Non-algebraically closed field in which every polynomial of degree $<n$ has a root

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Let $k$ be a field, $\bar k$ an algebraic closure of $k$. Fix $n>1$ natural. Consider the family $\mathcal{K}_n$ of fields $K$, $k\subset K\subset \bar k$ with the property: there exists a family of intermediate fields $$k = K_0 \subset K_1 \subset \ldots K_s= K$$ so that $[K_{i+1}\colon K_i]< n$ for all $1\le i \le s$. It is easy to check the following:

1. $K \in \mathcal{K}_n$, $K\subset L \subset \bar k$, $[L\colon K]< n$ implies $L \in \mathcal{K}_n$

2. $K$, $K'\in \mathcal{K}_n$ implies $K K'\in \mathcal{K}_n$.

3. $K \in \mathcal{K}_n$, $k \subset K'\subset K $ implies $K'\in \mathcal{K}_n$.

It is easy to see now that the union of the subfields in $\mathcal{K}_n$ is a subfield $k^{(n)}$ and every polynomial of degree $<n$ with coefficients in $k^{(n)}$ splits completely in $k^{(n)}$.

Note that the degree over $k$ of every element in $k^{(n)}$ has all its prime factors $<n$. Therefore, if $k$ is such that there exist elements in $\bar k$ whose degree over $k$ is a prime factor $>n$ (many examples here) then $k^{(n)}\ne \bar k$, that is $k^{(n)}$ is not algebraically closed.

Note: for $n=3$ we get $k^{(n)}$ are the constructible elements of $\bar k/k$.