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Is $\mathbb{R}^n$ a field for all $n$?

I suppose for n=1 and 2 the result is clear. What about higher values of $n$.

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  • $\begingroup$ That is trivial. My interest is to understand for higher values of $n$. $\endgroup$ – Fukuzita Jun 27 '14 at 3:21
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    $\begingroup$ How do you define addition and multiplication? $\endgroup$ – Asaf Karagila Jun 27 '14 at 3:22
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    $\begingroup$ You can map $\mathbb{R}^n$ one-to-one with $\mathbb{R}$, so any such set could be given the field operations that make an isomorphic copy of the reals (or the complex numbers, if you prefer). You need to specify the operations. There is a negative result in this direction, that $\mathbb{C}$ cannot be embedded in a finite-dimensional field extension (except itself of course). $\endgroup$ – hardmath Jun 27 '14 at 3:25
  • $\begingroup$ @hardmath: If you assume the axiom of choice, then you can even preserve the usual addition. $\endgroup$ – Asaf Karagila Jun 27 '14 at 3:32
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    $\begingroup$ @hardmath: Just look at both of these as vector spaces over $\Bbb Q$. They have the same dimension, so they are isomorphic, so the addition can stay the same. But without the axiom of choice there might not be an isomorphism (and it is indeed consistent that there is none). $\endgroup$ – Asaf Karagila Jun 27 '14 at 3:36
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If a set does not have arithmetic operations (addition and multiplication) defined for it, the question of whether that set is a field is simply answered no, because fields must have those operations defined.

Asaf has outlined one method for defining these operations on $\mathbb{R}^n$ that keeps the addition the same as in the usual vector space (adding coordinatewise) but defines multiplication in a fashion that makes $\mathbb{R}^n$ a field isomorphic to $\mathbb{R}$. So if we have flexibility in defining arithmetic operations, then yes, these sets can be fields.

Perhaps it is of interest to ask whether we can give these sets field operations that preserve the usual field operations on the subset $\mathbb{R} \subset \mathbb{R}^n$, i.e. that create a real field extension. The answer is that this can be done only for $n = 1,2$, because a finite dimensional real field extension is a simple algebraic extension.

We should find an irreducible real polynomial $f(x)$ of degree $n$ and form the quotient ring $\mathbb{R}[x]/(f(x))$, where $(f(x))$ is the principal prime ideal of $\mathbb{R}[x]$ generated by $f(x)$, i.e. all of its polynomial multiples. Such a prime ideal is maximal by virtue of the irreducibility of $f(x)$, and thus $\mathbb{R}[x]/(f(x))$ is a field.

Since $(f(x))$ contains no constants besides zero, a copy of $\mathbb{R}$ sits inside $\mathbb{R}[x]/(f(x))$ in a natural way, and $\mathbb{R}[x]/(f(x))$ is an $n$-dimensional vector space like $\mathbb{R}^n$ in a natural way as well (use coordinates as the coefficients of powers $1,x,\ldots,x^n$). Addition is done coordinatewise, and multiplication is performed modulo $f(x)$.

But over the field of real coefficients, the only irreducible polynomials are of degree 1 or 2. A real root $r$ would make irreducible $f(x) = x-r$. Any nonreal roots occur in conjugate pairs. If $f(x)$ has a conjugate pair of roots $\alpha, \overline{\alpha}$, then $f(x)$ will be divisible by real polynomial $(x - \alpha)(x - \overline{\alpha})$, so an irreducible real polynomial has degree at most 2.

For example the polynomial $f(x)=x^2+1$ gives an extension isomorphic to $\mathbb{C}$.

However $\mathbb{C}$ is algebraically closed, so there are no irreducible polynomials over $\mathbb{C}$ of degree $n \gt 1$, so the only irreducible polynomials over complex numbers are degree 1. Thus $\mathbb{C}$ has no proper field extensions of finite dimensions over $\mathbb{C}$ (although it does have field extensions of all infinite dimensions).


Added: For the Reader disappointed that there is no more to this story and willing to entertain algebras not quite so nice as fields, the older Question What lies beyond the Sedenions? may prove interesting.

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