I have read in an introduction to quaternions that it had been proven before that one can not structure the Euclidian $\mathbb{R}^3$ as a field.

Where can one find a proof thereof?

For which n, could we bestow $\mathbb{R}^n$ with a field structure?

  • $\begingroup$ The inventor of the quaternions tried for a long time to do it before going on to $\mathbb R^4$, maybe he found a proof why it was impossible. en.wikipedia.org/wiki/William_Rowan_Hamilton $\endgroup$ Sep 24, 2017 at 21:13
  • $\begingroup$ What would count as a good field structure on $\mathbb{R}^n$? I mean, in theory one could fix a bijection $\mathbb{R}^n \to \mathbb{R}$ and define operations on $\mathbb{R}^n$ by 'pulling back' the normal field operations from $\mathbb{R}$. $\endgroup$
    – Bib-lost
    Sep 24, 2017 at 21:13
  • $\begingroup$ @mathreadler no according to my textbook a proof was found long after he passed. $\endgroup$
    – Averroes
    Sep 24, 2017 at 21:20
  • $\begingroup$ It is provable that a division algebra over the real numbers must have dimension a power of two. $\endgroup$
    – Arthur
    Sep 24, 2017 at 21:22
  • $\begingroup$ See also: Hurwitz's theorem. $\endgroup$
    – Randall
    Sep 24, 2017 at 21:35

2 Answers 2


Just as a set, $\mathbb{R}^n$ has the same cardinality as $\mathbb{R}$. We can therefore fix a bijection $f : \mathbb{R}^n \to \mathbb{R}$ and define the field operations and constants on $\mathbb{R}^n$ by pulling back the known operations from $\mathbb{R}$ (e.g. for $x, y \in \mathbb{R}^n$, define $x + y = f^{-1}(f(x) + f(y))$). This makes $\mathbb{R}^n$ into a field, for any $n$.

Since you mentioned quaternions, I assume your question might have been: "For which $n$ can we make $\mathbb{R}^n$ into a field which respects the natural structure as an $n$-dimensional $\mathbb{R}$-vector space?" The answer is given by the Frobenius theorem: $n=1, 2$ if we need the field to be commutative, $n=4$ is also possible if we allow non-commutative fields.

Wikipedia: https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

  • 1
    $\begingroup$ If you're only interested in the commutative case, the complicated arguments of the Frobenius theorem are unneeded, since it follows immediately from the fact that $\mathbb{C}$ is algebraically closed that no other finite field extensions of $\mathbb{R}$ can exist. $\endgroup$
    – Bib-lost
    Sep 25, 2017 at 11:44

The only $n$ for which $\mathbb R^n$ is a field are $n=1,2,4$ the case of quaternions being the first example known of non-commutative field. A proof of the impossibility for $n=3$ you can read in "Algèbre linéaire et géométrie élémentaire" of Jean Dieudonné, p. 197, Hermann,Paris, 1964. There is english translation of this book.

  • $\begingroup$ Thank you for the reference $\endgroup$
    – Averroes
    Sep 24, 2017 at 21:28
  • $\begingroup$ You are welcome. $\endgroup$
    – Piquito
    Sep 25, 2017 at 4:13

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