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.

There are these finite fields of characteristic $p$ , namely $\mathbb{F}_{p^n}$ for any $n>1$ and there is the algebraic closure $\bar{\mathbb{F}_p}$. The only other fields of non-zero characteristic I can think of are transcendental extensions namely $\mathbb{F}_{q}(x_1,x_2,..x_k)$ where $q=p^{n}$.

Thats all! I cannot think of any other fields of non-zero characteristic. I may be asking too much if I ask for characterization of all non-zero characteristic fields. But I would like to know what other kinds of such fields are possible.

Thanks.

share|improve this question
    
Every field $F$ of characteristic $p$ is the image of some polynomial ring $\mathbb Z_p[X]$ where $X$ is some set of variables. That does not determine the field uniquely, of course. :) –  Thomas Andrews Nov 2 '11 at 18:43
    
@Thomas Andrews You mean ring-homomorphic image? –  Dinesh Nov 2 '11 at 18:46
    
Yes. So, for every $F$, there is a set $X$ (you can always find $|X|\leq |F|$,) and a maximal ideal $I\subset\mathbb{Z}_p[X]$ so that $F\cong \mathbb{Z}_p[X]/I$ –  Thomas Andrews Nov 2 '11 at 18:48
1  
No, two such ideals can give the same $F$. You can see that with $F_{p^k}=\mathbb{Z}_p[x]/<\pi(x)>$ where $\pi(x)$ can be any prime polynomial of degree $k$. That's why it is trickery - different ideals can give the same field. –  Thomas Andrews Nov 2 '11 at 19:09
2  
@Dinesh: You have also fields of power series $\mathbb F_p((x_1,...,x_n))$. –  user18119 Nov 2 '11 at 19:51

3 Answers 3

up vote 9 down vote accepted

There are finite extensions of the transcendental fields you've written down. Indeed, since $k(x_1,\ldots,x_n)$ is not algebraically closed when $n \geq 1$, no matter what field $k$ of coefficients you choose, it has non-trivial finite extensions.

The classification of these fields is not a simple matter; in fact, it is one of the main topics of algebraic geometry. (One can think of it as being the problem of classifying $n$-dimensional varieties up to birational equivalence.)

In any case, I would say that these fields, for some choice of $n$ (possibly $0$), and with $k$ equal to $\mathbb F_q$ or $\overline{\mathbb F}_p$, are the characteristic $p$ fields that arise the most often in practice.

[Also: one reason that you can't think of other examples is that any field of char. $p$ which is finitely generated over its prime subfield $\mathbb F_p$ is a finite extension of $\mathbb F_p(x_1,\ldots,x_n)$ for some $n$; that is also why these tend to be the examples that arise most often.]

share|improve this answer
    
I believe the only obvious class of fields I missed in the question is the finite extensions of transcendental extensions of finite fields(and its algebraic closure). My question is these are the only fields I can think of and I want some not so obvious fields other than them if they exist. Thanks. –  Dinesh Nov 2 '11 at 18:18
1  
@Dinesh: Dear Dinesh, You can always take the algebraic closure of, e.g. $\mathbb F_p(x)$, then form transcendental extensions of these, then take finite extensions of those, and then perhaps the algebraic closure of those, then take transcendental extensions of those, and continue to iterate, even transfinitely if you like. Nevertheless, the "obvious" examples are the ones that tend to come up in practice, at least in my experience. Regards, –  Matt E Nov 2 '11 at 18:21
1  
Also, it ought to be possible to show using Zorn's lemma that every field of characteristic $p$ is produced by some (possibly transfinite) chain of finite and/or transcendental extensions starting with $\mathbb F_p$. –  Henning Makholm Nov 2 '11 at 18:31
1  
No -- the "ought" indicates that I made it up on the spot, but it sounded plausible. I imagine using Zorn's Lemma on the set of all extension chains of subfields, ordered by sequence prefixes, and get a maximal extension chain. It should be easy to see that a maximal extension chain must end with the entire field. –  Henning Makholm Nov 2 '11 at 18:40
2  
A simpler statement is true: every field is an algebraic extension of a purely transcendental extension on its prime subfield. This can proven using Zorn to get a maximal algebraically independent set. Every algebraic extension can be expressed as a (transfinite) chain of finite extensions. –  Chris Eagle Nov 2 '11 at 18:50

No need to limit yourself to a finite number of transcendentals... So $\mathbb F_q(x_1,x_2,\dots,x_n,\dots)$ is another example. You can also use $\bar{\mathbb{F}_p}$ as the coefficient field. Many combinations are possible. What characterization are you after?

share|improve this answer
2  
Yeah of course, I was asking whether there are any other fields other than these(which I mentioned in the question) fields of non-zero characteristic.If yes, I would like to see them –  Dinesh Nov 2 '11 at 18:14

The basic structure theory of fields tells us that a field extension $L/K$ can be split into the following steps:

  1. an algebraic extension $K^\prime /K$,
  2. a purely transcendental extension $K^\prime (T)/K^\prime$,
  3. an algebraic extension $L/K^\prime (T)$.

The field $K^\prime$ is the algebraic closure of $K$ in $L$ and thus uniquely determined by $L/K$. The set $T$ is a transcendence basis of $L/K$; its cardinality is uniquely determined by $L/K$.

A field $L$ has characteristic $p\neq 0$ iff it contains the finite field $\mathbb{F}_p$. Hence you get all fields of characteristic $p$ by letting $K=\mathbb{F}_p$ in the description of field extensions, and by chosing $T$ and $K^\prime$ and $L/K^\prime (T)$ as you like. Of course in general it is then hard to judge whether two such fields are isomorphic - essentially because of step 3.

share|improve this answer
1  
Why do you include step 1? Every field extension is an algebraic extension of a purely transcendental extension. –  Chris Eagle Nov 2 '11 at 19:08
    
Thanks. This is a very systematic way of seeing it. –  Dinesh Nov 2 '11 at 19:09
    
I included step 1 because transcendental extensions $L/K$, where $K$ is algebraically closed in $L$, are easier to handle than the general case. Moreover the algebraic extensions of $K$ might be much simpler than those of $L$ -- like in the case we discuss here. –  Hagen Nov 2 '11 at 22:14

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.