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apologies if this is a naive question. Consider two Galois extensions, K and L, of the rational numbers. For each extension, consider the set of rational primes that split completely in the extensions, say Split(K) and Split(L).

If Split(K) = Split(L), then is it necessarily true that K and L are isomorphic as Galois extensions of the rationals?

If so, for a given set of rational primes, S, is there a way to construct the extension over which S is the set of completely split primes?

References welcomed! Thanks, Martin

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    $\begingroup$ There are uncountably many sets of rational primes, but of those only countably many can possibly be the set of completely split primes of a Galois extension of the rationals. So I don't understand the use of the phrase "the extension." $\endgroup$ – Qiaochu Yuan Sep 8 '14 at 6:15
  • $\begingroup$ Wow - these answers very helpful, although both answers made me realize how weak my intuition is on this topic. At the risk of outlasting your patience, can I ask a couple of follow-up questions? $\endgroup$ – user304582 Sep 9 '14 at 3:53
  • $\begingroup$ In general follow-up questions should be asked separately, perhaps with a link to the original question. (Short follow-ups like "can you provide a reference for this claim you made?" can be asked in comments though.) $\endgroup$ – Qiaochu Yuan Sep 9 '14 at 3:56
  • $\begingroup$ (a) In the trivial case, where we think of Q as a Galois extension over Q, do we say that every prime splits completely? I guess so. (b) Consider, Q < K < L, where both K and L are Galois extensions of Q, then at least one rational prime must split completely over K, but not over L. Is it easy to identify one such prime? $\endgroup$ – user304582 Sep 9 '14 at 4:02
  • $\begingroup$ a) Yes. b) What do you mean by "easy"? Consider $K = \mathbb{Q}(\zeta_n), L = \mathbb{Q}(\zeta_m)$, where $n | m$. A prime splits completely in $\mathbb{Q}(\zeta_d)$ iff it is congruent to $1 \bmod d$, so we are searching for a prime congruent to $1 \bmod n$ but not congruent to $1 \bmod m$. Is this easy? $\endgroup$ – Qiaochu Yuan Sep 9 '14 at 5:57
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This result is due to Bauer and dated to 1916:

$\textbf{Theorem}:$ Let $K$ be an algebraic number field, $L/K$ and $M/K$ finite Galois extensions, and $\text{Spl}(L/K)$, $\text{Spl}(M/K)$ the set of prime ideals of $K$ which split completely in $L$ and $M$, respectively. Then $L \subseteq M$ if and only if $\text{Spl}(M/K) \subseteq \text{Spl}(L/K)$.

The case of $K=\mathbb{Q}$ answers your first question in the affirmative.

As for the second question, not every subset of rational primes is $\text{Spl}(K/\mathbb{Q})$ for a number field $K$, simply by cardinality arguments. Class field theory gives a description of $\text{Spl}(L/K)$ when $\text{Gal}(L/K)$ is abelian, but the general problem is open and very hard.

An excellent source for this material is Keith Conrad's notes on the history of class field theory, which you can find under the Expository Notes section on his website here: http://www.math.uconn.edu/~kconrad

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For your first question, it is true that $L$ and $K$ are isomorphic. This follows from the Chebotarev density theorem, for example.

For the second question, it's not really known how to do this in general. But class field theory lets you describe the sets $S$ which arise from abelian extensions (over any number field even, not just $\mathbb{Q}$). It's the non-abelian extensions that are harder to understand.

A reference for all of this is the introduction to Milne's notes on class field theory.

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