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I have the following question:

Is there a simple way to see that if we put a multiplication $*$ on $\mathbb{R}^2$ (considered as a vector space over $\mathbb{R}$) such that with usual addition and this multiplication $\mathbb{R}^2$ becomes a field, then there exists a nonzero $(x,y)$ such that $(x,y)*(x,y)=-1$?

Remark:

  1. What I mean by "a simple way to see" is that I really don't want to refer to Frobenius's Theorem on real finite dimensional division algebras.

  2. I haven't said this in the problem but I'm also assuming that with this multiplication $\mathbb{R}^2$ becomes an algebra meaning $x*(\alpha y)=\alpha(x*y).$

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    $\begingroup$ It's not true, even if we demand $(1,0)\ast (x,y) = (x,y)$. You can let $(0,1)$ correspond to any $z \in \mathbb{C}\setminus\mathbb{R}$ and the corresponding multiplication makes $\mathbb{R}^2$ a field, but $(0,1)\ast(0,1) = -1$ only if you choose $z = \pm i$. $\endgroup$ – Daniel Fischer Aug 23 '13 at 17:57
  • $\begingroup$ I see your point. Now I edited my question. $\endgroup$ – Abelvikram Aug 23 '13 at 18:05
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    $\begingroup$ No. Assuming the axiom of choice, you can define multiplication to make $\mathbb{R}^2$ isomorphic to $\mathbb{R}$ (with the usual multiplication). Just consider dimensions as vector spaces over $\mathbb{Q}$. $\endgroup$ – George Lowther Aug 23 '13 at 18:27
  • $\begingroup$ I'm considering $\mathbb{R}^2$ as a vector space over reals. $\endgroup$ – Abelvikram Aug 23 '13 at 18:30
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    $\begingroup$ In that case, the answer is yes. As its an extension of degree 2, it is quadratic, hence obtained by appending a single square root of some non-square (hence negative) element of $\mathbb{R}$. As the square root of a negative number is a square root of $-1$ multiplied by a real, the extension is generated by appending the square root of $-1$. $\endgroup$ – George Lowther Aug 23 '13 at 18:47
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Sorry if I make it too elementary: If $1\in\mathbb R^2$ denotes $1$ of your field, and if $x\in\mathbb R^2$ is not its real multiple: $1,x,x^2$ are linearly dependent (over $\mathbb R$), i.e. $ax^2+bx+c=0$ for some $a,b,c$, and $a\neq 0$ (as $x$ is not a multiple of $1$), so we can suppose it's 1. If we complete squares, we get $(x+p)^2+q=0$ for some $p,q\in \mathbb R$. Now $q$ must be positive - otherwise $(x+p+\sqrt{-q})(x+p-\sqrt{-q})=0$, so you don't have a field (we found divisors of $0$). So finally $(x+p)/\sqrt{q}$ is the element you want.

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  • $\begingroup$ Thanks for the answer. It's really nice. $\endgroup$ – Abelvikram Aug 24 '13 at 4:46
  • $\begingroup$ +1, if only for the first sentence of the post, which apologizes for making it too elementary... $\endgroup$ – Did Aug 26 '13 at 18:36
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The usual addition forces $\Bbb R$ to be a two dimensional subfield of your field $F$ (Consider $\Bbb R1_F)$. If you assume the fundamental theorem of algebra, and have some background in field theory, then it is relatively straightforward. Assume no root of $x^2+1=0$, then $$F(i)=F[x]/(x^2+1)\ \text{is degree $2$ over $F$}$$ so $F(i)$ is a degree four finite (therefore algebraic) extension of $\Bbb R$ as $$[F(i):F][F:\Bbb R]=[F(i):\Bbb R] $$ But this cannot happen as any finite extension of $\Bbb R$ is contained inside a field isomorphic to $\Bbb C$. (since $\Bbb C$ is the algebraic closure of $\Bbb R$ and is of degree two over $\Bbb R$).

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  • $\begingroup$ sorry, but the point of the question is to find a simple proof. If you see the proof of Frobenius Theorem, this itself uses only the basic linear algebra. I was wondering that in case of $\mathbb{R}^2$ may be it's more simple. But anyway thanks for your answer. $\endgroup$ – Abelvikram Aug 23 '13 at 18:34
  • $\begingroup$ The other point being that I wanted to make understand this to undergrad. students who have done only a basic course in linear algebra so that's why the question. $\endgroup$ – Abelvikram Aug 23 '13 at 18:42
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You are looking at a field extension of $\Bbb R$ of degree two, but by the fundamental theorem of algebra, that field sitting above $\Bbb R$ is unique up to isomorphism and is isomorphic to $\Bbb C$.

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    $\begingroup$ But I fear now the question will change to "and without referring to the FTA..." :) $\endgroup$ – rschwieb Aug 23 '13 at 18:18

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