Normal field extensions

Question

Let $E/F$ be a finite extension and $F \leqslant K \leqslant E$ where $K/F$ is a normal extension.If $a,b \in E$ have the same minimal polynomial over $F$ and for some $n \in \mathbb{N}$ , $a^n \in K$ then prove that $b^n \in K$.

Here is my approach:

If $a^n \in K $ then $a^n=s \in K$ thus $a$ is a root of $f=X^n-s \in K[X]$

Also $a$ is the root of $min(a,F)$

We konw that $min(a,F)$ is irreducible and it cannot be coprime with $f$ because both of them have $a$ as a root,thus $min(a,F)|f$.

But $b$ is also a root of $min(a,F)$ and $min(a,F)$ divides $f$ thus $b$ is a root of $f$.

There from hypothesis that $K/F$ is a normal extension we have that $b \in K$ and $b^n \in K$($K$ is a field)

Is this proof valid or do i have to take a different approach?

Thank you in advance!

Answer 1

Your proof has two unwanted side affects: firstly that $b\in K$; secondly that $a^n = b^n$, since both are roots of $X^n-s$. Neither of these assumptions is true. For example, take $F=\mathbb Q, K=\mathbb Q(\sqrt2)$ and $E = \mathbb Q(i,\sqrt[4]2)$. Both $a=\sqrt[4]2$ and $i\sqrt[4]2$ have the same minimal polynomial over $F$ and $a^2\in F$, but $a^2\ne b^2$.

There are two problems here:

1. Even though $K/F$ is a normal extension, this doesn't tell you that $b\in K$. In order to deduce this, you would have to know that $f$ is irreducible (which it need not be), and that it has a root in $K$ (which it might not have).

However, from what you've written, you can easily deduce that $b^n\in K$: since $b$ is a root of $f$, in particular, $b^n =s\in K$. The question now becomes: where did you need the assumption that $K/F$ was normal? This leads to the second problem:

2. You can't simply deduce that $\min(a,F)\mid f$. The problem is that whilst $\min(a, F)$ is irreducible over $F$, it need not be irreducible over $K$.

In the above example, $\min(a,F) = X^4-2$ which splits as $(X^2+\sqrt2)(X^2-\sqrt2)$ in $F$. Here, $f=X^2-\sqrt2$, so certainly $\min(a,F)\not\mid f$.

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Here's what I would do instead. Since $a,b\in E$ have the same minimal polynomial over $F$, there is an $F$-automorphism $\sigma$ of $E$ such that $\sigma(a) = b$. So $\sigma(a^n) =b^n$.

Since $K$ is normal, the restriction $\sigma|_K$ is an $F$-automorphism of $K$. Since $a^n\in K$, it follows that $b^n=\sigma|_K(a^n)\in K$.