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A study guide for Algebraic Number Theory, a major topic in Algebra. It covers topics such as number fields, integrality, ideals, lattices, and class field theory. The guide includes references to Neukirch's Algebraic Number Theory and Cassels & Frohlich's Algebraic Number Theory, as well as additional references such as Poonen's Summary of the Statements of CFT and MIT Course Notes on Global CFT and Chebotarev Density Theorem. The guide also includes a chapter on memorization of key terms and concepts.
Typology: Study notes
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References: Neukirch, Algebraic Number Theory, Ch I.1-10, II.1-8, Cassels & Frohlich, Algebraic Number Theory, Ch VI, VII
Additional References:
15 OK , integral basis, and discriminant of Q(
Algebraic Number Theory Quals Questions (– best questions –) 21
Chapter 1 - Algebraic Integers 21 1.1 Gaussian Integers...................................... 21 1 Show that the units of Z[i] are precisely those with N (α) = 1.............. 21 2 Compute the units of Z[
−d] for any integer d > 1.................... 21 3 Show that Z[i] is a UFD................................... 21 4 Determine the prime elements of Z[i]............................ 21 1.2 Integrality........................................... 22 5 Show that every UFD is integrally closed.......................... 22 6 Is Z[
29] a PID?....................................... 22 7 What are some properties of integral elements/extensions?............... 22 8 Find an integral basis for the quadratic field Q(
D) where D is a square-free integer (D 6 = 0, 1). Use these to compute the discriminant..................... 22 9 When can we garuantee an integral basis exists? What are cases where it does not?. 23 10 How is the discriminant defined when OL is not a free OK module (no integral basis)? 23 11 Let K = Q(
−5), find OK and dK............................. 23 12 Show that { 1 , 3
2 } is an integral basis for K = Q( 3
13 The ring of integers OK is finitely generated as a Z-module, how would you show this? 24 14 Let f (x) = x^3 − x^2 − 2 x + 1. Show that f is irreducible over Q. Then let K = Q[x]/f and show that K is abelian (Hint: discriminant of f is 49)................. 24 1.3 Ideals.............................................. 24 15 Give an example of a ring of integers without unique factorization............ 24 16 Sketch of unique factorization for ideals in a Dedekind Domain............. 25 17 Show that 18 = 2 · 3 · 3 = (1 +
−17) are two different decompositions into irreducibles in OK for K = Q(
18 Decompose 33 + 11
−7 into integral irreducibles in Q(
19 In Z[
−3] let a = (2, 1 +
−3). Show that a 6 = (2), but a^2 = 2a. Conclude that ideals in Z[
−3] do not factor uniquely into prime ideals.................... 25 20 Given a number field K, what dedekind domains are contained in OK? What other dedekind domains (not necessarily inside OK ) can be constructed out of OK?..... 26 1.4 Lattices............................................. 26 21 Consider Γ = Z[i] ⊂ C. What is it’s fundamental region? Is it complete? Volume?.. 26 22 Give an example of a (finitely generated) subgroup which is not a lattice........ 26 23 State the Minkowski’s Lattice Point Theorem. Can the bound be improved?..... 26 1.5 Minkowski Theory...................................... 27 24 In what way is an ideal of OK a lattice? How can we compute the volume of an (integral) ideal?.............................................. 27 25 State the Minkowski Bound. How is it derived?...................... 27 1.6 The Class Number...................................... 27 26 What is the class number of a number field? Prove that it is finite............ 27 27 Show that the magnitude of the discriminant, |dK |, goes to ∞ as [K : Q] → ∞.... 28 28 Show that the quadratic field with discriminant dK ∈ { 5 , 8 } has trivial class group... 28 29 Show that the quadratic field with discriminant dK ∈ {− 3 , − 4 , − 7 , − 8 } has trivial class group.............................................. 28 30 How would you compute the class group for a (quadratic) number field?........ 28 31 Give a number field with non-trivial class group. How do you compute its class group? 29 32 Compute the class group for Q(
