Can we prove that every division algebra over $R$ or $C$ is a normed division algebra? Or is there any example of division algebra in which it is not possible to define a norm? Definition of normed division algebra is in here. Thanks!
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1A transcendental extension field? – Jyrki Lahtonen Apr 22 '13 at 04:39
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sorry. I cant understand. can you explain me little more clearly. – GA316 Apr 22 '13 at 05:39
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I am guessing a transcendental extension field is a division algebra but not normed division algebra. but can you tell me why it is true? thanks. – GA316 Apr 22 '13 at 05:45
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1Related: Existence of norm for C*-Algebra – ˈjuː.zɚ79365 May 29 '13 at 05:42
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Frobenius theorem for (finite-dimensional) asscoative real division algebras states there are only $\mathbb{R},\mathbb{C},\mathbb{H}$, and the proof is elementary (it is given on Wikipedia in fact).
If you don't care about finite-dimensional, then the transcendental field extension $\mathbb{R}(T)/\mathbb{R}$, where here $\mathbb{R}(T)$ is the field of real-coefficient rational functions in the variable $T$, is a divison algebra (it is a commutative field) but cannot carry a norm. Indeed, there are no infinite-dimensional normed division algebras.
Finally, there are real division algebras that are not $\mathbb{R},\mathbb{C},\mathbb{H}$ or $\mathbb{O}$ (which are the only normed ones), which means there are division algebras that cannot carry a norm. It is still true they all must have dimension $1,2,4$ or $8$ (see section 1.1 of Baez's The Octonions), but there are (for example) uncountably many isomorphism classes of two-dimensional real division algebras (but unfortunately I don't have a reference for this handy).
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I am interested to hear more about what you mentioned in the last paragraph, so I made a separate question over here https://math.stackexchange.com/questions/2892818/finite-dimensional-division-algebras-over-the-reals-other-than-mathbbr-math – PPenguin Aug 24 '18 at 09:18