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Algebraic Number Theory Lecture Notes
Abstract The following notes were taking during a course on Algebraic Number Theorem at the University of Washington in Fall 2015. Please send any corrections to [email protected]. Thanks!
- January 4, Lecturer: Bianca Viray; written, partially edited by Josh Swanson
- September 30th, 2015: Introduction—Number Fields, Integrality, Discriminants Contents
- October 2nd, 2015: Rings of Integers are Dedekind Domains
- October 5th, 2015: Dedekind Domains, the Group of Fractional Ideals, and Unique Factorization
- October 7th, 2015: Dedekind Domains, Localizations, and DVR’s
- October 9th, 2015: The ef Theorem, Ramification, Relative Discriminants
- October 12th, 2015: Discriminant Criterion for Ramification
- October 14th, 2015: Draft
- October 16th, 2015: Draft
- October 19th, 2015: Draft
- October 21st, 2015: Draft
- October 23rd, 2015: Draft
- October 26th, 2015: Draft
- October 29th, 2015: Draft
- October 30th, 2015: Draft
- November 2nd, 2015: Draft
- November 4th, 2015: Draft
- November 6th, 2015: Draft
- November 9th, 2015: Draft
- November 13th, 2015: Draft
- November 18th, 2015: Draft
November 20th, 2015: Draft........................................ 43
November 23rd, 2015: Draft........................................ 46
November 25th, 2015: Draft........................................ 47
November 30th, 2015: Draft........................................ 50
December 2nd, 2015: Draft........................................ 52
December 7th, 2015: Draft........................................ 54
December 9th, 2015: Draft........................................ 56
List of Symbols 60
Index 62
September 30th, 2015: Introduction—Number Fields, Integrality,
Discriminants
1 Remark This is a course in algebraic number theory. An undergraduate course in elementary number theory studies Z and primes–for instance, there are infinitely many primes, even of the form 4k + 3, 8k + 5, etc., and in fact: 2 Theorem (Dirichlet) For any a, n ∈ Z+^ with gcd(a, n) = 1, there are infinitely many primes congruent to a mod n.
Dirichlet’s theorem essentially identifies linear polynomials in one variable (namely a+nx) which produce infinitely many primes. Question: what are the primes of the form x^2 +ny^2 = (x+
−ny)(x−
−ny) ∈ Z[
−n]? This naturally leads to the study of rings of the form Z[
−n]. Algebraic number theory is primarily interested in the following objects: 3 Definition A number field is a finite degree field extension of Q. If K is a number field, the ring of integers OK of K is the set of elements of K integral over Z, i.e. those which satisfy a monic polynomial over Z[x].
Course texts: Osserman’s notes; supplemented with some material from Lang, Neukirch, and Stein.
4 Aside There are some closely related objects that won’t make much of an appearance in this class. Let R be a commutative domain with fraction field F and suppose A is a finite-dimensional F -algebra. An R-order O in A is a subring of A that is a finitely generated R-module with F O = A. Claim: OK is the (unique) maximal Z-order in K. For instance, this claim requires OK to be a ring, which we haven’t proved yet. 5 Definition A global field is either a number field or a finite extension of Fp(t). Many of the things that hold for number fields also hold for global fields. For instance, rings of integers of number fields are analogous to the integral closures of Fp[t] in finite extensions of Fp(t).
Proof If M = (mij )i,j in the given basis, then M αk =
k mkiαi^ by definition, so we compute Tr(M (αi)M (αj )) = Tr(
k,`
mkiαim`j αj )
k,`
mkim`j Tr(αiαj )
= M (Tr(αiαj ))ij M T^. Now take determinants.
11 Aside The preceding proposition relates DiscS/R( SpanR{α 1 ,... , αn}), DiscS/R( SpanR{M α 1 ,... , M αn}), and the index of B ⊂ A—more details on the homework.
12 Proposition If L/K is a separable field extension of degree n, then let σ 1 ,... , σn be the n embeddings L ↪→ K which are the identity on K. Then
∏n i=1 σi(α),
∑n i=1 σi(α), and
- DiscL/K (α 1 ,... , αn) = det(σi(αj ))^2.
13 Corollary The norm, trace, and discriminant of integral elements coming from a separable field extension are all integral. Proof Integrality is preserved by the σi, so the above quantities are all polynomial combinations of integral elements.
October 2nd, 2015: Rings of Integers are Dedekind Domains
Review of norm, trace, discriminant from last lecture:
Let R ⊂ S be rings, S ∼= Rn, α ∈ S, mα : S → S by x 7 → αx, NS/R(α) := det mα, TrS/R(α) := Tr mα, and DiscS/R((αi)i) := det(Tr(αiαj )i,j ).
If x 1 ,... , xn ∈ S are an R-basis for S, and M : S → S is R-linear, then Disc (M (xi)) = ( det M )^2 Disc ((xi)). Moreover, if L/K is a separable field extension of degree n, and if σ 1 ,... , σn : L ↪→ K are the n embeddings of L into the algebraic closure of K which fix K, then NL/K (α) =
∏n i=1 σi(α), TrL/K (α) =
∑n i=1 σi(α), DiscL/K^ (α) = det(σi(αj^ ))
Indeed, if L/K is inseparable, Tr(α) = 0 for all α ∈ L and Disc ((αi)i) = 0 whenever αi ∈ L. More details and proofs in the homework and Osserman’s notes.