33 Compute the class group for Q(
2.8 Extensions of Valuations.................................. 38 61 Let K = Q[α] where α is a root of xn^ − 2 for n ≥ 2. What is [K : Q]? How many ways can the archimedean absolute value on Q be extended? What about the 2-adic absolute value? What are the rank and torsion subgroup of O K∗?................. 38 62 Write down a polynomial f over Q 3 such that Q 3 [x]/(f ) is a totally ramified quartic extension of Q 3........................................ 38 63 What are all the valuations of Q(i)?............................ 38
Class Field Theory Statements 39 64 State Local Class Field Theory. What properties uniquely determine the map?..... 39 65 State Global Class Field Theory. How does it relate to the local maps? What needs to be checked to show that the composition is well defined on CK?............. 39 66 What is Artin Reciprocity? How is Quadratic Reciprocity a special case?........ 39 67 Let L/K be an extension of number fields in which almost all primes (all but finitely many) in K split completely in L. What can we conclude about L? Hint: Chebotarev Density............................................. 40 68 How many quadratic extensions of Q 2 are there? Q 5?................... 40 69 In the case of K = Q, how does the global artin map simplify?.............. 40 70 How do the idele class group and the ideal class group relate?.............. 41 71 What is the Hilbert Class Field? How can we see that it has that galois group?.... 41
Memorization (– key terms –)
Chapter 1: Algebraic Integers
1 units, irreducible elements, prime elements, associated elements
units are invertible, irreducible cannot be written as a product of two non-units, primes p | ab =⇒ p | a or p | b, associated elements differ by a unit
2 F [α] = F (α) for field F and algebraic element α
F [x] is Euclidean Domain, so if f is minimal polynomial for α, then for g(a) ∈ F [a] with deg(g) < deg(f ) then f (x)h(x) + g(x)k(x) = 1 so then g(a)k(a) = 1 and so g(a) has an inverse.
3 Euclidean domain, UFD
Euclidean Domain: There is a ϕ : R − { 0 } → N such that for any α, β, we can find q, r such that α = qβ + r and either r = 0 or ϕ(r) < ϕ(β)
UFD (unique factorization domain) - every nonzero nonunit element has a unique factorization into prime (equiv to irreducible) elements
4 Noetherian, separable
Noetherian - ideals finiteily generated, ACC on ideals, nonempty collections of ideals have a maximal element,
separable - polynomials when no repeat roots, extensions when all elements have separable min polys
Note: all K/Q are separable, because repeat root means that x − α | f (x), f ′(x) so min poly for α divides f ′^ (so its degree is less than f ) and divides f (so not irreducible!)
5 primitive element theorem
finite separable extensions are primitive, i.e. L = K(α) for some α.
In particular, all number fields (finite ext over Q) are primitive!
6 structure theorem for finitely generated abelian groups
If G is a finitely generated (or just finite) abelian group then
G ∼= Z/n 1 Z ⊕ Z/n 2 Z ⊕ · · · ⊕ Z/nmZ ⊕ Zk
where Z/n 1 Z ⊕ Z/n 2 Z ⊕ · · · ⊕ Z/nmZ is the torsion part and Zk^ is the torsion free part, G has rank k. Can assume that n 1 | n 2 | · · · | nm.
7 structure theorem for modules over Dedekind Domains/PIDs
R a PID (or DD) and M a finitely generated R-module. Then there are nonzero ideals such that
M ∼= R/I 1 ⊕ R/I 2 ⊕ · · · ⊕ R/Im ⊕ Rk
where Rk^ is the free part of the decomposition.
8 algebraic number field, algebraic numbers and integers
algebraic number field = finite field extension K over Q
15 OK , integral basis, and discriminant of Q(
D), D square-free
√D 2 ] Z[
{α 1 , α 2 } =
√D 2 } { 1 ,
dK =
D D ≡ 1 mod 4 4 D D ≡ 2 , 3 mod 4
Key example to recall: −1+√− 3 2 is a cube root of unity, hence minimal polynomial divides^ x
(^3) − 1 and is in OK for Q(√−3).
