How to prove that pZ is a maximal ideal for the ring of integers?

mathwonk

Thank you morphism. Now what about the non galois case? If we take any finite extension of Q, and normalize the integers in there we get a ring of integers R, such that R is a module of finite type over Z, and the map spec(R)-->spec(Z) should again have degree equal to the degree of the field extension, but not necessarily of the "Galois" group.

But every prime in Z should again have the same number of preimages in some sense.

e.g. although now the exponents in the factorization of p in R need not be equal, they should all add up to the same number, or rather give the same number after multiplication by the degree of the residue field extension.

just guessing......???

OOps, am I assuming unique factorization? there seems always to be a notion of preimages, but they may not correspond to a unique factorization in some cases? do theyn give in some sense a substitute for that notion?

morphism

Yup, that's right. If there are r distinct primes lying over p, with exponents e_i and inertial degrees f_i, then ##\sum_{i=1}^r e_i f_i = [K : \mathbb Q]##.

re: unique factorization: The map specR -> specZ is explicitly given by ##P \mapsto P \cap \mathbb Z##. It's not obvious that this map is surjective -- this is the "going up" lemma in this setting. Its counterpart, the going down lemma, says (amongst other things) that the primes in the preimage of p are in fact the primes that occur in the unique factorization of pR. This is easy to see: it follows from the fact that P contains pR iff it divides it. (And of course ##P \cap \mathbb Z = p\mathbb Z \iff P \supset pR##.)

mathwonk

apparently i am using the terminology wrong. i was calling any prime ramified if it had fewer than the maximal number of distinct prime factors upstairs. the traditional terminology seems to restrict that word to the case where this is caused by exponents higher than one, not where, as above, it is caused by degree of residue field extension greater than one. thus it seems the only traditionally "ramified" prime is 2. those of form 4k+3 are perhaps called "inertial". But again I am a beginner who has never studied this topic.

morphism

Right.

There are many good reasons for distinguishing between primes that don't "split completely" and those which are ramified. Let me give you one reason that's motivated by geometric considerations. The context is the analogy between extensions of number fields and extensions of function fields (i.e., maps of algebraic curves). To be concrete (and perhaps a little dishonest - forgive me), the relevant part of this analogy states that corresponding to the notion of a finite extension K/Q of number fields is the notion of a finite extension C(z,w)/C(z) of function fields, "i.e." a finite branched cover f:X->P^1 of the Riemann sphere by a compact Riemann surface X. The primes of (the ring of integers of) K correspond to points of X (this is a lie, but not a big one). The splitting of primes specR->specZ corresponds to the pulling back of divisors f*:Div(P^1)->Div(X) (another minor lie). The exponents in the prime factorization correspond to the ramification indices in pullback.

Now here's the punchline: the primes in K that ramify (in the sense that they have e_i > 1 for some i) are completely detected by an ideal in the ring of integers (called the different). Under the number field <-> function field analogy, the different corresponds to the ramification divisor of the covering map X->P^1.

Hurkyl

For another description, look at what happens if you don't assume

(Assume k is algebraically closed if need be.)

Consider the real polynomial x²+1. Being degree 2, the zero set of this polynomial "ought" to consist of two points on the real line, counting multiplicity. But, alas, its factorization only includes a single polynomial; a single point of Spec R[x]! But we 'know' that the zero set is really two points: i and -i.

This is one of the intuitions you want for higher degree ideals, I think: a degree n prime ideal corresponds in some sense to n individually indistinguishable points.

Going back to your Gaussian integer example, the ideal (7) corresponds to two "geometric" points. One way to manifest them is as the two different residue maps to F₄₉ (or to [itex]\overline{\mathbf{F}_7}[/itex], if you prefer). e.g. the "function" i has the value [itex]\sqrt{-1}\in \mathbf{F}_{49}[/itex] at one point, and [itex]-\sqrt{-1}\in \mathbf{F}_{49}[/itex] at the other point (after choosing a square root of -1 in F₄₉, of course). The "function" 7, of course, vanishes at both points.

The ideal (2), however, corresponds to a double point.

mathwonk

Hurkyl, So these geometric points sound like the geometric "k valued points" where k is the algebraic closure of Z/(7), in the language of Mumford's redbook.

so although spec(k) only has one point, there are two maps of spec(k) to spec(Z[i]) taking that point to the point (7) in spec(Z[i]). hence there are two ways to pullback "functions" in Z[i] from spec(Z[i]) to spec(k). I.e. two ring maps from Z[i] to k. Under both maps, 7 goes to zero, but the two maps correspond exactly to the two choices for the image of i in k.

mathwonk

morphism, does the "different" ideal allow us to define a canonical divisor in spec(Z[i]) by analogy with Hurwitz' formula for the ramification divisor of a map?

but i guess first we would need a canonical divisor in spec(Z).

mathwonk

So is there an analog of the genus of a curve, for a number field?

presumably those rings of integers with non trivial class group would be candidates for curves of positive genus.

what would be the analog of a curve of genus one?

if quadratic extensions are examples analogous to hyperelliptic curves, they should have arbitrary genus. can the class group of a quadratic extension be arbitrarily large?

morphism

There probably are analogues of the genus and canonical divisor for number fields. I don't know what they could be though. Someone more well-versed in Arakelov geometry would be able to answer.

At least I can tell you that the class group of a quadratic field can indeed be arbitrarily large: Gauss conjectured, and Heilbronn proved, that the class number of the imaginary quadratic field ##\mathbb{Q}(\sqrt{-d})## tends to infinity as ##d \to \infty##.