Does multiplying two intermediate fields give a subfield?

Question

I have the following inquiry : If $K$ is a field extension of $k$, and $F_1$ and $F_2$ are intermediate fields such that $k \subseteq F_1 \subseteq K$ and $k \subseteq F_2 \subseteq K$, is the set $F_1F_2 = \{a_1a_2 : a_1 \in F_1, a_2 \in F_2\}$ always a subfield of $K$ ? It seems to me that this is true. If not, what would be a counterexample to this?

Here are some thoughts about searching for a possible counterexample:

• I don't think a counterexample will arise out of letting $K = \mathbb{C}$. $\mathbb{C}$ is in some sense the "largest" field, so it seems any intermediate fields $F_1$ and $F_2$ will still give you a subfield $F_1F_2$ of $\mathbb{C}$, regardless of what $k$ is.
• A counterexample cannot arise out of letting $K = \mathbb{Q}$, since $\mathbb{Q}$ does not contain any proper subfields.
• The first counterexample I tried was to let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, so that it has intermediate subfields, for example, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$, but no such finite extension of $\mathbb{Q}$ by adjoining irrationals seems to break the claim.
• I next thought that perhaps $K = \mathbb{Q}$ could provide a counterexample. But, $\mathbb{R}$ is also fairly "large" in some sense, so it seems the only way that a counterexample could arise is if multiplying two subfields somehow gave us $\mathbb{C}$. But I don't know of any way to achieve this.

I'm starting to suspect that a counterexample may lie in letting $K$ be some extension of a finite field, like $\mathbb{F}_4 = GF(2^2)$. But, how to construct such an example ?

Thanks!

Answer 1

Here's an example for $F_1,F_2$ finite extensions of $k$. Take $k = \mathbb{Q}, F_1 = \mathbb{Q}(\sqrt{2}), F_2 = \mathbb{Q}(\sqrt{3}), K = \mathbb{C}$. I claim $F_1F_2$ is not a field. Indeed, $\sqrt{2} = \sqrt{2}\cdot1 \in F_1, \sqrt{3} = \sqrt{3}\cdot1 \in F_2$, but $\sqrt{2}+\sqrt{3} \not \in F_1F_2$ since if $\sqrt{2}+\sqrt{3} = (a+b\sqrt{2})(c+d\sqrt{3})$ for $a,b,c,d \in \mathbb{Q}$, then we must have $bd = 0$; if $b = 0$, then we'd need $ac+ad\sqrt{3} = \sqrt{2}+\sqrt{3}$, impossible, and similarly if $d = 0$.

Answer 2

Let $F_1$ be the field of rational functions in the variable $x$ (over any field) and $F_2$ the rational functions in the variable $y$. They are both subfields of the field of rational functions in $x$ and $y$.

$x$ and $y$ are in $F_1F_2$ but $x+y$ is not, so $F_1F_2$ is not a field.

This isn't an answer if you're looking for a finite extension of the base field.