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

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.

• It's a little unclear: is $n$ fixed? – Qiaochu Yuan Dec 23 '11 at 19:24
• 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. – Joel Cohen Dec 23 '11 at 19:24
• Yes, n is a fixed natural number. I have edited the statement in order to make it more clear. – Charlie Dec 23 '11 at 19:30
• 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. – Jack Schmidt Dec 23 '11 at 19:32
• @Jack Schmidt: Thank you very much. If you post your comment as an answer, I'll gladly accept it. – Charlie Dec 23 '11 at 19:47

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.

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.

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 $$ with coefficients in $$k^{(n)}$$ splits completely in $$k^{(n)}$$.

Note that the degree over $$k$$ of every element in $$k^{(n)}$$ has all its prime factors $$. Therefore, if $$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 $$\bar k/k$$.