How many field structures does $\mathbb{R}\times \mathbb{R}$ have? I know that in $\mathbb{Q}\times \mathbb{Q}$ there are several ways to define a field structure. For example, if $p$ is a prime number, then if we define addition as $(a,b)+(c,d)=(a+b,c+d)$ and multiplication as $(a,b)\star(c,d)=(ac+bdp,ad+bc)$, then $(\mathbb{Q}\times \mathbb{Q},+,\star)$ is a field. I know that $\mathbb{R}\times \mathbb{R}$ has one field structure. My question is: How many field structures does $\mathbb{R}\times \mathbb{R}$ have?
Thanks for you kindly help.
 A: As Arturo says in the comments, the key word here is transport of structure. Given any field $F$ with the same cardinality as $S = \mathbb{R} \times \mathbb{R}$ (which has the cardinality of the continuum), there exists a bijection $\phi : S \to F$, and using this bijection we can define field operations
$$r \oplus s = \phi^{-1}(\phi(r) + \phi(s))$$
$$r \otimes s = \phi^{-1}(\phi(r) \phi(s))$$
on $S$ (where $r, s \in S$). So a more precise way to phrase your question (after modding out by isomorphism) is as follows:

How many fields, up to isomorphism, have the cardinality of the continuum?

The answer is lots. Some examples:


*

*$\mathbb{R}$.

*$\mathbb{C}$.

*$\mathbb{Q}_p$, the field of $p$-adic numbers, for any prime $p$.


Some mechanisms for producing families of examples:


*

*If $F$ is a field with the cardinality of the continuum, then so is any algebraic extension of $F$, and so is $F(x)$.

*If $F$ is a field which is at most countable, then $F((t))$, the field of formal Laurent series over $F$, has the cardinality of the continuum, and so does $F(x_i : i \in \mathbb{R})$. 


Some examples of fields which are at most countable:


*

*The finite fields $\mathbb{F}_q$.

*$\mathbb{Q}$. 


Some mechanisms for producing fields which are at most countable:


*

*If $F$ is a field which is at most countable, then so is any algebraic extension of $F$, and so is $F(x)$. 



Here are some other questions you could have asked. Perhaps you wanted to preserve the structure of $\mathbb{R} \times \mathbb{R}$ as an abelian group. Now, assuming the axiom of choice, $\mathbb{R} \times \mathbb{R}$ is just a vector space over $\mathbb{Q}$ of dimension continuum, so another version of your question is:

How many fields, up to isomorphism, have an underlying abelian group which is a vector space over $\mathbb{Q}$ of dimension continuum?

The answer is that such fields are precisely the fields with the cardinality of the continuum which, in addition, have characteristic zero.
In the strongest version of your question, perhaps you want to preserve the structure of $\mathbb{R} \times \mathbb{R}$ as an $\mathbb{R}$-vector space. Then the corresponding version of your question is:

How many field extensions of degree $2$ does $\mathbb{R}$ have, up to isomorphism?

The answer is $1$: $\mathbb{C}$ is the unique such field extension, and this is a straightforward corollary of the fundamental theorem of algebra. 
