Does there exist a field which has infinitely many subfields? Does there exist an enormous supply of such fields?

I don't know how to begin.

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    $\begingroup$ $\mathbb Q(\pi)$. $\endgroup$ – Andrés E. Caicedo Jul 17 '13 at 5:00
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    $\begingroup$ A good one @Andres! The only "shortcoming" being that all the intermediate fields are isomorphic to each other (Lüroth). Not that the example in my answer would be any better in that respect :-) $\endgroup$ – Jyrki Lahtonen Jul 17 '13 at 7:27

The complex numbers $\mathbb{C}$ is an example of such a field. It has infinitely many subfields, since you can adjoin family of irrational numbers (pick your favourite ones!) to $\mathbb{Q}$. My favourites would be roots of $\sqrt[n]{2}$ for each $n\in\mathbb{N}$. So in this case, the infinite family of subfields would be $\{\mathbb{Q}(\sqrt[n]{2})\}_{n=1}^{\infty}$


Recall the notion of prime subfield $K_0$ of a field $K$: this is the unique smallest subfield of $K$. If $1+\ldots + 1 = 0$ for some $n$, then the least such $n$ is a prime number $p$ and the prime subfield is $\mathbb{F}_p$. Otherwise the prime subfield is the rational numbers $\mathbb{Q}$. Now here is the full answer to your question:

A field $K$ has finitely many subfields iff $K/K_0$ has finite degree.

Proof: $K/K_0$ has finite degree, then $K$ is either a finite field or a number field. The first case is obvious; the second case can be done e.g. by replacing $K$ with its Galois closure over $\mathbb{Q}$ (showing that a larger field has only finitely many subfields will be sufficient!) and using the Galois correspondence.

Now suppose $K/K_0$ has infinite degree.

Case 1: If $K/K_0$ is algebraic, then choose $x_1 \in K \setminus K_0$. Then $K_0(x_1)$ has finite degree over $K_0$ so is proper in $K_1$. Repeating this argument generates an infinite ascending chain of finite degree subfields of $K$.

Case 2: If $K/K_0$ is not algebraic, let $t \in K$ be transcendental over $K_0$. Then $\{K_0(t^n)\}_{n=1}^{\infty}$ are infinitely many subfields of $K$.

Afterthought: Let $L/K$ be any field extension, and consider the lattice $\operatorname{Sub}(L/K)$ of subextensions, i.e., fields $F$ with $K \subset F \subset L$. Then the OP's original question asks about when for a field $K$, $\operatorname{Sub}(K/K_0)$ is infinite, and my answer characterizes when this happens. But upon further reflection, in each of Cases 1 and 2 I am showing the failure of a different weaker finiteness property. Namely, if $K/K_0$ is infinite degree algebraic, then $\operatorname{Sub}(K/K_0)$ contains an infinite ascending chain of subfields, so it is not Noetherian (this is just the definition of a Noetherian partially ordered set: it doesn't contain infinite ascending chains). If $K/K_0$ is transcendental, then $\operatorname{Sub}(K/K_0)$ contains an infinite descending chain -- if $t$ is transcendental over $K_0$, take $F_n = K_0(t^{2^n})$ -- so it is not Artinian (again, by definition).

In fact this part of the answer doesn't use any special properties of $K_0$. In fact it shows that if $L/K$ is infinite degree algebraic then $\operatorname{Sub}(L/K)$ is not Noetherian, and if $L/K$ is transcendental then $\operatorname{Sub}(L/K)$ is not Artinian. (The converse works when $K_0$ is perfect but not in general; see Jyrki's answer and my comments on it below.)

You can check that it doesn't work the other way around.

Claim: Let $K_0$ be any prime field (i.e., $\mathbb{F}_p$ or $\mathbb{Q}$). Then:
a) There is an infinite algebraic field extension $K/K_0$ such that $\operatorname{Sub}(K/K_0)$ is Artinian.
b) There is a transcendental field extension $K/K_0$ such that $\operatorname{Sub}(K/K_0)$ is Noetherian.

I leave the verifications to you (if you care to do them), with the following hints: a) take a $\mathbb{Z}_p$-extension. b) Use Luroth's Theorem.

Maybe we should ask for characterizations of field extensions $L/K$ such that $\operatorname{Sub}(L/K)$ is Noetherian, or Artinian?

