$\alpha$ algebraic over field $F$ implies $[F(\alpha):F]=\text{degree of minimal polynomial over $\alpha$}$

Question

Let $\alpha$ be some element algebraic over a field $F$. Then $F(\alpha)$ is isomorphic to $F[x]/\langle m_{\alpha,F}\rangle$, where $m_{\alpha,F}$ is the minimal polynomial with root $\alpha$ over $F$. Moreover, $[F(\alpha) : F] = \deg(m_{\alpha,F})$.

The first part of this theorem appears to be clear, but what about the second part (after the word "moreover")? Why does the index of $F(\alpha)$ over $F$ equal the degree of the minimal polynomial in $F[x]$ with root $\alpha$? I think this should be almost obvious, but perhaps I'm lacking some theory.

Answer 1

Given a field extension $K/F$, the degree of the extension $K/F$ is defined as the dimension of $K$ viewed as a vectorial space over $F$. In other words $$[K:F]=\dim_FK.$$

For your question, I let you to prove (if you haven't already proved it) that if $\alpha$ is algebraic over $F$, then $$F(\alpha)=\{g(\alpha): g(x)\in F[X]\; \text{and}\; g\equiv 0\; \text{or}\; \deg(g)<\deg(m_{\alpha,F})\}.$$

Now let's set $m_{\alpha,F}=p$, then if $p(x)=a_0x^n+\cdots +a_{n-1}x+a_n$ we claim that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a basis for $F(\alpha)$ over $F$. Indeed, every element of $F(\alpha)$ has the form $b_0\alpha^{k}+\cdots +b_{n-1}\alpha+b_n$ for some $ g(x)=b_0x^k+\cdots +b_{n-1} x+b_n\in F[x] $ with $k<n$, so it follows easily that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a generating set of $F(\alpha)$ as a vectorial space over $F$. To prove that the above set is l.i. we write $$c_0+c_1\alpha+\ldots c_{n-1}\alpha^{n-1}=0.$$ Then $q(x)=c_0+c_1x+\ldots c_{n-1}x^{n-1}\in F[x]$ satisfies $q(\alpha)=0$, so either $q\equiv 0$ or $p(x)\mid q(x)$. But the later is impossible because that would mean that $n=\deg(p)\le deg(q)=n-1$, an absurd. Therefore $q\equiv 0$, which implies that $c_0=c_1=\ldots =c_{n-1}=0$.

Hence $\{1,\alpha,\ldots \alpha^{n-1}\}$ is l.i. and thus it's a basis for $F(\alpha)$ over $F$. From this is immediate that $$[F(\alpha):F]=n=\deg(p)=\deg(m_{\alpha,F}).$$

Answer 2

The minimal polynomial gives you a way of expressing $\alpha^n$ (n the degree of the minimal polynomial) as a combination of the other powers of $\alpha$ and the coefficients of the minimal polynomial. Therefore the dimension of $F(\alpha)$ over $F$ is not infinite.

Answer 3

It seems the other answer and comment are about $F[\alpha]$ being a $F$-vector space of dimension $deg(p)$ ($F[\alpha]$ is the smallest ring containing $F$ and $\alpha$, i.e. $F[\alpha] = \{ \sum_{n =0}^d c_n \alpha^n, c_n \in F\}$).

For showing that $F[\alpha] = F(\alpha)$ you need to prove that $\varphi : F[x]/(p(x)) \to F[\alpha],\ \ \varphi(f(x)) = f(\alpha)$ is an isomorphism of rings ($\varphi$ is clearly surjective, and it is injective by definition of the minimal polynomial)

Hence $F[\alpha]$ is a field, so $F[\alpha] = F(\alpha)$ and $[F(\alpha):F]= deg(p)$