This is question 4.5 of D.J.H Garling's Galois theory and this post I reckon already quoted it, but the autor quoted it differently (maybe there is some sort of typo in the copy of the book of the author), and no answer is provided. I thought that I would post it here "again", as the two questions are essentially different, and I'm struggling with finding an answer:

Suppose $L(\alpha):L:K$ is a field extension, and $[K(\alpha):K]$ and $[L:K]$ are relatively prime. Then the minimal polynomial of $\alpha$ over $L$ has has its coefficients in $K$.

Can anyone give me a hint on how to prove this? Help is much appreciated!

Edit: I forged some sort of proof, but I'm not sure whether it's correct:

Proof: For $[K(\alpha):K] = n$ and $[L:K] = m$ we have $(n,m) = 1$. Now denote with $l_\alpha$ the minimal polynomial of $\alpha$ over $L$; it is of the form $l_\alpha(t) = \sum_{j=0}^{r}\lambda_jt^j$ for some $r>0$ and $\lambda_j \in L$. Similarly, the minimal polynomial $k_\alpha$ of $\alpha$ over $K$ is of the form $k_\alpha(t) = \sum_{j=0}^{n-1} \kappa_jt^j$ where $\kappa_j \in K$. As $K\subseteq L$, clearly $k_\alpha \in (l_\alpha) \subseteq L[x]$. As $l_\alpha$ is minimal, we have $r = \deg(l_\alpha) \leq \deg(k_\alpha) = n$. Now note that $L(K(\alpha)) = L(\alpha)$, and $[L(\alpha):K] = [L(\alpha):L][L:K] = [L(K(\alpha)): K(\alpha)][K(\alpha):K]$, and hence $m[L(\alpha):L] = n[L(\alpha):K(\alpha)]$. As $m$ and $n$ are coprime, we conclude that $n$ divides $[L(\alpha):L]$ and therefore $n \leq \deg(l_\alpha)$, as $\deg(l_\alpha) = [L(\alpha):L]$. Putting together, this gives $\deg(l_\alpha) = n$ and we must hence have $l_\alpha = k_\alpha$ as $k_\alpha$ is the unique monic polynomial of smallest degree in $(l_\alpha)$ and is hence the minimal polynomial.

Are all of these steps valid? And if so, is there perhaps a more direct way to prove the proposition? Again, I will appreciate any critique or adivce! Thank you!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.