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Can $\mathbb{R}^3$ be turned into a field just like $\mathbb{R}^2$ by defining $$(a,b)(c,d):=(ac-bd,ad+bc)$$ $$(a,b)+(c,d):=(a+c,b+d)$$ And if so, what about $\mathbb{R}^n$ for $n\ge 4$? This question seems unapproachable to me, but it's such a common thing to ask that it seems to me that most mathematicians know the answer.

user829347
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Manuel Ocaña
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1 Answers1

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The answer is no -- you can see why if you've had a course in field theory: any finite dimensional vector space over $\mathbb{R}$ would be a finite degree field extension of $\mathbb{R}$, but by the fundamental theorem of algebra, the only such extension is $\mathbb{C}$.

If we drop the requirement that multiplication be commutative, the answer is still no for $n = 3$ or $n \geq 5$, as was first asked and answered by the inimitable Frobenius. https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

hunter
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    To be extremely pedantic, $\mathbb{R}$ is a finite degree extension over itself. – Mark Saving Feb 10 '22 at 01:03
  • Sorry to ask this, but in your answer did you consider defining addition and multiplication in a way so that R is not a subfield of R^n? – Manuel Ocaña Feb 10 '22 at 02:26
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    But, @ManuelOcaña if you drop that requirement of being a subfield, the answer is trivially "yes". $\mathbb R^n$ is of the same cardinality as $\mathbb R$, so take any bijection $f:\mathbb R^n\to\mathbb R$ and use it to " transport" field structure from $\mathbb R$ to $\mathbb R^n$, e.g. $u+v:=f^{-1}(f(u)+f(v))$ and similar for $\cdot$. –  Feb 10 '22 at 03:04
  • In fact, you can even choose a multiliplication on $\mathbb{R}^3$ together with the componentwise addition which makes it a field. The idea being that $\mathbb{R}^3$ and $\mathbb{R}$ are both $\mathbb{Q}$-vector spaces of dimension $\mathfrak{c}$ and therefore are isomorphic, so you can use that to transport the multiplication on $\mathbb{R}$ to a multiplication on $\mathbb{R}^3$. – Daniel Schepler Feb 10 '22 at 04:07
  • @ManuelOcaña no apologies needed, I should have mentioned this! As the other commenters say, we don't require that $\mathbb{R}$ be a subfield then there are lots and lots of (silly) field structures taken by picking a bijection of $\mathbb{R}^n$ with your favorite field of cardinality the same as $\mathbb{R}$. – hunter Feb 10 '22 at 05:17