As a corollary, if L/K is an extension of number fields and K contains a subring R, and if α ∈ L is integral over R, then N (α), Tr(α) are integral over R. Similarly if α 1 ,... , αn ∈ L are integral over R, then Disc ((αi)) is integral over R. In particular, if R = Z and K = Q, then the discriminant of elements of OL are in OQ = Z.
14 Lemma Let L/K be an extension of fields. Assume L = K(α) for some α. Let f (x) be the minimal polynomial of α over K (in this class, f is thus assumed monic). Then
DiscL/K (1, α,... , αn−^1 ) =
i<j
(αi − αj )^2 =: Disc f (x) ,
where α = α 1 , α 2 ,... , αn are the roots of f (x).
Proof In this case, (σi(αj^ )) is a Vandermonde matrix. (If the extension is inseparable, both sides are zero, so it still works.)
15 Lemma Let L/K be a field extension of degree n with α 1 ,... , αn ∈ L. Then DiscL/K (α 1 ,... , αn) 6 = 0 if and only if α 1 ,... , αn is a K-basis for L and L/K is separable.
Proof Assume L/K is separable, so L = K(β) for some β.
⇐ By the previous lemma, Disc (1, β,... , βn−^1 ) 6 = 0. We can use an (invertible) change of basis matrix to send βi^ to αi and an earlier proposition then says that Disc ((βi)) and Disc((αi)) differ by a unit (in R).
⇒ If {α 1 ,... , αn} is K-linearly dependent, then {σj (αi)}ni=1 is K-linearly dependent with the same dependence relation, for each j. Hence det(σj (αi))^2 = 0.
The preceding discussion was quite general, though here are some special properties in the case R = Z involving rings of integers.
16 Notation Let K be a number field.
17 Proposition Let I ⊂ OK be a nonzero ideal. Then I contains a Q-basis for K, and among the Q-bases for K contained in I, any basis with minimal absolute value of the discriminant is a Z-basis for I. In particular, I is a free Z-module of rank n.
Proof Let n = [K : Q] and let α 1 ,... , αn be any Q-basis for K. Notice that for all i, there exists some di ∈ Z such that diαi ∈ OK (essentially, take di large enough to clear all denominators of the minimal polynomial). Then if 0 6 = β ∈ I, βd 1 α 1 ,... , βdnαn is a Q-basis for K contained in I. By the comments in the above review, the discriminant of (αi) is in Z.
Now let α 1 ,... , αn be a Q-basis for K contained in I of minimal | Disc (αi)|. If β ∈ I, then β =
i aiαi^ for some^ ai^ ∈^ Q. Assume to the contrary that^ a^1 6 ∈^ Z. Write^ a^1 =^ b^ +^ ^ where b ∈ Z and 0 < < 1. Let α′ 1 := β − bα 1 and consider {α′ 1 , α 2 ,... , αn}. This is clearly a Q-basis for K and is contained in I. The change of basis matrix from {α 1 ,... , αn} to {α′ 1 , α 2 ,... , αn} in the {α 1 ,... , αn} basis is lower triangular with , 1 ,... , 1 on the main diagonal, which has squared determinant 0 < ^2 < 1, giving a contradiction.
(Aside: the above change of basis proposition says that the discriminant of any two bases have the same sign, since they differ by the square of the determinant of a matrix over Q.)
18 Definition The preceding proposition says that ideals in OK have Z-bases, so the discriminant of any such ideal can be well-defined up to a unit (in Z) as the discriminant of any Z-basis. Precisely, d(OK ) = D(OK ) = Disc(OK ) is the discriminant of any Z-basis of OK , which is well-defined up to a sign.
19 Proposition If 0 6 = I ⊂ OK , then OK /I is finite.
20 Lemma Let R ⊂ S be integral domains and let I ⊂ S be a non-zero ideal. Suppose 0 6 = α ∈ I satisfies a non-zero polynomial f (x) ∈ R[x]. Then I ∩ R 6 = 0.
Proof WLOG assume f (0) 6 = 0 by canceling enough powers of x. Consider f (α) − f (0) ∈ I, but f (α) = 0, so 0 6 = −f (0) ∈ R.
26 Definition If R is an integral domain with K := Frac(R), then a subset I ⊂ K is a factional ideal of R if (i) I is an R-submodule;
(ii) There exists c ∈ R such that cI ⊂ R.
A fractional ideal I is a principal fractional ideal if I = αR for some α ∈ K.
(The idea behind (ii) is that the “denominators don’t get too large” and can all be canceled simultaneously. “Fractional ideal” naively would suggest a subset of ideals satisfying special properties, but in fact fractional ideals are more general than ideals.)
27 Remark Dedekind domains are not necessarily UFD’s, despite OQ = Z being the first example. For instance, OQ(√−5) = Z[
− 5 ] is not a UFD since 2 · 3 = 6 = (1 +
− 5 ). (This is the standard counterexample).
