Is the standard complex multiplication the only one that makes $\Bbb R^2$ a field?

Question

If $(a_1,a_2)*(b_1,b_2)=(a_1b_1-a_2b_2,a_1b_2+a_2b_1)$ then $\Bbb R^2$ is a field with pointwise addition and $\ast$

Is there another multiplication $\ast$' that has this property?

Answer 1

The answer is "yes if you add few extra natural conditions", but no otherwise.

Here is an example. Pick the field $K= \mathbb R(X)$ of rational functions on $\mathbb R$.

Then $(\mathbb R^2,+)$ and $(\mathbb R(X), +)$ have the same dimension as $\mathbb Q$ vector spaces, therefore they are isomorphic as vector spaces.

In particular, they are isomorphic as abelian groups.

Let $f : (\mathbb R^2,+) \to (\mathbb R(X), +)$ be any isomorphism.

Define $$a\star b := f^{-1} (f(a) \cdot f(b))$$

This induces a field structure on $(\mathbb R^2,+)$ and as a field $(\mathbb R^2, + \star)$ is isomorphic to $(\mathbb R(X), +, \cdot)$ which makes it very different than $\mathbb C$.

The same construction works for any extension of $\mathbb R$ of finite transcendence degree.

But if you want your $\star$ to be "compatible" with real multiplication, in the sense that, as a field $(\mathbb R^2, + , \star)$ contains a copy of $\mathbb R$ and has a finite basis, then the only possibility is, up to isomorphism, complex conjugations.