Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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. –  Arturo Magidin Feb 3 '12 at 22:00
4  
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? –  André Nicolas Feb 3 '12 at 22:04
1  
@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. –  Arturo Magidin Feb 3 '12 at 22:04
4  
@spohreis: Look up 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. –  Arturo Magidin Feb 3 '12 at 22:16
2  
@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, ...). –  Qiaochu Yuan Feb 3 '12 at 22:31

1 Answer 1

up vote 6 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.