28 Theorem Let R be a Dedekind domain. Then (1) Every ideal of R factors uniquely into prime ideals.
(2) The set of fractional ideals forms a group under multiplication, with identity R. 29 Remark The product of two fractional ideals I, J is the set of finite sums of pairwise products ij for i ∈ I, j ∈ J.
(1) says that while Dedekind domains need not be UFD’s, their primes at least factor uniquely. If every prime were principal, then we would have unique factorization on the level of elements. Hence (2) says we can measure the failure of unique factorization by considering the “ideal class group” of K, ClK , which is the quotient of the fractional ideals by non-zero principal ideals. More on this later.
Proof We follow the proof from Lang, which proves (2) and deduces (1). Neukirch does (1) implies (2); Osserman does (2) implies (1) in a different way.
We first outline the proof of (2) as a series of claims: 30 Definition If I is an ideal, define I−^1 := {x ∈ K : xI ⊂ R} ⊂ K. (This is a generalized ideal quotient.)
(I−^1 is indeed a factional ideal: let 0 6 = a ∈ I, so (a) ⊂ I, so (a)I−^1 ⊂ II−^1 ⊂ R.)
Claim 1: If I ⊂ R is a nonzero ideal, there exist non-zero prime ideals p 1 ,... , pr such that p 1 · · · pr ⊂ I.
Claim 2: Every maximal ideal p is invertible with inverse p−^1.
Claim 3: Every nonzero ideal is invertible and its inverse is a fractional ideal.
Claim 4: If 0 6 = I ⊂ R is an ideal and J ⊂ K is an inverse of I, then J = I−^1.
Assuming the claims, we first prove (2). Claim 3 says every ideal is invertible. To amplify this up to all fractional ideals, suppose I is a fractional ideal with c ∈ K such that cI ⊂ R. An inverse for I is the product of (c) and an inverse for cI. Claim (4) says inverses are unique.
For claim (1), suppose not. Take a maximal counterexample I (using the Noetherian hypothesis). Since I is not prime, we have b 1 , b 2 6 ∈ I such that b 1 b 2 ∈ I. Note that
(I + b 1 )(I + b 2 ) ⊂ I ( I + (bi).
Since I is a maximal counterexample, I + (bi) cannot be counterexamples, so they each contain a product of non-zero primes. But then the product of those products of primes is contained in (I + b 1 )(I + b 2 ), hence in I, a contradiction.
For claim (2), from the definition we see that p ⊂ pp−^1 ⊂ R is an ideal. Since p is maximal, either pp−^1 is either p or R. So, assume to the contrary pp−^1 = p. From the finitely generated Z-module definition of integrality above, this says p−^1 ⊃ R consists of integral elements, so since R is integrally closed, p−^1 = R. We next exhibit an element of p−^1 − R, giving a contradiction and proving the claim. Pick 0 6 = a ∈ p. By claim (1), there exist non-zero primes p 1 ,... , pr such that p 1 · · · pr ⊂ (a) ⊂ p. Take r minimal with this property. Since p is prime, some pi ⊂ p, say i = 1. Primes are maximal here, so we have p 1 = p. Since r is minimal, p 2 ,... , pr 6 ⊂ (a). Take b ∈ p 2 · · · pr 6 ∈ (a). Note that bp ⊂ (a), so ba−^1 ∈ p−^1 and ba−^1 6 ∈ R. Note also that we have shown p−^1 ) R in general.
For claim (3), again assume not and take a maximal counterexample I. By claim (2), I is not prime, so I ( p for some prime p. Now I ⊂ Ip−^1 ⊂ R, and p−^1 has non-integral elements, so I 6 = Ip−^1. But then Ip−^1 has inverse J, so Jp is an inverse for I.
For claim (4), take 0 6 = I ⊂ R and J ⊂ K such that IJ = R. By definition J ⊂ I−^1. On the other hand, for x ∈ I−^1 , xI ⊂ R, so xR = xIJ ⊂ J, so x ∈ J and J = I−^1.
We finally prove unique factorization (1). For existence, assume it fails and take a maximal counterexample I. Obviously I cannot be prime. Let p ) I be a maximal ideal. Now I ( Ip−^1 ⊂ R, so Ip−^1 has a prime factorization, and we can multiply it by p to get a factorization for I.
For uniqueness, first define: 31 Definition If I, J ⊂ K are fractional ideals, say I | J if there is an ideal I′^ ⊂ R such that J = II′. Equivalently (using the existence of inverses), J ⊂ I.
Now assume p 1 · · · pr = q 1 · · · qs. Then p 1 | q 1 · · · qs. Hence qi ⊂ p 1 for some i, say i = 1. Then q 1 = p 1 may be canceled from both sides, giving the result by induction.
32 Example Let’s return briefly to the motivating question from the first day. Let n ∈ Z and pick a prime p. We want to know if there exist x, y ∈ Z such that p = x^2 + ny^2 = (x +
−ny)(x −
−ny) ∈ Z[
−n]. Assume for the moment that Z[
−n] is integrally closed (equivalently, −n is squarefree and −n ≡ 4 2 , 3). Then p = x^2 + ny^2 if and only if (p) ⊂ Z[
−n] factors as the product of two principal prime ideals. (They have to be prime since the norm of (p) is p^2 , and since norm is multiplicative, the norms of the two factors above are each p, which implies they were prime.) Factoring (p) ⊂ Z[
−n] into not-necessarily principal primes is well-understood, but fuguring out when they are principal is not well understood since class groups are complicated.
