# Minimal polynomials in a tower of field extensions

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.

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

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 $$p(\alpha) = 0$$, $$q\mid p$$.
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 $$\deg(q)<\deg(p)$$.