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My problem is to build, for every prime $p$, a field of characteristic $p$ in which every polynomial of degree $\leq n$ ($n$ a fixed natural number) has a root, but such that the field is not algebraically closed.

If I'm not wrong (please correct me if I am) such a field cannot be finite, by counting arguments. But on the other hand, the union of all finite fields (or of any ascending chain of finite fields) of characteristic $p$, which is what I get if I start with $F_p$ and add a root to each polynomial of degree $\leq n$ in each step, is the algebraic closure of $F_p$, hence algebraically closed. I don't see how I can control this process so that in the end I get a field that is not algebraically closed.

Any hint will be welcome. Thanks in advance.

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    $\begingroup$ It's a little unclear: is $n$ fixed? $\endgroup$ – Qiaochu Yuan Dec 23 '11 at 19:24
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    $\begingroup$ Yes, such a field cannot be finite: if $k$ is a field whose characteristic is not $2$, since $X^2-\lambda$ has a root for all $\lambda \in k$, then every element of $k$ is a square (actually this is equivalent to having a root for all polynomial of degree $2$ in $k$). But since the square map $x \mapsto x^2$ has non trivial kernel ($\{-1, 1\}$), then $k$ is infinite. $\endgroup$ – Joel Cohen Dec 23 '11 at 19:24
  • $\begingroup$ Yes, n is a fixed natural number. I have edited the statement in order to make it more clear. $\endgroup$ – Charlie Dec 23 '11 at 19:30
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    $\begingroup$ If you start with a field of size q, and adjoin a root of any (all) irreducible quadratic, you get a field of size q^2. Start with q=2, and you get 2, 4, 16, 256, etc. None of these fields contains a root of an irreducible cubic over the original field (with q=2, that would require a field whose size was a power of 8). In other words, you don't get the algebraic closure, since for any prime r bigger than n, you don't get the roots of any irreducible polynomials of degree r. $\endgroup$ – Jack Schmidt Dec 23 '11 at 19:32
  • $\begingroup$ @Jack Schmidt: Thank you very much. If you post your comment as an answer, I'll gladly accept it. $\endgroup$ – Charlie Dec 23 '11 at 19:47
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If you start with a field of size q, and adjoin a root of any (all) irreducible quadratic, you get a field of size $q^2$.

Start with $q=2$, and you get 2, 4, 16, 256, etc. None of these fields contains a root of an irreducible cubic over the original field (with $q=2$, that would require a field whose size was a power of 8).

In other words, you don't get the algebraic closure, since for any prime r bigger than n, you don't get the roots of any irreducible polynomials of degree r.

As Lubin mentions, this is equivalent to taking a Sylow pro-2-subgroup of the Galois group of the algebraic closure, and I guess in general you want a Hall pro-n-subgroup of the Galois group, but I prefer just thinking about repeatedly squaring a number.

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I’ll give you a hint, and not an answer. Best route to understanding here is to use Galois Theory. The total Galois group of a finite field $k$, i.e. the group of the algebraic closure over $k$, is $\hat{\mathbb Z}$, the profinite completion of the integers. It’s topologically generated by the single automorphism, the Frobenius of $k$. To understand $\hat{\mathbb Z}$, use Chinese Remainder Theorem, and you see that it’s the direct product of all the groups ${\mathbb Z}_p$, with $p$ running through all the primes. You take it from there.

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