$[F(a):F] \leq [\mathbb{Q}(a):\mathbb{Q}]$ - proof?

Question

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

I'm trying to prove that $[F(a):F] \leq [\mathbb{Q}(a):\mathbb{Q}]$ (where $F(a) := \mathbb{Q}(F\cup\left\{ {a}\right\}))$ and deduce from this that the set of complex numbers which are algebraic over the rationals is a subfield of $\mathbb{C}$.

For the first part, I guess $a$ has to be algebraic to ensure the two degrees are finite. Now as $F$ is a field, certainly $\mathbb{Q} \subseteq F$. Then I've tried thinking in terms of minimal polynomials but I get completely confused and dont really see how to provide a rigorous proof... Also, I don't really see how I could use such a result to show the last part.

I've been going round and round about this and got no result, could you hint me please? It would really help me as I'm only starting to read about field extensions and am not familiar with them!

Thank you!

Answer 1

If $K\subset \Omega$ is a field extension, and $a \in \Omega$, then

$$[K(a) : K] = \left\{ \begin{matrix} \mbox{degree of the minimal polynomial of $a$ over $K$} & \mbox{if $a$ is algebraic} \\ \infty & \mbox{otherwise} \end{matrix} \right. $$

Given this preliminary lemma, the exercise now is trivial.

If $a \in \mathbb{C}$ is algebraic (over $\mathbb{Q}$), then it is algebraic over $F$. Let $m \in \mathbb{Q}[X]$ be its minimal polynomial over $\mathbb{Q}$ , and $n \in F[X]$ be its minimal polynomial over $F$.

But $\mathbb{Q}[X] \subset F[X]$, so $n$ divides $m$, and in particular its degree is smaller.

As for the second part: let $a,b \in \mathbb{C}\setminus \{ 0\}$ algebraic. Then $a+b, a-b, ab, a/b \in \mathbb{Q}(a,b)$ We want to show that they are algebraic. It is sufficient to show that $[\mathbb{Q}(a,b) : \mathbb{Q}] < \infty$. And now $$[\mathbb{Q}(a,b) : \mathbb{Q}] = [\mathbb{Q}(a,b) : \mathbb{Q}(a)] \cdot [\mathbb{Q}(a) : \mathbb{Q}] \leq [\mathbb{Q}(b) : \mathbb{Q}] \cdot [\mathbb{Q}(a) : \mathbb{Q}]$$ hence it is finite.

Answer 2

Let $\;P_F(x)\,,\,\,P_{\Bbb Q}(x)\;$ be the minimal polynomial of $\;a\;$ over $\;F\,,\,\,\Bbb Q\;$ , resp., so that

$$[F(a):F]=\deg P_F(x)\;\;,\;\;\;[\Bbb Q(a):\Bbb Q]=\deg P_{\Bbb Q}(x)$$

But we certainly have that also

$$\;P_{\Bbb Q}(x)\in F[x]\;\implies\;\; P_F(x)\mid P_{\Bbb Q}(x)\implies \deg P_F(X)\le\deg P_{\Bbb Q}(x)$$

adn we're done.