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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.

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    Lots (at least as many cardinals as there are cardinals strictly less than $2^{\aleph_0}$). Given any field $F$ of cardinality $2^{\aleph_0}$ (for example, any polynomial ring in $\kappa$ variables, with $1\leq\kappa\leq 2^{\aleph_0}$;), biject $\mathbb{R}\times\mathbb{R}$ with $F$, and use transport of structure. Different cardinalities of sets of variables will yield non-isomorphic fields. (Same thing with $\mathbb{Q}\times\mathbb{Q}$ and countable fields). Somehow, I suspect that you are not asking what you *really* mean to ask.2012-02-03
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    Do you want the number of field structures _up to isomorphism_? Then the answer is $1$ and this follows readily from the fundamental theorem of algebra.2012-02-03
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    We need to make some restrictions, otherwise the question becomes the uninteresting how many fields are there with the cardinality of the continuum. Is the *additive* structure the usual one, as in your $\mathbb{Q}\times\mathbb{Q}$ example?2012-02-03
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    @Qiaochu: I don't see that; you can biject $\mathbb{R}\times\mathbb{R}$ with both $\mathbb{R}$ and $\mathbb{C}$, so that will give you at least two nonisomorphic field structures. Of course, if you respect the *additive* structure then I may agree with you.2012-02-03
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    @AndréNicolas´: Is there another way to define an new addition and a new multiplication in $\mathbb{R}\times\mathbb{R}$ different from the usual one? Sorry, but it is an interesting question for me, that's the reason I've decided to ask that question.2012-02-03
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    @Arturo: okay, so I assumed that the OP wanted a degree-2 field extension of $\mathbb{R}$.2012-02-03
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    @spohreis: it depends very strongly on what properties of $\mathbb{R} \times \mathbb{R}$ you care about preserving. Do you only want to think of it as a set? As an abelian group? As a vector space over $\mathbb{R}$? (I assumed implicitly that you wanted this last option based on your example.)2012-02-03
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    @spohreis: Look up [transport of structure](http://en.wikipedia.org/wiki/Transport_of_structure). Since $\mathbb{R}\times\mathbb{R}$ can be bijected (as a set) with, say, $\mathbb{R}(x,y)$ (field of rational functions on two variables), then given any bijection $f\colon \mathbb{R}\times\mathbb{R}\to \mathbb{R}(x,y)$ will give you a field structure on $\mathbb{R}\times\mathbb{R}$ by $(a,b)\oplus(c,d) = f^{-1}(f(a,b)+f(c,d))$ and $(a,b)\odot(c,d) = f^{-1}(f(a,b)f(c,d))$. This structure may have *nothing* to do with the usual additive structure.2012-02-03
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    @QiaochuYuan: I want to define a field structure on $\mathbb{R}\times \mathbb{R}$. I mean, define an addition and a multiplication which makes $\mathbb{R}\times \mathbb{R}$ a field. But it is not the same as the complex multiplication.2012-02-03
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    @ArturoMagidin: Thank you very much! I was looking for this kind of answer, but I think a mess up my own question.2012-02-03
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    @spohreis: that doesn't answer my question. What structure already on $\mathbb{R} \times \mathbb{R}$ do you care to preserve? As a set it's just $2^{\omega}$ and there are many fields with this cardinality (Arturo gives an example).2012-02-03
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    @QiaochuYuan: None. I just want to define a new field structure on $\mathbb{R}\times \mathbb{R}$. What's wrong with that?2012-02-03
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    @spohreis: it's confusing to call it $\mathbb{R} \times \mathbb{R}$ if you only care about its cardinality; that notation carries with it connotations of lots of other structure (additive, vector space, topological, ...).2012-02-03
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    @QiaochuYuan: Sorry, I didn't know that. Thanks for bear with me!2012-02-03
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    If you want to conserve the additive structure, I would say this becomes more a question of topological groups and would have to see with Pointriaguin's and Kovalski's results. So if you can show that weak topology induced by sum and multiplication is usual topology, or at leat contained in it, then the only possible structure would be that of complex numbers. Also, I have to see that in $\mathbb{Q}^2$ the coincidence is exactly with the algebraic extension of $\mathbb{Q}$ of order two.2012-02-03

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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:

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.