Prove that there exists an infinite field $F$ such that $F$ is algebraic over a finite field, and $F$ is not algebraically closed

Question

Prove that there exists a field $F$ such that

$F$ is infinite

$F$ is algebraic over a finite field

$F$ is not algebraically closed.

I can see that if condition (3) is removed, then the algebraic closure of the finite field is the answer. But, I can't see how to keep all the conditions in place and still make this work.

$\mathbb{Z}/p^n\mathbb{Z}$ isn't a field. Do you mean $\mathbb{F}_{p^n}$? — Qiaochu Yuan, Sep 19 '22 at 20:09
In that case it's better to write $\mathbb{F}_q$. What do you mean by "$\mathbb{F}_q[x]$ is finite"? That's the polynomial ring which is infinite but not a field. — Qiaochu Yuan, Sep 19 '22 at 22:01
@QiaochuYuan thanks. That analysis was wrong. I was thinking that the choice of coefficients is finite and so must be the ring (I forgot about the powers) — Sayan Dutta, Sep 20 '22 at 02:42

score 4 · Answer 1 · answered Sep 19 '22 at 20:20

This can be done using a variation of the construction of the algebraic closure of a finite field $\mathbb{F}_q$. Namely, instead of adjoining roots of all irreducible polynomials we can fix a prime $\ell$ and adjoin only roots of irreducible polynomials of degree a power of $\ell$. Equivalently, in the algebraic closure we consider the increasing union of the subfields

$$\mathbb{F}_q \subset \mathbb{F}_{q^{\ell}} \subset \mathbb{F}_{q^{\ell^2}} \subset \dots .$$

This is the maximal $\ell$-extension of $\mathbb{F}_q$, and it is infinite and algebraic but not algebraically closed because it does not contain roots of irreducible polynomials of any other degrees except powers of $\ell$. (This is an exercise: more generally, $\mathbb{F}_{q^n}$ consists of $\alpha \in \overline{\mathbb{F}_q}$ whose minimal polynomials have degree dividing $n$.)

In Galois theory terms this is the subfield of $\overline{\mathbb{F}_q}$ corresponding to the quotient map $\widehat{\mathbb{Z}} \to \mathbb{Z}_{\ell}$, where $\widehat{\mathbb{Z}} \cong \text{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ is the profinite integers and $\mathbb{Z}_{\ell}$ is the $\ell$-adic integers.

I don't think $\ell$ needs to be prime. It just needs to be a natural number greater than or equal to $2$. Of course, picking it to be a prime will work. — Arthur, Sep 19 '22 at 21:29
Yes, that's right. We get the subfield of $\overline{\mathbb{F}q}$ corresponding to the quotient map $\widehat{\mathbb{Z}} \to \mathbb{Z}{\ell}$ where $\mathbb{Z}_{\ell}$ is the product of the $p$-adic integers over all $p \mid \ell$. — Qiaochu Yuan, Sep 19 '22 at 21:52
You don’t need Galois theory to answer your question from this response. Just read the first two paragraphs. — Arkady, Sep 20 '22 at 03:25

score 3 · Accepted Answer · answered Sep 19 '22 at 20:19

3

Consider the quadratic closure of any finite field that does not have characteristic $2$. This is an infinite field as no finite field of characteristic other than $2$ can be quadratically closed. See here. However it is not algebraically closed as the minimal polynomial of any element over the finite field has degree $2^n$ for some $n$.

answered Sep 19 '22 at 20:19

David Lui

6,295

Thanks for the answer (+1), but I don't understand the last argument... Can you please help me in that a little bit... – Sayan Dutta Sep 20 '22 at 04:11
1

@SayanDutta Any element in any algebraic extension has a minimal polynomial. In this particular extension all minimal polynomials have degree $2^n$ for some $n$. Yet there are elements in the algebraic closure whose minimal polynomial have degrees like $3$ or $10$. Thus this extension cannot be the algebraic clusure. – Arthur Sep 20 '22 at 07:14

Prove that there exists an infinite field $F$ such that $F$ is algebraic over a finite field, and $F$ is not algebraically closed

2 Answers2