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Let $L = K(a)$ be an algebraic extension. Let $E \subset L$ be a sub-field containing $K$. Let $m_a(x) = x^n + b_1x^{nāˆ’1} + \cdots + b_n$ be the minimal polynomial of $a$ over $E$. Prove that $E = K(b_1, \ldots , b_n)$.

Here $K \subset E \subset L$.

$K(b_1, \ldots , b_n) \subset E$ is trivial; we have to prove the other inclusion.

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    Use the fact that if $p(x) \in F[x]$ is the minimal polynomial of some $a\in \overline{F}$, and $q(x) \in F[x]$ is another polynomial for which $q(a)=0$, then $p(x)\mid q(x)$. ā€“ 2017-01-28

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Let $f(x)$ be the minimal polynomial of $a$ over $K(b_1,\ldots,b_n)$. Since $f(x)\in E[x]$ vanishes at $a$, we have that $m_a(x)$ divides $f(x)$. On the other hand, we also have that $m_a(x) \in K(b_1,\ldots,b_n)[x]$ vanishes at $a$. Thus $f(x)$ divides $m_a(x)$, and then $f(x)=m_a(x)$. Therefore $$[L: K(b_1,\ldots,b_n)]=\deg f(x)=\deg m_a(x)=[L: E]$$ and $$K(b_1,\ldots,b_n)\subseteq E \subseteq L$$

gives $E=K(b_1,\ldots,b_n)$.