Let K/F an extension if $b\in K $ is algebraic of degree n over $F$ then $[F(b):F]=n$

Question

If $b\in K $is algebraic of degree n over $F$ then $[F(b):F]=n$

My try

I call $m(x)$ the minimal polynomial of $b$ in $F[x]$ so $m(b)=0 $

Then for any $p(x) \in F[x]$ where $b$ is a root

$m(x) $ divides $p(x)$

so $p(x)=m(x)q(x)$ with $q(x)\in F[x]$ and degree of $m(x)=n$ but im stuck here

Answer 1

Exercise: Take $f \in F[x]$ and call $n$ the degree of $f$. Prove that $K := F[x]/(f)$ is an $n$-dimensional $F$-vector space. Moreover, the cosets of $1,x,\dots,x^{n-1}$ form a basis for it. Conclude that if $f$ is irreducible, then $K$ is a field extension of $F$ of degree $n$.

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Now, for your problem, consider the ring homomorphism $\operatorname{ev}_b \colon F[x] \to F(b)$ that evaluates a polynomial at $b$. From what you said, $\ker(\operatorname{ev}_b) = (m)$. Thus, the image $\operatorname{im}(\operatorname{ev}_b)$ [being isomorphic to $F[x]/(m)$] is a field extension of $F$ of degree $n$, and since it contains $\operatorname{ev}_b(x) = b$, it follows that $\operatorname{im}(\operatorname{ev}_b) = F(b)$.