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Suppose that $q$ is a rational prime, and $K$ is a $p$-extension over $\Bbb{Q}$ where $p$ is prime, then $q$ ramifies in $K$ if and only if $q$ is $p$ or $q$ is congruent to $1 \mod p$. I don't know why this is true, can someone convince me of this?

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3 Answers 3

If $K/\mathbb{Q}$ is a degree $p$ extension of $\mathbb{Q}$ then any particular prime can ramify: consider $K=\mathbb{Q}(q^{1/p})$. Suppose that $K$ is a Galois and so cyclic degree $p$ extension. Then only finitely many primes ramify in $\mathbb{Q}$ and they can only have the form $q=p$ or $q\equiv1$ (mod $p$). (I'm sure this is the result you are thinking of).

The same is true for $p$-power degree Galois extensions (consider the ramification groups at their localizations).

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This can't possibly be true: by a famous theorem of Dirichlet, there are infinitely many prime numbers $q$ such that $q \equiv 1 \pmod p$. However, only finitely many primes ramify in any finite extension $K/\mathbb{Q}$: those which divide the discriminant.

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For the case $K=\mathbb{Q}(q^{1/p})$ a $p$-extension over $\mathbb{Q}$ the discriminant of $K/\mathbb{Q}$ is $p^{3}q^{p-1}$ so both $p$ and $q$ ramify. I don't understand why $q$ should be congruent to 1 mod $p$.

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By definition a "p-extension" is Galois. This is crucial in the proof of the result you are referencing. –  Cam McLeman Jan 2 '12 at 16:10

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