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    Probably not, he'd have proved the entire classification of (associative) real division algebras then. But it seems one has to dig up an edition of Weierstrass' collected works to find out. – Daniel Fischer Jul 24 '13 at 01:56

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I think the author meant Frobenius clasiffication of real division algebras (see for example Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?) that its an stronger result, so if you want to add commutativity you loose the quaternions.

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