October 7th, 2015: Dedekind Domains, Localizations, and DVR’s
38 Definition Let R be an integral domain with fraction field K := Frac(R). Let I ⊂ K be a fractional ideal and suppose p ⊂ R is a non-zero prime ideal. Then Ip := IRp ⊂ K. (Note that we may assume all localizations S−^1 R are literal subsets of K.)
39 Lemma In the notation of the preceding definition, Ip is a fractional ideal of Rp and I = ∩pIp.
Proof I is a fractional ideal, so pick c ∈ R such that cI ⊂ R, so cIp ⊂ Rp. It follows that Ip is indeed a fractional ideal of Rp.
For I = ∩pIp, the ⊂ direction is clear, so let x ∈ ∩pIp ⊂ K and write x = a/b for a, b ∈ R. Let J := {y ∈ R : ya ∈ bI} ⊂ R. We claim J = R, in which case 1 ∈ J says a ∈ bI, so a/b ∈ I. For the claim, pick p prime, so x ∈ Ip says a/b = c/d for some c ∈ I, d 6 ∈ p. Hence da = bc ∈ bI and d ∈ J − p. Therefore J 6 ⊂ p for all p, forcing J = R.
40 Remark Osserman proves the theorem on unique factorization in Dedekind domains by building up the local picture and using this lemma repeatedly.
41 Proposition Let R be a noetherian integral domain. Then R is a Dedekind domain if and only if Rp is a DVR for all primes p.
Proof In the ⇒ direction, 37 shows that Rp is a (local) Dedekind domain, hence is a DVR. For ⇐, we need to show that non-zero primes are maximal and that R is integrally closed.
If 0 6 = p ⊂ m ⊂ R, then localize at m. By assumption, Rm is a DVR, so p = m are both the unique prime in Rm. Since primes in Rm are in bijection with primes in R contained in m, we have p = m in R. Now suppose x ∈ K is integral over R. Then x is integral over Rp for all p. Since Rp is a DVR, hence is a UFD, so x ∈ Rp for all p. By the lemma with I = R, x ∈ R.
42 Remark Let K be a number field with an order O ⊂ K. The maximal order OK is a Dedekind domain, but O need not be. More details in Neukirch, Chapter 1. We can still talk about fractional ideals with respect to O, but they are not necessarily invertable. 43 Definition Let a ⊂ K be a fractional ideal of O. Say a is invertible if there exists another fractional ideal b ⊂ K of O such that ab = O.
If a is invertible, then a−^1 = {x ∈ K : xa ⊂ O}. This follows from claim (4) in the theorem from last time. 44 Proposition a is invertible if and only if ap := aOp is a principal fractional ideal for all p.
Proof Assume a is invertible with b := a−^1. Then 1 =
∑r i=1 aibi^ for some^ ai^ ∈^ a, bi^ ∈^ b. At least one of the products aibi 6 ∈ p, say with i = 1. We claim ap = (a 1 ). Let x ∈ ap, so xb 1 ∈ apb = Op. Write x = (xb 1 )(a 1 b 1 )−^1 a 1 ∈ OpO× p a 1. Ran out of time to finish it off—see Neukirch for the rest.
Hence the failure of invertibility of fractional ideals is intimately related to the failure of localizations being PID’s.
October 9th, 2015: The ef Theorem, Ramification, Relative
Discriminants
Summary To determine if there exists a solution to p = x^2 + ny^2 , we want to know how p factors in Z[
−n]. So, the next few days will be concerned with figuring out how (p) factors over rings of integers.
We’ll use the following background notation in the next few lectures:
45 Notation Let R ⊂ S be Dedekind domains. Suppose S is finitely generated as an R-module. Let K := Frac(R), L := Frac(S), where L/K is an extension of degree n.
46 Theorem (“ef Theorem”) Let 0 6 = p ⊂ R be a prime ideal and let pS =
∏r i=1 q
ei i where the^ qi^ are distinct non-zero prime ideals, and ei ∈ Z> 0. Then S/qi is a finite dimensional R/p-vector space and n =
∑r i=1 eifi^ where fi := dimR/p S/qi. 47 Proposition RpS is a free Rp-module of rank n.
Proof Since S is finitely generated as an R-module, RpS is a finitely generated Rp-module. Note that RpS has finitely many prime ideals, so is a local Dedekind domain, hence is a PID. Using the structure theorem for finitely generated modules over a PID, we just need to show that RpS is torsion free and of rank n. Since RpS ⊂ L is a subset of a field, it is obviously torsion-free. As for the rank, let x 1 ,... , xm be free generators for RpS over Rp. Then x 1 ,... , xm is also a K-basis for L, essentially by canceling denominators. Hence m = n.