Hence −3 has half integers and gives the 1 mod 4 condition.
16 Dedekind domain
Noetherian, integrally closed domain where every (nonzero) prime ideal is maximal
17 ideal operations
a | b ⇐⇒ b ⊆ a
a + b = {a + b | a ∈ a, b ∈ b} =smallest ideal containing a and b = gcd(a, b)
a ∩ b = lcm(a, b)
ab = {
i aibi^ |^ ai^ ∈^ a, bi^ ∈^ b} 18 Chinese Remainder Theorem
Given ideals a 1 ,... , an in a Dedekind domain O, pairwise coprime (ai + aj = gcd(ai, aj ) = (1) = O).
a := ∩ai O/a ∼=
i
O/ai
19 fractional ideals, integral ideals, and ideal inverses
fractional ideal is finitely generated O-submodule of K (field of fractions) (i.e. gen’d by finitely many elements from K with coefficients in OK )
integral ideals of K are the usual ring ideals of O
a−^1 = {x ∈ K : xa ⊆ O} inverse ideal
fractional ideals are quotients of 2 integral ideals
20 ideal group, JK
the abelian group of fractional ideals of K, with (1) identity and ideal inverses.
by unique factorization of fractional ideals (from integral ideals) JK is freely generated by prime ideals.
21 ideal class group, ClK
PK is the subgroup of fractional principal ideals
ClK = JK /PK
22 lattice
subgroup of an n dimensional R-vector space of the form Zv 1 + · · · + Zvm with linearly independent vi’s in V.
23 complete lattice, fundamental region
a lattice is complete if it has the same dimension as the vector space it lives in, i.e. |{v 1 ,... , vm}| = dim V.
fundamental region/mesh = coeffs in [0, 1) = {x 1 v 1 + · · · xmvm | 0 ≤ xi < 1 } = Φ
24 discrete subgroup
a subgroup of a vector space is discrete if every point is isolated, i.e. has a neighborhood in V where it is the only point in the subgroup in that neighborhood.
subgroup = lattice ⇐⇒ subgroup is discrete
25 volume of a lattice
given a lattice spanned by v 1 ,... , vn
vol(Γ) = vol(Φ) = | det(〈vi, vj 〉)|^1 /^2
Example : Γ = Z[i] = Z + Zi
vol(Γ) =
∣∣det
〈 1 , 1 〉 〈 1 , i〉 〈i, 1 〉 〈i, i〉
∣∣det
1 / 2 = |− 1 |^1 /^2 = 1
26 centrally symmetric
defn: if x ∈ X then −x ∈ X
examples and nonexamples :
examples: unit circle
{(x, y) | x ∈ [− 1 , 1]} strip is centrally symmetric, {(x, y) | x ∈ [0, 1]} off-center strip is not
27 convex subset
defn: if x, y ∈ X then (x + y)/ 2 ∈ X (their midpoint)
examples and nonexamples
example: unit circle, squares, rectangles, circles, triangles.
non-example: things that fold in on themselves or have gaps, like the union of two strips
28 Minkowski Lattice Point Theorem
Theorem Let Γ be a complete lattice in the Euclidean vector space X and X a centrally symmetric, convex, subset of V. Suppose vol(X) > 2 n^ vol(Γ)
then X contains at least one nonzero lattice point γ ∈ Γ.
29 Minkowski Space
K/Q number field, with n embeddings τ : K ↪→ C
KC =
τ C^ (with^ K^
j −→ KC by α 7 → (τ (α))τ )
Then complex conjugation acts on the indices τ 7 → τ as well as the elements, call this F
KR ⊆ KC, the Minkowski Space is the fixed subspace under F
38 Minkowski Bound on Ideal Norms in Class Group
Every class [a] ∈ ClK has an ideal with absolute norm
N(a) ≤ n! nn
π
)s (^) √ |dK |
in particular focus on powers of primes less than the bound.
39 Minkowski Lower Bound for Discriminant
K/Q with [K : Q] = n and s is the number of complex embedding pairs
nn n!
