We can turn $\mathbb{R}^2$ into a field by using the product $(a ,b) \times (c, d)= (ac - bd, ad + bc)$ since this is just multiplication of complex numbers.
Is there such a product on $\mathbb{R}^n$ for $n > 2$?
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1See https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras) and https://math.stackexchange.com/a/1300963/589 – lhf Apr 16 '20 at 00:21
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2Yes, but not in a way which is nice or explicit. You can pick a bijective map from $\mathbb{R}^n$ to $\mathbb{R}$ and define operations on $\mathbb{R}^n$ by using operations on their images from $\mathbb{R}$. – Conifold Apr 16 '20 at 00:53
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You didn't say but I suppose you want to retain the standard operation of addition in $\mathbb R^n$? – bof Apr 16 '20 at 00:56
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1@bof Yes I want the addition structure to be the standard one. – Mustafa Said Apr 16 '20 at 00:58
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No. A famous theorem by Frobenius says:
The only finite-dimensional associative division algebras over the real numbers are $\mathbb R$, $\mathbb C$, and $\mathbb H$ (the quaternions).
The quaternions are not commutative and so are not a field.
lhf
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1The OP is not asking for a division algebra over $\mathbb{R}$, but for field operations on $\mathbb{R}^n$. And they most certainly exist, e.g. by using a bijection between $\mathbb{R}^n$ and $\mathbb{R}$ to pull operations from $\mathbb{R}$. – Conifold Apr 16 '20 at 00:49
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This was not stated explicitly in the question, but I suppose you want to keep the standard operation of addition in $\mathbb R^n$. (Confirmed by the OP in a comment.)
Yes. For $n\gt0$, since $(\mathbb R^n,+)\cong(\mathbb R,+)\cong(\mathbb C,+)$, there is a product on ($\mathbb R^n,+)$ which turns it into a field isomorphic to the field of real numbers, and another product which turns it into a field isomorphic to the field of complex numbers. (In fact you can make it isomorphic to any field of characteristic zero and cardinality $2^{\aleph_0}$.)
To see that the groups $(\mathbb R^n,+)$ and $(\mathbb R,+)$ are isomorphic, note that they are isomorphic as vector spaces over $\mathbb Q$, each having a Hamel basis of cardinality $2^{\aleph_0}$.
Of course this requires the axiom of choice.
P.S. See the accepted answer to this question, which says more or less the same thing.
bof
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Such a field will be obtained as a quotirnt of the polynomial ring $\mathbf{R}[X]$ by an ideal generated by an irreducible polynomial $f(X)$ of degree $n$.
Its is a known fact that for real numbers degree 3 onwards all polynomials are reducible. Reducibility of polynomials with odd degree $n>2$ is immediate from intermediate value theorem (in the limit $f(\infty)$ and $f(-\infty)$ would be of opposite sign and so it will have a real root showing reducibility).
A relevant theorem (Fundamental Theorem of Algebra) is that all polynomials with complex coefficients have root within the complex field (See Lang's algebra, or a book on Complex Analysis)
P Vanchinathan
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