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The question is as follows.

Let $K = \mathbb{Q}(\sqrt[m]{a},\sqrt[n]{b}) $, where $m,n,a,b$ are positive integers such that they are pairwise coprime. Assume that $[K:\mathbb{Q}]=mn$/ Prove that no prime numbers can totally ramify in $K/\mathbb{Q}$.

I assume we would need to find such prime numbers that are ramified in $\mathbb{Q}(\sqrt[m]{a})/\mathbb{Q}$, $\mathbb{Q}(\sqrt[n]{b})/\mathbb{Q}$ respectively. I know if $p|\text{disc}(L)$, then it is ramified. The discriminate can be found using the following:

- Euler's $\phi(n)$ function
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$$

\text{disc}(\alpha) = (-1)^{\frac{k(k-1)}{2}}\left[(-1)^{1-k}(k-1)^{k-1}c^k + k^kd^{k-1}\right]

$$

Where $f(x) = x^k + cx + d$. And since $a,b,n,m$ are pairwise coprime, then disc($K$)=disc$(\mathbb{Q}(\sqrt[m]{a}))^n$disc$(\mathbb{Q}(\sqrt[n]{b}))^m$.

Now if there is a prime number that will totally ramify in $K/\mathbb{Q}$, then there is only one ramification index, $e_i$, such that $e = mn = [K:\mathbb{Q}]$. All I would need to do is show that there is more than one ramification index via transitivity. This is where I am stuck. How would I go about showing there is more than one ramification index, such that no prime number that totally ramifies $K/\mathbb{Q}$.

Thanks in advance for any feedback.

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