# Minimal polynomial of an element over a subfield when the degrees of extensions are coprime

This is a question from D.J.H. Garling's "A course in Galois Theory", and I am struggling to find an answer. Can anyone help me?

Suppose $K\leq L\leq L(\alpha)$ are field extensions, and that $|L(\alpha):L|$ and $|L:K|$ are relatively prime. Show that the minimal polynomial of $\alpha$ over $L$ has its coefficients in $K$.

Let $|L(\alpha):L| = n$ and $|L:K| = m$ where $(m, n) = 1$. I can prove that the minimal polynomial of $\alpha$ over $K$ must have degree which is a multiple of $n$. Hence, $|K(\alpha):K|=nt$ for some integer $t$. Let $|L(\alpha):K(\alpha)|=s$.

Now by the tower law, we have $nts = mn \implies ts =m$.

If I can show that $t=1$, then we are done. But why can't we have $ts=m$ where $t$ and $s$ are proper factors of $m$?

I know I need to use the fact that $(m,n) = 1$ somewhere, but I can't see where.

It's easy if we are given two quantities on 'different sides of the ladder' as coprime, but here $m$ and $n$ are on the same side of the ladder. If this is a typo in the book, is there a counterexample?

• Did you mean Galois extensions ? Jan 5 '17 at 13:14
• Not necessarily. Just finite extensions. Jan 5 '17 at 13:24
• @Globe Theatre I know your post is already older, but I stumbled across the same problem in Garling's and in my version its $[K(a):K]$ and $[L:K]$ are coprime, not $[L( a):L$ and $[L:K]$ . Do you have any hints for this question? Aug 25 '17 at 11:23
• GlobeTheatre, I started a new post with the corrected version of Garling's question here: math.stackexchange.com/questions/2405708/… Aug 25 '17 at 14:53

Try $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha=\sqrt[6]{2}$.

Another counterexample to the exercise as written is to take $$\alpha \in L \setminus K$$. Then $$|L(\alpha):L|=1$$ is relatively prime to everything, but the minimal polynomial for $$\alpha$$ over $$L$$ is simply $$x - \alpha$$ which clearly does not have coefficients in $$K$$.

I think it is a typo and that Garling meant $$|K(\alpha):K|$$ and not $$|L(\alpha):L|$$. Another statement for the exercise, which is probably more to the point and has essentially the same proof, is:

Suppose that $$K \subset L \subset L(\alpha)$$ are field extensions, and that $$|K(\alpha):K|$$ and $$|L:K|$$ are relatively prime, then $$|L(\alpha):L| = |K(\alpha):K|$$.

A proof would be:

Since $$|K(\alpha):K|$$ and $$|L:K|$$ are relatively prime, and both divide $$|L(\alpha):K|$$, their product must divide $$|L(\alpha):K|$$, implying $$|L(\alpha):K| \geq |K(\alpha):K| \cdot |L:K|$$.

But then, the general fact that $$|K(\alpha):K| \geq |L(\alpha):L|$$ implies that $$|K(\alpha):K| \cdot |L:K| \geq |L(\alpha):L| \cdot |L:K| = |L(\alpha):K|$$.

So the inequalities collapse, forcing $$|L(\alpha):L| = |K(\alpha):K|$$.

The exercise in this form can be used, for example, in exercise 5.9 to show that the positive $$p$$-th roots of $$2$$, as $$p$$ varies over the primes, are linearly independent over $$\mathbb Q$$.