( (^) π 4
)s ≤
|dK |
40 Dirichlet’s Unit Theorem
O K∗ ∼= μ(K) × Zr+s−^1 where μ(K) is the roots of unity in K (finite group) and r is the number of real embeddings, s is number of complex embeddings pairs.
41 fundamental units
The r + s − 1 units in K that generate the unit group of OK.
42 Multiplicative Minkowski set up
Hyperplane in
τ R^ is the kernel of the trace map
K∗^ K C∗ =
τ C
τ R
NK/Q
j:a 7 →(τ a)τ N
`:(aτ )τ 7 →(log |aτ |)τ T r log |·|
43 Dedekind Kummer Theorem
Dedekind Kummer Theorem: If K = Q(α) and OK = Z[α] (or general L/K) with f (x) the minimal polynomial of α. Then however f (x) factors mod p is how p splits in OK.
Note: Generalizes when OK 6 = Z[α] as long as p - [OK : Z[α]]
44 ramification index and inertia degree
Given L/K and OL over OK with p a prime in OL that splits as p = qe 11 · · · qe rr
ei is the ramification index of qi and fi = [OL/qi : OK /p] is the inertia degree.
45 split (completey), ramified/unramified, inert
split - multiple primes lying over (completely - n distinct primes lying over)
ramified - at least one dividing prime divides to a power, unramified - all primes divide only once
inert - remains prime (maximal inertia degree)
46 State Quadratic Reciprocity.
Quadratc Reciprocity: Given two distinct odd primes p and q, ( p q
q p
p− 1 (^2) (−1)
q− 1 2
Proof Idea:
Work in the field Q(ζp) and look at quadratic gauss sums, use these to express a quantity in two ways, where equating gives the desired expression.
47 Legendre Symbol Formulas
For odd primes:
( − 1 p
p− 1 2
p
p^2 − 1 8
Also in general (^) ( a p
≡ a
p− 1 (^2) mod p
48 Proof that Gal(L/K) acts transitively on the primes
If not,take p a prime lying over q, and suppose p 6 = σp′^ for all σ ∈ Gal(L/K) then by CRT choose x ∈ p but not in σp′^ for all σ ∈ Gal(L/K) (hence σ(x) ∈/ p′^ for all σ)
Taking norm of x, NL/K (x) =
σ σ(x). Since^ x^ ∈^ p^ and^ N^ (x)^ ∈ OK^ ,^ x^ ∈^ p^ ∩ OK^ =^ q. But p′^ ∩ OK = p ∩ OK = q and none of σ(x) ∈ p′^ which is prime, contradiction!
49 ramification degree/inertia index in Galois extensions
since Gal(L/K) acts transitively on the primes, they have the same inertia index and ramification degrees, so ei = e and fi = f and n = ef r where r is the number of primes lying over p.
50 decomposition group
Given a prime p ∈ OL and G = Gal(L/K),
Gp = {σ ∈ G | σ(p) = p}
Properties: |Gp| = ef and Gσp = σGpσ−^1
51 inertia group
Ip = ker(Gp → Gal((OL/p)/(OK /p))
|Ip| = e
52 decomposition and inertia subfields
L/K Galois extension, with LD^ the decomposition subfield and LI^ the inertia subfield
[L : LI^ ] = e [LI^ : LD] = f [LD^ : K] = r
K → LD^ the prime splits completely
LD^ → LI^ the prime is inert
LI^ → L the prime is totally ramified
60 p-adic valuation and absolute value
vp(a) = vp(pm bc ) = m where (bc, p) = 1 |a|p = (^) pvp^1 (a) = (^) p^1 m
61 product formula for Q
For any a ∈ Q∗^ (nonzero)
p |a|p^ = 1 where^ p^ =^ ∞,^2 ,^3 ,^5 ,^7 ,...^ (all primes plus infinity)
62 (multiplicative) valuation (properties and equivalence)
| · | : K → R satisfying
i |x| ≥ 0 and |x| = 0 ⇐⇒ x = 0
ii |xy| = |x||y|
iii |x + y| ≤ |x| + |y|
Equivalent: | · | 1 , | · | 2 give same topology (d(x, y) = |x − y|) ⇐⇒ |x| 1 = |x|s 2 for some s > 0.
