Suppose that $L(\alpha)/L/K$ and that $[K(\alpha):K]$ and $[L:K]$ are relatively prime.

Question

Suppose that $L(\alpha)/L/K$ and that $[K(\alpha):K]$ and $[L:K]$ are relatively prime. Show that the minimal polynomial of $\alpha$ over $L$ has its coefficients in $K$.

I tried an approach but I got stuck:

We have that the following field extensions: $L(\alpha)/L$ and $L/K$ and we have the following fact that $[L:K]$ and $[K(\alpha):K]$ are relatively prime.

From this: it is safe to assume that $[K(\alpha): K] < \infty$ thus $\alpha \in K$ is algebraic over K.

Also: $[L(\alpha): K] = [L(\alpha): L][L : K]$ and $[L:K] = [L:K(\alpha)][K(\alpha):K]$ (I feel like this isn't true because $[L:K]$ and $[K(\alpha):K]$ are relatively prime.)

Now I've tried to do things but I couldn't progress anywhere. How would I show that the minimal polynomial of $\alpha$ over $L$ has coefficients in $K$?

Answer 1

Assume $[L:K]$ and $[K(\alpha):K]$ are finite and relatively prime. The minimal polynomial $p$ of $\alpha$ over $L$ divides the minimal polynomial $q$ of $\alpha$ over $K$. The degree of $q$ is $[K(\alpha):K]$, so the degree of $p$ divides the degree of $q$. Furthermore, $\deg p=[L(\alpha):L]$, and since $K(\alpha)\subset L(\alpha),\deg q$ divides $[L(\alpha):K]$. Set $n=\deg q/\deg p$. Then $n$ divides $[L(\alpha):K]/[L(\alpha):L]=[L(\alpha):K]/\deg p=[L:K].$ But $n$ also divides $\deg q$, which is relatively prime to $[L:K]$, so $n=1$.

Answer 2

the basic numerical law for finite extensions is: $$ [L(\alpha):K] = [L(\alpha):K(\alpha)][K(\alpha):K] = [L(\alpha):L][L:K] $$ thus because of the data on relative primality: $$ [K(\alpha):K]|[L(\alpha):L] $$ and in particular, therefore: $$ [K(\alpha):K] \le [L(\alpha):L] \le [L(\alpha):K] $$ which makes the required point, since, because $K \subset L$ we must have $$ [L(\alpha):L] \le [K(\alpha):K] $$