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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.

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    @QiL Interestin$g$ explicit example2011-11-02

3 Answers 3

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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.]

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    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.2011-11-02
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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.

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    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.2011-11-02
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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?

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    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 them2011-11-02