Ramification Definition Confusion

Ramification Definition Confusion
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I recently out of curiosity came across this paper summarizing Tate's thesis which, in its introductory section, claims that, given a number field $K/\mathbb{Q}$, a prime ideal $\wp$ of $O_K$ is ramified if, when it lies over the prime $p$ of $\mathbb{Q}$, $\wp^2\cap\mathbb{Z}\neq(p).$ But this clashes with other definitions of ramification I've seen, because I thought it was usually used to describe the primes of $\mathbb{Q}$, or in general the base field of the extension, rather than $K$.

Also, all other definitions I've seen seem to imply that, for example, the prime 2 ramifies in $\mathbb{Q}(i)/\mathbb{Q}$, as $(1+i)^2=(2)$, so that the prime ideal $(1+i)$ appears as a repeated factor in the factorization. However, by the definition given in this paper and the fact that $(1+i)^2=(2)$, the fact that $(2)\cap\mathbb{Z}=(2)$ would seem to imply that $(1+i)$ is not ramified in this situation.

I'm somewhat of a beginner to algebraic number theory, and I figure I must be making some sort of mistake or misinterpreting the definition. Are these just two definitions of differing concepts, or have I made an error?

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2 Answers

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Maybe there is a typo. One way to characterize ramification is to say that $\wp^2 \cap \mathbb Z = (p)$. So the condition you've written is precisely the condition to be unramified. (And you are correct that $(1+i)$ is a ramified prime in $\mathbb Z[i]$.)

Note also that the adjective ramified can be applied either to primes in the base field (which is $\mathbb Q$ in your discussion) or to primes in the top field ($K$ in your discussion). So we say that $\wp$ is ramified is $\wp^2 \cap \mathbb Z = (p)$, and we say that $p$ is ramified if there is at least one ramified prime ideal $\wp$ lying over it.

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That's a mistake in the article; the correct condition should be $P^2 \cap \mathbb{Z} = (p)$.

But regarding your other point: in general, ramification is a condition on a prime on the upper field of the extension, not the base field. When an extension $L/K$ is not Galois, it's possible to have two primes $Q_1, Q_2$ of $L$ lying over a single prime $P$ in $K$ where $P$ factors as something like $Q_1^2 Q_2$. In that case, $Q_1$ would be ramified but $Q_2$ would not be, even though they both lie over the same prime of $K$. (In a Galois extension, this cannot happen - all exponents in the factorization of a base prime in the upper field must be the same.)

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James H. Sterling
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James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.