Let $F\subset E\subset K$. Let $\alpha\in K$ be algebraic both over $F$ and $E$.

I want to show that $\alpha\in K/F$ be separable means $\alpha\in K/E$ is separable.

As an intermediate result, I'm trying to show that if $p(x)\in F[x]$ and $q(x)\in E[x]$ are the minimal polynomials, then $q(x) \mid p(x)$.

Intuitively, I feel this should be true, since $F$ lies inside $E$ and therefore its minimal polynomial should be some reduced (or not) form of $p(x)$. However, I don't know how to go about this.

Edit: The minimal polynomial for $\alpha\in E/F$ is the unique monic irreducible polynomial $p\in F[x]$ with $\alpha$ as a root.

  • 1
    $\begingroup$ What is your definition of "minimal polynomial"? $\endgroup$ Mar 4, 2022 at 2:40
  • $\begingroup$ Will update post with this info. But, the minimal polynomial of $\alpha$ is the unique irreducible monic polynomial containing $\alpha$ as a root. $\endgroup$
    – user975734
    Mar 4, 2022 at 2:42
  • $\begingroup$ And as to definitions, to clarify your own ideas, you might look into the definition(s) of separability of an element over a field. $\endgroup$
    – Lubin
    Mar 4, 2022 at 5:02

1 Answer 1


For $\alpha\in K$ and $F\subseteq K$, I would define the minimal polynomial of $\alpha$ over $F$ to be the unique monic polynomial $p\in F[x]$ such that (1) $p(\alpha) = 0$ and (2) for all $q\in F[x]$ such that $q(\alpha) = 0$, $p\mid q$.

With this definition, the result you want is obvious: Suppose $p$ is the minimal polynomial of $\alpha$ over $F$ and $q$ is the minimal polynomial of $\alpha$ of $E$. Since $F\subseteq E$, $p\in E[x]$. And since $p(\alpha) = 0$, $q\mid p$.

Ok, why is your definition equivalent to mine?

Let $p$ be a monic irreducible polynomial in $F[x]$ with $\alpha$ as a root. Suppose $q\in F[x]$ and $q(\alpha) = 0$. Let $r = \gcd(p,q)$. Since $r$ can be written as a linear combination of $p$ and $q$, $\alpha$ is a root of $r$, so $r\neq 1$. Since $p$ is irreducible, $r = p$. Thus $p\mid q$. So $p$ is the minimal polynomial of $\alpha$ over $F$.

Conversely, suppose $p$ is the minimal polynomial of $\alpha$ over $F$ (by my definition). If $p$ were reducible, we could write $p = qr$ where both $q$ and $r$ have degree less than $\deg(p)$. Since $\alpha$ is a root of $p$, it is a root of $q$ or of $r$, say $q(\alpha) = 0$ without loss of generality. Then $p\mid q$, contradicting the fact that $\deg(q)<\deg(p)$.


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