Note that Jyrki's answer also fits into this framework: he shows that $L/K$ having finite degree need not imply that $\operatorname{Sub}(L/K)$ is finite (although it is evidently Noetherian and Artinian, i.e., of finite length). As I like to mention, I got a PhD in arithmetic geometry / number theory before I learned about this example: up until not so very long ago I was sure that basic field theory showed that such examples could not exist.

  • $\begingroup$ I like complete answers :) Thanks for teaching me this theorem that I didn't know (+1) $\endgroup$ – Prism Jul 17 '13 at 5:07
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    $\begingroup$ @Prism: I didn't know the "theorem" until about 10 minutes ago either. $\endgroup$ – Pete L. Clark Jul 17 '13 at 5:12
  • $\begingroup$ I checked your field theory notes, and quick search showed that it wasn't there. I then concluded that you had some secret theorems you hid from us :) Didn't know you figured that out instantaneously... $\endgroup$ – Prism Jul 17 '13 at 5:14

Adding an example of infinitely many intermediate fields of a finite extension $L/K$.

Let be $F$ be an algebraic closure of $\mathbb{F}_2$, so $F$ is infinite. Let $x$ and $y$ be algebraically independent trascendental elements over $F$. Let $L=F(x,y)$ and $K=F(x^2,y^2)$. We easily see that $[L:K]=4$. A basis for this extension consists of $\{1,x,y,xy\}$. Let $\alpha\in F$ be arbitrary. Consider the intermediate space $L_\alpha$ spanned over $K$ by $\{1,x+\alpha y\}$. We have $$ L_\alpha\cap L_{\alpha'}=K $$ whenever $\alpha\neq\alpha'$ (linear algebra), so the spaces $L_\alpha$ are distinct. The key is that $L_\alpha$ is also a subfield. This is because $$ (x+\alpha y)^2=x^2+2\alpha xy+\alpha^2y^2=x^2+\alpha^2y^2\in K. $$

This does not contradict the argument in Pete L. Clark's answer, because the extension $L/K$ is purely inseparable, and thus cannot be contained in a Galois extension. And that was the key in his argument in the case of number fields.

  • $\begingroup$ A similar construction is available in characteristic $p>2$ as well. $\endgroup$ – Jyrki Lahtonen Jul 17 '13 at 7:24

Let $K$ be any field and let $X$ be any infinite set. For each choice of subset $Y\subseteq X$, we get a different subfield $K(Y)\subseteq K(X)$.

Let $K$ be any field and let $t$ be an indeterminate. For each choice of integer $n\geq 1$, we get a different subfield $K(t^n)\subseteq K(t)$.

In case you're not familiar with it, here is the definition of $K(X)$: it is the field of fractions of $K[X]$, the ring of polynomials in the indeterminates $X$ over $K$. That is, the elements of $K[X]$ are polynomials where the coefficients are elements of $K$, and the "variables" or "indeterminates" are the elements of $X$. Thus for example, if $K=\mathbb{Q}$ and $X=\{s,t\}$, then $K[X]=\mathbb{Q}[s,t]$ consists of polynomials such as $\frac{1}{2}+5s+t^2+3st$, and $K(X)=\mathbb{Q}(s,t)$ consists of rational functions such as $$\frac{\frac{1}{2}+5s+t^2+3st}{7-3s^3+4st^2}.$$

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    $\begingroup$ This is surely the most conceptually economical answer, and also the most elementary. $\mathbb{C}$ is more natural and more mathematically important, but proving that adjoining algebraically independent irrational numbers gives infinitely many distinct subfields is a bit more work than simply observing that $K(Y) \neq K(Y')$ for distinct $Y, Y' \subseteq X$. $\endgroup$ – Peter LeFanu Lumsdaine Jul 17 '13 at 14:11
  • $\begingroup$ You should probably make the assumption that $X$ is an infinite algebraically independent set. $\endgroup$ – tomasz Jul 28 '13 at 14:21
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    $\begingroup$ @tomasz: Any set can be used as a set of indeterminates; there would be nothing wrong with using the set $X=\{2,5\}\subset\mathbb{Q}$ as a set of indeterminates over $\mathbb{Q}$, making a field $\mathbb{Q}(X)$ of transcendence degree $2$ over $\mathbb{Q}$ (though notationally it would certainly be confusing, since of course $\mathbb{Q}(2,5)=\mathbb{Q}$). It is this sense in which I intend the notation $K(X)$. $\endgroup$ – Zev Chonoles Jul 28 '13 at 15:17

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