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With this product:

$$(a,b)\cdot (x,y):=(ax-by,ay+bx)$$

$\,\Bbb R^2$ becomes a field, actually isomorphic to $\,\Bbb C$.

Are there other interesting products that turn $\,\Bbb R^2$ into a field?

user420196
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  • It depends on what is "interesting" to you. Abstractly, $\mathbb R^2$ is just some particular set, and so one can give it the same field structure as any other field with cardinality $|\mathbb R^2|$ e.g. it has the same cardinality as $\mathbb Q_p$ for any prime $p$, and so $\mathbb R^2$ can be given a field structure making it isomorphic to $\mathbb Q_p$. –  Apr 12 '20 at 07:58
  • A product is called alternative for an algebra $A$, if $x(xy)=(xx)y$ and $(yx)x=y(xx)$ for all $x,y\in A$. For example, the octonions form a non-associative alternative algebra. So the term "alternative product" may be misleading. – Dietrich Burde Apr 12 '20 at 08:36
  • @TokenToucan I think he means give $\mathbb R^2$ a multiplication that is compatible with the already existing $+$ which makes it a field. Giannakos, you might be interested in the Frobenius theorem https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras) although it doesn't exactly answer your question. – Noel Lundström Apr 12 '20 at 11:47
  • There are other products that give you a field isomorphic to $\Bbb C$, but with $i$ placed elsewhere than $(0,1)$. However, $\Bbb C$ is the only finite-dimensional extension field of $\Bbb R$. So the only other alternative is a field structure that doesn't contain a copy of the reals. – Paul Sinclair Apr 12 '20 at 17:29
  • Thanks everybody. This was my line of thought / motivation: – user420196 Apr 13 '20 at 08:47

  • $,\Bbb R^2$ is isomorphic to $,\Bbb C$ as vectors spaces, but not as fields with the 'natural' product, since $,\Bbb R^2$ is not even a field with that product.

  • However, one can give $,\Bbb R^2$ the product above, which makes them also isomorphic as fields.

  • Are there any other examples of products that turn $,\Bbb R^2$ into a different field?

  • I guess not: by Frobenius Theorem, since fields are division algebras, the resulting field would be isomorphic to $,\Bbb C$, and thus fundamentally the same.

    • – user420196 Apr 13 '20 at 08:55
  • Since $(\mathbb R^2,+)$ is isomorphic to $(\mathbb R,+)$, there is product which turns $(\mathbb R^2,+)$ into a field isomorphic to the field of real numbers, but it's not interesting. – bof Apr 13 '20 at 10:20