63 Approximation Theorem
Given | · | 1 , | · | 2 ,... , | · |n be pairwise inequivalent valuations on a field K
and a 1 , a 2 ,... , an ∈ K
Idea: We can approximate these arbitrarily well with respsect to each valuation
for all ε > 0 there exists some x ∈ K such that |x − ai|i < ε for all i = 1, 2 ,... , n
64 nonarchimedean and archimedean valuations
nonarchimedean: |n| is bounded for all n ∈ Z
(should be bounded by 1, since | 1 y| = | 1 ||y| so | 1 | = 1 and |n| = |1 + · · · + 1| ≤ max{| 1 |} = 1)
archimedean: |n| is not bounded for all n ∈ Z
65 strong triangle inequality
Normal Triangle Inequality: |x + y| ≤ |x| + |y|
Strong Triangle Inequality: |x + y| ≤ max{|x|, |y|}
Consequence: |x| 6 = |y| then |x + y| = max{|x|, |y|}
Valuation is nonarchimedean ⇐⇒ satisfies strong triangle inequality
66 Valuations on Q
The only (nontrivial) valuations are | · |p and | · |∞.
Proof Sketch:
Case: Nonarchimedean (will yield | · |p)
|n| ≤ 1 for all n ∈ Z, and for some prime p, |p| < 1 (otherwise trivial valuation)
Then pZ ⊂ {x ∈ Z : |x| < 1 } but pZ maximal, so equality holds.
|a| = |pmb| = |pm||b| = |p|m^ = |a|sp for some s.
Case: Archimedean (will yield | · |∞)
Claim |m|^1 /^ log(m)^ = |n|^1 /^ log(n)^ for all n, m > 1.
So C = |n|^1 /^ log(n)^ = es^ implies |n| = Clog(n)^ = es^ log(n)^ = |n|s ∞ and extend to all positive rationals.
67 exponential (additive) valuations (properties and equivalence)
v : K → R ∪ {∞} such that
i v(x) = ∞ ⇐⇒ x = 0 ii v(xy) = v(x) + v(y) (is additive) iii v(x + y) ≥ min{v(x), v(y)}
two valuations are equivalent if there is some s > 0 such that v(x) = su(x) for all x.
68 relationship between additive/multiplicative valuations
v(x) =⇒ |x| = q−v(x)^ for some q > 1
|x| =⇒ v(x) = − log |x|
69 valuation ring
O in K is valuation ring if for all x ∈ K either x ∈ O or x−^1 ∈ O
maximal ideal is {x ∈ O : x−^1 ∈ O}/
70 discrete valuation, normalized valuation
discrete if there is a smallest positive value s, that is v(K∗) = sZ. normalized if s = 1
71 prime elements (w.r.t. normalized additive valuation)
if v(K∗) = Z then π ∈ O = {x ∈ K : v(x) ≥ 0 } is prime if v(π) = 1
72 principal units and nth higher unit groups
U (1)^ = 1 + p are the principal units, U (n)^ = 1 + pn^ nth higher unit group
U (n+1)/U (n)^ ∼= O/p.
73 complete valued field
(K, | |) complete if every cauchy sequence (with respect to d(x, y) = |x − y|) converges to an element in K.
74 completion w.r.t. a valuation
Given K with valuation | |, take R to be the ring of cauchy sequences in K with respect to | |, and the maximal ideal m of nullsequences (converges to 0) then K̂ = R/m
K → K̂ by a 7 → (a, a, a,.. .)
extend the valuation | | to K̂ by defining |(xn)| = limn→∞ |xn|.
completions are unique (up to isomorphism)
75 Ostrowski’s Theorem
The only complete fields with respect to archimedean valuations are R and C (up to isomorphism)
76 Hensel’s Lemma
Hensel’s Lemma If f ∈ Zp[x] with some a 0 ∈ Z/pZ such that f (a 0 ) ≡ 0 mod p but f ′(a 0 ) 6 = 0 mod p then there is a lift α ∈ Zp of a 0 such that f (α) = 0.