Proof of theorem: write Sp := RpS. Notice that S/pS ∼= Sp/pSp and Sp ∼= Rn p by the proposition, so∏ S/pS is an R/p-vector space of dimension n. By the Chinese Remainder Theorem, S/pS ∼= r i=1 S/q
ei i. Now we need only show^ dim^ S/q
ei i =^ eifi. For any^ m,^ S/q m i ∼=^ Sqi /q m i Sqi. But^ Sqi is a DVR since S is a Dedekind domain. Then we have
S/qi ∼= Sqi /qiSqi^ ∼= qi/q^2 i Sqi^ ∼= · · · ∼= qmi /qm i +1Sqi.
The claim now follows by induction and the observation (Sqi /qm i +1Sqi )/(qmi /qm i +1Sqi ) ∼= Sqi /qmi Sqi.
48 Definition We say that a prime ideal q ⊂ S lies above a prime ideal p ⊂ R if q ∩ R = p. (Of course, this is just saying the induced map Spec S → Spec R sends q to p.)
49 Proposition Every non-zero prime ideal of S lies above a unique non-zero prime ideal of R. The following are equivalent: a) q lies above p
b) p ⊂ q (equivalently, pS ⊂ q)
c) q occurs in the factorization of pS.
Proof We’ve already shown q ∩ R 6 = 0. If x, y ∈ R such that xy ∈ q ∩ R, then since q is prime, x or y ∈ R, so x or y is in q ∩ R, so q ∩ R is prime.
We essentially showed (b) ⇔ (c) when we defined divisibility of fractional ideals. For (b) ⇒ (a), we have p = p ∩ R ⊂ q ∩ R is a maximal ideal, so p = q ∩ R. Note that (a) ⇒ (b) is by definition.
If S is free over R then DS/R is a principal ideal generated by DS/R(x 1 ,... , xn) where (xi) is any R-basis for S.
If L is a number field, then DOL/Z = 〈DL〉.
Proof The two discriminants in fact agree for all tuples, giving (1). For (2), use the change of basis proposition for discriminants above. Hence (3) follows from (2) using the definition of DL.
October 12th, 2015: Discriminant Criterion for Ramification
Summary Last time we defined relative discriminants in two contexts: A) when S/R was free of rank n, and B) when R ⊂ S was a domain with fields of fractions K ⊂ L of degree n, where R is integrally closed in K and for all x ∈ S, x is integral over R.
Our main goal today is to prove a standard result relating ramification and relative discriminants.
(The “B” context was originally stated without the two integral hypotheses, but then isn’t not clear the result is an ideal in R. The additional hypotheses have been incorporated into the previous lecture’s notes.)
55 Notation Today, S and R will be as in cases A) or B) above. Again write Sp := RpS.
56 Lemma Let S 1 , S 2 be free R-modules of finite rank and let S = S 1 ⊕ S 2. Then DS/R = DS 1 /RDS 2 /R.
Proof Note that xy = 0 for all x ∈ S 1 , y ∈ S 2. Hence mx,S : S → S given by (s 1 , s 2 ) 7 → (xs 1 , 0) is essentially mx,S 1 , and similarly with y. It follows that if z 1 , z 2 ∈ S 1 ∪ S 2 , then
Tr(z 1 z 2 ) =
0 if zi ∈ S 1 , zj ∈ S 2 TrS 1 (zizj ) if zi, zj ∈ S 1 TrS 2 (zizj ) if zi, zj ∈ S 2
Since S is a free R-module, DS/R is a principal ideal generated by DS/R((z 1 ,... , zn)) for any R-basis for S. The resulting matrix of traces is block diagonal, and the result follows.
57 Lemma (Case B.) If R is a field and Nil(S) 6 = 0, then DS/R = 0.
Proof Let 0 6 = x ∈ Nil(S) and let y ∈ S. Consider mxy and let z ∈ S be an eigenvector of mxy with eigenvalue λ. Then 0 = xnynz = (xy)nz = mnxy (z) = λnz for sufficiently large n. Hence λn^ = 0, so λ = 0. While there may be no eigenvectors in S itself, we may pass to an algebraic closure and deduce that the minimal polynomial of mxy is of the form tk, so that Tr(xy) = 0 for all y ∈ S. Since R is a field, there exists a basis containing x, so that DS/R = 0.
58 Lemma (Case A.) Let I ⊂ R be an ideal. Then D(S/IS)/(R/I) ≡ DS/R mod I.
Proof If x 1 ,... , xn is an R-basis for S, then x 1 ,... , xn is an R/I-basis for S/IS. The result now follows from the definitions.
59 Lemma (Case B.) Let p ⊂ R be a prime ideal. Then DSp/Rp = RpDS/R.
Proof ⊃ is clear from the definitions, since computing discriminants of tuples of elements in S can be done either before or after localizing without affecting the answer. For the ⊂ inclusion, let xi = yi/zi ∈ Sp, so yi ∈ S and 1/zi ∈ Rp. Then using the change of basis formula gives
Disc((xi)) = Disc((yi/zi)) =
i
z i^2
Disc((yi)) ∈ RpDS/R.
60 Theorem Let R ⊂ S be a finitely generated extension of Dedekind domains. Then p ⊂ R is ramified in S if and only if p | DS/R.
