A proof needed in minimal polynomial.

Question

How do you prove this statement?

If a, b are different prime numbers, the minimal polynomial of $\sqrt[n]{b}$ over the extension field $\mathbb{Q} (\sqrt[m]{a})$ of the rational number field $\mathbb{Q}$ is $g(y)=y^{n}-b$. $(m,n\in \mathbb{N}, and\ m,n\geq2)$ (marked as equation ① in following)

The obstacle here is how to use following equation to launch contradiction according to the equation ①：

$\left ( \sqrt[n]{b}\right )^{t} = l_{1}+l_{2}\cdot \sqrt[m]{a}+\cdots+l_{m}\cdot \sqrt[m]{a^{m-1}}\ (l_{i}\in\mathbb{Q})$(marked as equation ② in following)

If equation ② is satisfied,then it can be inferred that n exact division m.

In isolation this could be an ok question. I do have a few reservations. It is recommended that you should give a bit more context in a question that looks like it might be a homework assignment (irrespective of whether it is!). Our guide for new askers has been designed to give a few pointers. For example, here some solvers may want to apply what they know about ramification of prime ideals in an extension of number fields. Given that you tagged the question with [tag:algebraic-number-theory] it is not unreasonable. — Jyrki Lahtonen, May 07 '19 at 14:25
(cont'd) However, earlier in a suitable course you might still be expected to do without and/or simply apply what you know about the theory of field extensions in general. Please, tell us such bits! That way answerers can take your background into account! Another possible reservation is that we might have already handled a question like this on our site. I cannot make a promise, but I also want to encourage everybody to spend a while searching. — Jyrki Lahtonen, May 07 '19 at 14:28
Seaching may be difficult without former experience on the site, so our expectations are reasonable. You seem to know a bit of LaTeX already. That's great. Unfortunately the standard search engines cannot grok TeX. Approach0 may help. But learning that is not the highest priority! — Jyrki Lahtonen, May 07 '19 at 14:30
Context can be added in a variety of ways: What makes the problem interesting? What makes it difficult? What level of study motivates the problem? Etc. — hardmath, May 07 '19 at 14:38
Here's how using discriminants may help. More hopefully generalizable ideas for a special case. More quality answers but still only a special case. — Jyrki Lahtonen, May 07 '19 at 15:19
Other primes for a change :-). Looks like I'm using wrong buzzwords, sorry :-( — Jyrki Lahtonen, May 07 '19 at 15:24
Do you know about Kummer theory ? If yes, I could provide an answer. — nguyen quang do, May 12 '19 at 20:40

A proof needed in minimal polynomial.

How do you prove this statement?

0 Answers0