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Hi!

Is it possible to define a product on R^n for n>2 such that R^n can be made into a field?

R is a field in its own right with the standard operations and R^2 can be made into a field by introducing the product (a,b)*(c,d) = (ac-bd,ad+bc) i.e. the product of the complex numbers.

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    $\begingroup$ Holy mackeral! Redundant answer party! I deleted mine to avoid making it worse. $\endgroup$
    – Ben Webster ♦
    Dec 15, 2009 at 18:39
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    $\begingroup$ If anyone else votes to close, I'd be happy to close this one. I don't think we need questions where the best answer is "No <link-to-wikipedia>". -1 $\endgroup$
    – Scott Morrison
    Dec 15, 2009 at 18:53
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    $\begingroup$ I thought so too, but I was persuaded in the Hilbert's Fifth problem question that the issue is complicated. Even Anton later posted a question whose answer was just there in Wikipedia. Certainly the shortest answer is the best one and should quickly be accepted. And maybe it's best to CW the posting. $\endgroup$
    – Greg Kuperberg
    Dec 15, 2009 at 19:00

5 Answers 5

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No.

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  • $\begingroup$ I endorse this answer. $\endgroup$
    – Greg Kuperberg
    Dec 15, 2009 at 18:57
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I guess it depends on what properties of $R^n$ you want. This is kind of a stupid answer but you can certainly pick a bijection $R^n \to R$ and define a field structure using that. It's not going to preserve anything nice about $R^n$ though.

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    $\begingroup$ This is the real answer to the question actually asked. The OP didn't say it has to be an algebra over the usual R. He didn't say the multiplication has to be continuous in the usual toplogy of R. So (with axiom of choice) the additive group R^n is isomorphic to the additive group R, and we get the "yes" answer. $\endgroup$
    – Gerald Edgar
    Dec 15, 2009 at 19:32
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    $\begingroup$ This is clearly not the intent of the question. What you should've done is commented and asked for clarification on what properties of R^n the OP wanted preserved. $\endgroup$
    – Qiaochu Yuan
    Dec 15, 2009 at 19:41
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    $\begingroup$ The way PW construed it is another standard problem: for instance, a variation on it it appears in Halmos' "automathography". It was honestly not clear to me what the OP intended. I think both answers are good. $\endgroup$
    – Pete L. Clark
    Dec 15, 2009 at 19:56
  • $\begingroup$ Downvoted for being obnoxious! $\endgroup$
    – Harry Gindi
    Dec 15, 2009 at 23:02
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    $\begingroup$ I think that too many people forget the existence of bijections between R^n and R, and the interesting paradoxes one can derive from them. $\endgroup$
    – Daniel Miller
    Oct 10, 2010 at 12:36
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No. $\mathbb{C}$ is algebraically closed, so any finite extension of $\mathbb{R}$ must be isomorphic to a subfield of $\mathbb{C}$, and therefore has dimension no more than $2$.

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No if you use the usual additive structure on $\mathbb{R}^{n}$ for the field addition; but if you give up commutativity of multiplication, you have the skew-field of hamiltonians, or quaternions, on $\mathbb{R}^{4}$, and if you then give up associativity of multiplication, you have the non-associative Cayley algebra, or octonians , on $\mathbb{R}^{8}$. The Cayley-Dickson process builds the complex field from the real field, the skew-field of hamiltonians from the complex field, the non-associative Cayley algebra from the hamiltonians, and in general a $2^{n+1}$-dimensional involutive Cayley-Dickson algebra from the $2^{n}$-dimensional involutive Cayley-Dickson algebra. A.A. Albert did much to articulate the state of affairs in the early part of the twentieth century, if memory serves.

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It is not possible to do this beyond a certain point. R^2 can be made into a field the complex numbers, then after that some of the structure is lost with each doubling the quaternions do not have commutative multiplication, the octonions do not have commutative multiplication or associativity and finally after that after doubling octonions there is zero division. John Conway has a book about octonions that discusses this. There is a review here

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