Simple field extension inequality proof

Question

Let $\alpha \in \mathbb{C}$ be algebraic over $\mathbb{Q}$ and $F\subseteq \mathbb{C}$ be a subfield.

Prove that $[F(\alpha):F]\leqslant [\mathbb{Q}(\alpha):\mathbb{Q}]$.

This looks like a problem in which I'm supposed to use the tower law, but I keep getting stuck.

I know $\mathbb{Q}\subseteq\mathbb{Q}(\alpha)\subseteq\mathbb{C}$ and $F\subseteq F(\alpha)=\mathbb{Q}(F\cup \left \{ \alpha \right \})\subseteq\mathbb{C}$ but the fact that $[\mathbb{C}:\mathbb{Q}]=\infty$ was proving quite problematic.

I then thought about using the fact that $[\mathbb{Q}(\alpha):\mathbb{Q}]=deg(f)$ and $[F(\alpha):F]=deg(g)$ where $f$ is the minimum polynomial of $\alpha$ over $\mathbb{Q}$ and $g$ is the minimum polynomial of $\alpha$ over $F$. But, we've only just been introduced to the minimum polynomial and I'm having trouble using it. I know $f$ is irreducible and such that $\mathbb{Q}(\alpha)\cong \mathbb{Q}[x]/ \left \langle f \right \rangle$ and $g$ is irreducible and such that $F(\alpha)\cong F[x]/ \left \langle g \right \rangle$ but I can't see where to go from here. Can anyone help?

Thanks in advance!

Answer 1

Hints:

We do know that $\;\Bbb Q\subset \Bbb F\subset\Bbb C\;$ , and suppose $\;f(x)\in\Bbb Q[x]\;$ is the minimal polynomial of $\;\alpha\;$ over the rationals, and $\;g(x)\in\Bbb F[x]\;$ is the min. pol. of $\;\alpha\;$ over $\;\Bbb F\;$ , thus

$$\Bbb Q[x]\subset\Bbb F[x]\implies f(x),g(x)\in\Bbb F[x]\implies g(x)\mid f(x)\;\;\text{within}\;\;\Bbb F[x]\;\text{(why?)}$$

and from here that

$$...=\deg g\le\deg f=...$$

Answer 2

Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $g$ the minimal polynomial over $F$.

Note that $f$ is a polynomial with coefficients in $F$ (because $\mathbb{Q}\subset F\subset \mathbb{C}$) with $f(\alpha)=0$.

Since $g$ is the minimal polynomial over $F$, we get that $f$ must divide $g$.

In particular, $[F(\alpha):F]=\deg f\le \deg g=[\mathbb{Q}(\alpha):\mathbb{Q}]$.

Answer 3

First of all, you've written something incorrect; it should be $$\mathbb{Q}(\alpha)\cong\mathbb{Q}[x]/(f),\qquad F(\alpha)\cong F[x]/(g)$$ Obviously you cannot mod out a ring (e.g., $\mathbb{Q}$) by something that is not an element of it (e.g., $f$).

Now that that's cleared up, you should use the fact about (or definition of) the minimal polynomial $g$ that $h\in F[x]$ has $\alpha$ as a root if and only if $g\mid h$. Note that $\mathbb{Q}[x]\subset F[x]$, so that $f\in F[x]$.