What are all different (non-isomorphic) field structures on $\mathbb R \times \mathbb R$ We know that $\mathbb R \times \mathbb R$ forms a field under addition and multiplication defined as $(a,b)+(c,d)=(a+c,b+d)$ ; $(a,b)*(c,d)=(ac-bd,ad+bc)$ ; is there any other way to make $\mathbb R \times \mathbb R$ into a field non-isomorphic to this usual field ?
 A: Yes, there are bijections between $\mathbb{R}$ and $\mathbb{R}\times \mathbb{R}$, and thus we can take any bijection $f:\mathbb{R}\rightarrow \mathbb{R}\times \mathbb{R}$ and declare it a field isomorphism.  That is, it forms a field with addition
$$ (a,b)+(c,d)=f(f^{-1}(a,b)+f^{-1}(c,d))$$
where the additive identity is $f(0)$, and multiplication
$$ (a,b)\cdot (c,d)=f(f^{-1}(a,b)\cdot f^{-1}(c,d))$$
where the multiplicative identity is $f(1)$.
These are all isomorphic (being isomorphic to $\mathbb{R}$), but not to the normal field structure on $\mathbb{R}\times \mathbb{R}$, since the latter is not isomorphic to $\mathbb{R}$.
A: There are many ways to do it while preserving addition as it is.
Note that if $F$ is any field of characteristics $0$ and cardinality $2^{\aleph_0}$ then as a vector space over $\Bbb Q$, $F$ and $\Bbb R$ and $\Bbb{R\times R}$ are isomorphic. So we can transport the multiplicative structure from $F$ to $\Bbb R$ or $\Bbb{R\times R}$ while preserving addition.
So really you're asking how many non-isomorphic fields of characteristics $0$ there are on a set of cardinality $2^{\aleph_0}$. Some non-isomorphic examples are $\Bbb R$, the different $p$-adic fields, any transcendental extension of $\Bbb R$ by at most $2^{\aleph_0}$ transcendental elements; any subfield of $\Bbb R$ of the right cardinality, $\Bbb C$, etc.
A: Assuming you want $\mathbb{R}$ to be a sub-field of the field $\mathbb{R}\times\mathbb{R}$, the answer is no. To see that, note that $\mathbb{C}$ is algebraically closed, and that it is the only possible algebraic extension of $\mathbb{R}$. In other words, once you extend $\mathbb{R}$ by a new number which is algebraic over $\mathbb{R}$, you automatically get something isomorphic to $\mathbb{C}$.