84 valuation extensions from minimal polynomial
If L = K(α) where α has minimal polynomial f ∈ K[x] then extension wi of v correspond to irreducible factors of f in Kv (e.g. R, C, Qp)
85 fundamental identity for valuations
[L : K] =
w|v[Lw^ :^ Kv] where^ w^ |^ v^ ranges over all valuations^ w^ extending^ v. For Kv = Qp (v is discrete), [L : K] =
w|v ewfw^ with^ ew^ = (w(L ∗) : v(K∗)) and fw = [λw : κ]
86 tame inertia
tame inertia is cyclic, that is when p - |Iq| in extension K/Qp, then Iq is cyclic with order e.
Class Field Theory
87 Local Class Field Theory Statements
Let K be a local field. Then there is a local artin map φK that is a continuous surjection (K∗^ with topology induced by valuation and Gal(·/·) with Krull topology)
K∗^ φK −−→ Gal(Kab/K)
where Kab^ is the maximal abelian extension of K. For any finite abelian extension L/K, the quotient map Gal(Kab/K) → Gal(L/K) composes to get a surjective map φL/K : K∗^ → Gal(L/K). If L/K is unramified and π is any uniformizer for K, then φL/K (π) = Frobp ∈ Gal(L/K). Furthermore, the kernel of φL/K is NL/K (L∗) and this is inclusion reversing by Galois theory.
As a consequence, φK induces an isomorphism when passed to the profinite completion. Furthermore, φL/K (O K∗ ) gives the inertia subgroup of Gal(L/K).
88 Global Class Field Theory Statements
Let K be a global field. Let CK be the idele class group (IK /K∗^ where IK are the ideles, the unit group of the adeles).
Then there is a global artin map φK that is a continuous surjection (CK with ideles topology and Gal(·/·) with Krull topology)
CK φK −−→ Gal(Kab/K)
where Kab^ is the maximal abelian extension of K. This again induces an isomoprhism on the profinite completions.
For any finite abelian extension L/K, the quotient map Gal(Kab/K) → Gal(L/K) composes to get a surjective map φL/K : CK → Gal(L/K), which has kernel NL/K (CL).
f L/K is unramified and π is any uniformizer for K, then φL/K (1,... , 1 , π, 1 ,.. .) = Frobp ∈ Gal(L/K).
Furthermore, φL/K (O∗ p ) gives the inertia subgroup for the ideal p of K in Gal(L/K).
89 Conductor
The conductor is defined for local fields as pn^ for the smallest n such that the local artin map φQ is trivial on 1 + pnZp. The global conductor is the product of the local ones. If p is unramified, then n = 0 so this is a finite product of the primes that ramify.
90 Hilbert Class Field
The Hilbert Class Field is the maximal unramified abelian extension of K, and if we denote it by H, we have ClK ∼= Gal(H/K) where the left hand side is the ideal class group.
91 Artin Reciprocity
Artin Reciprocity Statement: Let K/Q be an abelian extension. The primes of Q the split com- pletely in K are determined by a congruence condition modulo the conductor fK/Q.
92 Adeles and Ideles
Let K/Q be a number field. Then adeles are AK =
ν Kν^ where^ ν^ ranges over all valuations of^ K, Kν is the completion of K with respect to the valuation ν, and the
indicates a restricted product, meaning if (αν ) ∈ AK then for all but finitely many ν, αν ∈ O∗ ν (i.e. lies in the valuation ring).
The ideles are the units within the adeles, i.e. IK = A∗ K =
ν K ∗ ν. 93 Idele Class Group
For each valuation ν, there is an embedding K ↪→ Kν so combining these maps we have K∗^ ↪→ IK. Quotienting by the image of this injection we define the idele class group CK = IK /K∗.