Proof Our rough strategy is to localize and use the last lemma to allow us to use the previous lemmas.
Let pS =
∏m i=1 q
ei i. Claim:
DS/R ⊂ p ⇔ p | DS/R ⇔ D(S/qei i)/(R/p) = 0 for some i.
Recall that (S/pS)/(R/p) falls in Case A since R/p is a field. Likewise Sp ⊃ Rp falls in Case A by the proposition from last lecture. By the Chinese Remainder Theorem, S/pS ∼= ⊕S/qe i i, so by the first lemma, the final clause of the claim occurs iff D(S/pS)/(R/p) = 0. Using the third lemma, this occurs iff DSp/Rp ⊂ p in Rp. Using the fourth lemma, this final condition is the same as saying p | DS/R, giving the claim.
Now suppose p is unramified. Then ei = 1 and S/qi is a separable field extension of R/p. Hence Disc(S/qi)/(R/p)(any basis) 6 = 0 for all i, so p - DS/R.
If p is ramified, then either ei > 1, so S/qe i iover R/p is an extension with nilpotents, so the discriminant of the quotient is zero, or ei = 1 and S/qi is an inseparable field extension. But we already showed the discriminant of any inseparable field extension is zero. Hence p | DS/R.
61 Corollary Under the assumptions of the theorem, if L/K is separable, then only finitely many prime ideals ramify in S/R.
Next we discuss factoring ideals in extensions of Dedekind domains.
62 Notation We now return to the notation from last class, namely R ⊂ S is a finitely generated extension of Dedekind domains, and the corresponding extension of fields of fractions K ⊂ L is of degree N.
63 Lemma Let α ∈ S. Assume that K(α) = L and let f (x) be the minimal polynomial of α. Then Disc (f (x)) ∈ DS/R.
Proof We showed in an earlier lemma that Disc (f (x)) = DiscL/K ((1, α,... , αn−^1 )). This latter quantity is in DS/R since each αi^ is in S.
The next theorem allows us to often determine how a prime factors in an extension (and more) using factorization over finite fields.
64 Theorem Let p ⊂ R be a non-zero prime ideal. Assume there exists α ∈ S such that Sp = Rp[α]. Let f (x) be the minimal polynomial of α over R. If f (x) factors as
∏m i=1 f^ i(x)
ei (^) in (R/p)[x] with the f i distinct and separable, then
70 Lemma S is the integral closure of R in L and so σ induces an automorphism of S for all σ ∈ Gal (L/K). Proof Not difficult; can see it in Osserman; fine exercise.
Proof of proposition: take σ ∈ Gal (L/K). Can check that σ(qi) is a prime ideal directly. Also, σ(qi) ∩ R = σ(qi ∩ R) = σ(p) = p, which says that σ(qi) = qj for some j. Applying σ to the factorization shows that ei = ej and σ induces an isomorphism S/qi ∼= S/qj , so fi = fj. Hence the full statement follows from transitivity, which we now turn to.
Suppose qj is a prime lying over p outside of the orbit of q 1. By the Chinese Remainder Theorem, there exists x ∈ S such that x ∈ qj , x 6 ∈ qi for all qi in the orbit of q 1. Consider
NL/K (x) =
∏^ n
i=
σi(x) σi ∈ Gal(L/K).
This is a product of integral elements in R, so is itself in R. Also, it is in qj (since the identity is in Gal (L/K)), but the product is not in q 1 since by assumption none of the factors is in q 1. But since qj ∩ R = p = q 1 ∩ R, this is a contradiction.
71 Definition Let p ⊂ R be a non-zero prime ideal, q ⊂ S lying over p. The decomposition group of q/p is Dq/p := {σ ∈ Gal(L/K) : σ(q) = q} and the inertia group of q/p is Iq/p := {σ ∈ Dq/p : σ(x) ≡q x, ∀x ∈ S}.
72 Theorem Let LD,q denote the fixed field of Dq/p and let LI,q denote the fixed field of Iq/p. Further write SD,q := S ∩ LD,q and SI,q := LI,q ∩ S. Diagrammatically, we have
q S L
q 1 SI,q LI,q
q 0 SD,q LD,q
p R K
Then (i) LD,q is the minimal subfield of L containing K such that q is the unique prime lying over q 0. Further eq 0 = ep, fq 0 = fp, and [L : LD,q] = epfp.
(ii) Assume that S/q is separable over R/p. Then there is a short exact sequence
0 → Iq/p → Dq/p → Gal((S/q)/(R/p)) → 0.
In particular, #Iq/p = ep and #Dq/p = fpep.
(iii) LI,q is the minimal subfield of L containing K such that q 1 is totally ramified. Further, eq 1 = ep and fq 1 = 1. (Totally ramified means q is a power of i, and the interia degree is always 1. Double check.) ([L : LI,q] = ep = eq 1 , fq 1 = 1.)
More intuitively, there is no splitting in L/LD,q, [LI,q : LD,q] = fp, [L : LI,q] = ep. The topmost extension is totally ramified, the middle is inert, and the extensions involving the top three fields are all Galois. However, LD,q/K is not necessarily Galois.
