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

  • 1
    $\begingroup$ So you want to prove $\{1,\dots,b^{n-1}\}$ is a $F$-basis for $F(b)$. $\endgroup$
    – Kenta S
    Sep 6, 2022 at 2:21
  • $\begingroup$ This fact is all the textbooks explaining algebraic extensions. No need to discuss it any more here in my opinion. We have covered it also. For example here. Search for more, if the answers there don't work for you. Or modify the question, and explain, what is giving you difficulties. $\endgroup$ Sep 7, 2022 at 4:55

1 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 $n$.

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 $\operatorname{im}(\operatorname{ev}_b) = F(b)$.


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