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 was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser & Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show that S' is a subfield of F'. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.

I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.

What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?

share|improve this question

4 Answers 4

up vote 4 down vote accepted

Is there an analogous classification for fields of characteristic zero?

Yes, but it is somewhat useless and nobody would call it a classification.

Every field of characteristic zero has the form $Quot(\mathbb{Q}[X]/S)$, where $X$ is a set of variables and $S$ is a set of polynomials in $\mathbb{Q}[X]$ (which you may replace by the ideal generated by $S$, which must be prime). This may be improved by the existence of transcendence bases: Every field of characteristic zero has the form $Quot(\mathbb{Q}[X])[T]/S$, where $X$ and $T$ are sets of variables and $S$ consists of polynomials, which have each only one variable of $T$.

share|improve this answer
3  
In particular, C is such a field. Now try writing down the variables and the polynomials! –  Qiaochu Yuan Sep 8 '10 at 19:58

If $F$ is any field, the rational functions over it form a field $F(t)$ with the same characteristic (and cardinality, if $F$ is infinite). This field consists of all rational functions $P(t)/Q(t)$ (considered as equivalence classes, i.e. if $P_1(t) Q_2(t) = P_2(t) Q_1(t)$ then $P_1(t)/Q_1(t)$ and $P_2(t)/Q_2(t)$ are identified).

You can also replace polynomials with formal power series to get a different field. And you can iterate the construction or just consider rational functions in several variables.

share|improve this answer

Yuval just posted the function field construction, which is an important example for your question.

You mentioned the algebraic numbers, but just to make it explicit, there are many algebraic number fields (often called just "number fields"). We have $ \bar Q $, the field of all algebraic numbers, but we also have $Q(\sqrt{2})$, $Q(\sqrt[5]{7})$, $Q(i,\sqrt{2},\sqrt[5]{7})$, and any other field you wish to generate over $Q$ by some tower of finite extensions.

(If we take $Q(\pi)$, we get a field isomorphic to the function field $Q(t)$, since $\pi$ is not algebraic.)

You can also take infinitely many different algebraic extensions of the p-adic fields, like $Q_3(i)$ or $Q_3(\sqrt[4]{2})$.

If you like big constructions, you can take the algebraic closure $\bar {Q_p}$of a p-adic field (which turns out not to be p-adically complete), then take the completion $\hat { \bar {Q_p}}$ of that to get a p-adcially complete and algebraically complete field which is isomorphic to the field of complex numbers (but not in any canonical way).

share|improve this answer

Yes, you are right in saying that field axioms are restrictive. Some other examples of the restriction [for finite fields] are:

1) You can't have field of arbitrary order. Only of the order $p^n$ are possible.

2) Non-zero elements of a field form a multiplicative group. When the field is finite, this group is cyclic [it's not straightforward to prove this starting with field axioms but you should try it]

3) Any finite division ring is a field [Wedderburn's theorem]. This is a very surprising result as a division ring need not by commutative. But finiteness imposes it.

For the question that you mentioned [from that book], you should note that when some set with some algebraic structure [like group, ring, field] is isomorphic [as that algebraic structure] to some other set, then that under that isomorphism, we are giving an algebraic structure to that other set. Hence in your case S' is a field.

share|improve this answer
    
The specific issue I'm having trouble with is this: consider the finite field of order 7. It is isomorphic to a subset of $Q$. But the first is not a subfield of the second. I think I'm missing something simple here. –  dls Sep 8 '10 at 16:08
    
All finite fields of the same order are isomorphic. That's quite restrictive. –  Yuval Filmus Sep 8 '10 at 19:56
    
In response to dls's comment: the source of confusion is a shift in meaning in the term isomorphism. There are many bijections from the finite field $F$ of order $7$ to subsets of $Q$. But none of these are field homomorphisms, or even homomorphisms of the additive groups of the fields. Group homomorphisms must send elements of finite order to elements of finite order. So if $\phi: (F, +) \to (Q, +)$ is a group homomorphism, every element of $\phi(F)$ has finite additive order. Every nonzero element of $(Q,+)$ has infinite order. So $\phi$ must be the zero map, and not a bijection. –  leslie townes Oct 23 '11 at 22:44

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.