Proof We begin with (i). By definition, Gal(L/LD,q) = Dq/p = {σ : σ(q) = q}. Hence the orbit of q under the Galois group is just q, but from the previous proposition the Galois group acts transitively on primes lying over q 0 , so q is the unique prime over q 0.
Now assume there were some K ⊂ E ⊂ LD/q with q the unique minimal prime above q ∩ E. Then Gal(L/E) ⊂ DL,q. Cardinality count gives equality.
We know that #Dq/p = eq 0 fq 0. We also know that [L : k] = mepfp = (#Dq/p) · # of cosets. The number of cosets is just the number of primes lying above p, which is m by definition. Hence epfp = eq 0 fq 0. We also know that inertia degrees and ramification indexes multiply in towers, so eq 0 eq/p = ep and likewise with f ’s. These three equations imply ep = eq 0 and fp = fq 0.
October 16th, 2015: Draft
We continue the proof of the theorem from last time. Recall that R ⊂ S was a finitely generated extension of Dedekind domains with fields of fractions K ⊂ L of degree n.
Proof (Continued.) The following lemma essentially assures us that subextensions of Dedekind domains preserve our background hypotheses. 73 Lemma Let E be a subfield extension of L/K (i.e. L/E/K). Set T := S ∩ E. Then T is a Dedekind domain with Frac(T ) = E. Further, T is a finitely generated R-module and S is a finitely generated T -module. Proof That Frac(T ) = E is a straightforward verification. It is clear that T is integrally closed in E. That T is noetherian and a finitely generated R-module follows since R is noetherian and S is a finitely generated R-module. S is a finitely generated T -module since it is a finitely generated module over R ⊂ T. To see that every non-zero prime in T is maximal, apply the going up theorem holds for S/T.
The lemma completes the proof of (i). Now we turn to (ii). We must show S/q is normal in R/p. Pick α ∈ S/f q and consider the minimal polynomial f ∈ R/p[x]. Lift α to α ∈ S and let f ∈ R[x] be its minimal polynomial. Since L/K is Galois, L, and therefore S, contains all roots of f. Viewing these in S/q, it follows that S/q contains all roots of f mod q, which f divides, so we have all roots of f. Since we’re assuming separability, S/q is Galois over R/p.
Recall Dq/p = {σ ∈ Gal (L/K) : σ(q) = q}. This definition gives a map Dq/p → Gal((S/q)/(R/p)) whose kernel by definition is Iq/p. Hence we need only show surjectivity. Since S/q is separable over R?p, there exists a primitive element α ∈ S/q. For τ ∈ Gal ((S/q)/(R/p)), let β be such that τ (α) = β. We can lift α, β to α, β ∈ S where f is the minimal polynomial of α and β is a root of f. By transitivity of the Galois group, there is some σ ∈ Gal (L/K) such that σ(α) = β. We claim σ fixes q—this gap will be filled in later, and it’s in Osserman’s notes. This proves surjectivity, giving the short exact sequence. From (i), we have #Dq/p = epfp. By definition, # Gal((S/q)/(R/p)) = fp. Hence #Iq/p = ep.
We turn to (iii). By definition of totally ramified and (i), LI,q ⊇ LD,q. From (ii), [L : LI,q] = eq. Suppose there were another field E ⊂ LI,q such that q′^ := q ∩ E were totally ramified. Then we could apply (i) and (ii) to L/E instead of L/K. Notice E ⊇ LD/q. Hence we have
LD,q ⊂ E = LD,q/q′ ⊂ ︸ ︷︷ ︸ fp
LI,q ⊂ L ︸ ︷︷ ︸ ep
Note that E ⊂ LI,q corresponds to Gal ((S/q)/(RE /q′)) which is 1 by the totally ramified assumption and the preceding two observations, so E = LI,q is indeed minimal.
81 Proposition Set λ := 1 − ζ.
- (λ) is a prime ideal of OK with interia degree f = 1
OK = (λ)φ(n)^ where φ is Euler’s totient function, φ(ν^ ) = ν−^1 ( − 1) = [Q(ζ) : Q].
Proof The minimal polynomial of ζ over Q is classically the nth cyclotomic polynomial Φn(x) , namely
Φn(x) =
σ∈(Z/nZ)×
(x − ζσ^ ) =
(xν− 1 )^ − 1 x`ν−^1 − 1
= (x`
ν− 1 )−^1 + · · · + x
ν− 1
Since there are factors, plugging in x = 1 gives =
σ∈(Z/nZ)×^ (1^ −^ ζ
σ (^) ). Now fix σ ∈ (Z/nZ)× and write 1 − ζσ^ = (1 − ζσ^ )/(1 − ζ)λ. We claim the fraction is a unit in the ring of integers. Using the geometric series and the fact that ζ is in OK , the fraction is also in OK. To show that its inverse is as well, pick σ ∈ (Z/nZ)×^ such that στ ≡n 1. Then its inverse is
1 − ζ 1 − ζσ^
1 − (ζσ^ )τ 1 − ζσ^
∈ OK.
Now = (units)λφ(n), soOK = (λ)φ(n)^ = (λ 1 · · · λr )φ(n). Since Q(ζ)/Q is Galois with Galois group (Z/nZ)×, from our earlier theorem we have [Q(ζ) : Q] =
eifi = ( # primes ) · e · f , and we’ve already accounted for φ(n) factors, we must have r = 1, f = 1, e = φ(n), so λ is prime with f = 1.
82 Notation Let B := { 1 , ζ, ζ^2 ,... , ζφ(n)−^1 } be the usual Q-basis for Q(ζ)/Q. We continue to have n = `ν^.
83 Proposition Disc B = ±`μ.
Proof Let ζ 1 ,... , ζφ(n) be the Gal (Q(ζ)/Q)-conjugates of ζ and again write Φn(x) =
∏φ(n) i=1 (x^ −^ ζi). From our earlier lemma, up to a unit we have Disc B is the Vandermonde determinant
i 6 =j (ζi^ − ζj ). If we apply the product rule, we find
Φ′ n(x) =
i
j 6 =i
(x − ζj ),
so that
± Disc B =
i 6 =j
(ζi − ζj ) =
φ ∏(n)
i=
Φ′ n(ζi) = NK/Q(Φ′ n(ζ)).
On the other hand, (xν (^) − 1 )Φn(x) = x ν − 1, so differentiating gives
Φ′ n(ζ) =
ν^ · ζ−^1 ζν−^1 − 1
Since the norm is multiplicative, we consider each piece. Note that NK/Q = ±since the constant term of Φn(1 − λ) is Φn(1) =. Similarly, if we replace ν with ν − s, then
NQ(ζs )/Q(1 − ζ
s ) = ±`.
We know that NM/K = NL/K ◦ NM/L and that NM/L(α) = α for α ∈ L. Putting it all together, NK/Q(1 − ζ`
s ) = ±m, so that Disc B = ±μ.
84 Remark [OK : Z[ζ]] = m. Hencem
′ OK ⊂ Z[ζ] for m′^ large enough. Note that Disc (OK /Z)[OK : Z(ζ)]^2 = Disc B.
85 Theorem If K = Q(ζ) = Q(ζn) = Q(ζ`ν ), then OK = Z[ζ].
Proof Again let λ := 1 − ζ. Since f = 1, [OK /λOK : F] = 1, so OK /λOK ∼= Z/Z. Hence OK = Z + λOK , so OK = Z[ζ] + λOK. Multiply this last expression by λ and substitute the resulting right-hand side in for λOK in this last expression to get OK = Z[ζ] + λZ[ζ] + λ^2 OK. We can of course ignore the middle term. Iterating this computation, we find
OK = Z[ζ] + λtOK t ≥ 1.
If we use the observation in the remark, if OK = (λ)φ(n), letting t = φ(n)m′^ gives OK = Z[ζ] +m
′ OK = Z[ζ].
86 Notation We now consider general n, so write n := ν 11 · · ·ν s s. For notational convenience, write ζi := ζ n/`νii n and let Bi := { 1 , ζi,.. .} be the corresponding basis.
87 Remark Notice that Q(ζ) = Q(ζ 1 ) · · · Q(ζs) since the `i’s are coprime, and Q(ζ 1 ) · · · Q(ζi) ∩ Q(ζi+1) = Q. (For the intersection identity, suppose p 6 = q are two primes and consider K ⊂ Q(ζp) ∩ Q(ζq ). Consider those subfields in which p and q are totally ramified–the result will follow immediately.)
We have Disc Bi = ±`μ i iare pairewise coprime. A technical lemma (Neukirch, 2-11) says that in this situation ζα 1 1 · · · ζ sα sas αi ∈ Z varies forms a Z-basis for the ring of integers of the product Q(ζ). Hence if α ∈ OK , α = f (ζ) for f ∈ Z[x]. Now deg f ≤ φ(n) − 1, which says { 1 , ζ,... , ζφ(n)−^1 } is a Z-basis for OQ(ζ).
October 21st, 2015: Draft
Francesca will continue lecturing today on cyclotomic extensions, the splitting behavior of their primes, and Fermat’s Last Theorem.
88 Theorem Let n =
p p
νp (^). For each p let fp be the smallest positive integer such that pfp (^) ≡ n/pνp^1. Then
pOQ(ζn) =
( (^) r ∏
i=
pi
)φ(pνp (^) )
where r := φ(n/pνp^ )/fp.
Proof Last time, we saw OQ(ζn) = Z[ζn], so trivially [OQ(ζn) : Z[ζn]]. By (what is sometimes called) Dedekind-Kummer, study Φn(x) mod p. Fix p and let n = pνp^ u for (p, u) = 1. Since ζn^ = 1 = ζp νp (^) u let ζu := ζp νp , which is a primitive uth root of unity. Likewise set ζpνp := ζu, which is a pνp^ th root of unity.
We know that Q(ζ) = Q(ζu)Q(ζpνp ), if we consider ζiu · ζpjνp for i ∈ (Z/uZ)×, j ∈ (Z/pνp^ Z)× varying, we get all the primitive roots of unity. Therefore
Φn(x) =
σ∈(Z/nZ)×
(x − ζσ^ ) =
i∈(Z/uZ)×
j∈(Z/pνp^ Z)×
(x − ζui · ζpjνp ).
