Algebraic Number Theory: summary of notes [Lecture notes]

64 Pages·2005·0.357 MB·English

by  Robin Chapman

Algebraic Number Theory
summary of notes
Robin Chapman
3 May 2000, revised 28 March 2004, corrected 4 January 2005
This is a summary of the 1999–2000 course on algebraic number the-
ory. Proofs will generally be sketched rather than presented in detail. Also,
examples will be very thin on the ground.
I first learnt algebraic number theory from Stewart and Tall’s textbook
Algebraic Number Theory (Chapman & Hall, 1979) (latest edition retitled
Algebraic Number Theory and Fermat’s Last Theorem (A. K. Peters, 2002))
and these notes owe much to this book.
I am indebted to Artur Costa Steiner for pointing out an error in an
earlier version.
1 Algebraic numbers and integers
We say that α ∈ C is an algebraic number if f(α) = 0 for some monic
polynomial f ∈ Q[X]. We say that β ∈ C is an algebraic integer if g(α) = 0
for some monic polynomial g ∈ Z[X]. We let A and B denote the sets
of algebraic numbers and algebraic integers respectively. Clearly B ⊆ A,
Z ⊆ B and Q ⊆ A.
Lemma 1.1 Let α ∈ A. Then there is β ∈ B and a nonzero m ∈ Z with
α = β/m.
Proof There is a monic polynomial f ∈ Q[X] with f(α) = 0. Let m be the
product of the denominators of the coefficients of f. Then g = mf ∈ Z[X].
Write g = Pn a Xj. Then a = m. Now
j=0 j n
n
X
h(X) = mn−1g(X/m) = mn−1+ja Xj
j
j=0
1
is monic with integer coefficients (the only slightly problematical coefficient
is that of Xn which equals m−1A = 1). Also h(mα) = mn−1g(α) = 0.
m
Hence β = mα ∈ B and α = β/m. (cid:3)
Let α ∈ A. Then there is a monic polynomial f ∈ Q[X] of least degree
such that f(α) = 0. This polynomial is uniquely determined.
Proposition 1.1 Let α ∈ A. Then there is precisely one monic polynomial
f ∈ Q[X] of minimum degree with f(α) = 0. This polynomial f has the
property that if g ∈ Q[X] and g(α) = 0 then f | g.
Proof Note first that if h ∈ Q[X] is a nonzero polynomial with deg(h) <
deg(f), then h(α) 6= 0 since otherwise h = a−1h is a monic polynomial,
1
where a is the leading coefficient of h, with the property that deg(h ) <
1
deg(f) and h (α) = 0. That would contradict the definition of f. Now
1
f is unique, since if f had the same degree as f and also satisfied the
1
same conditions then h = f − f , if nonzero, has h ∈ Q[X], h(α) = 0 and
1
deg(h) < deg(g) which is impossible.
Now let g ∈ Q[X] and suppose that g(α) = 0. By the division algorithm
(Proposition A.1), g = qf +h where q, h ∈ Q[X], and either h = 0 (which
means that f | g as we want) or h 6= 0 and deg(h) < deg(f). But as
h(α) = g(α)−f(α)q(α) = 0 this is impossible. (cid:3)
We call this f the minimum polynomial of α and call its degree the degree
of α. Minimum polynomials are always irreducible.
Lemma 1.2 Let f be the minimum polynomial of α ∈ A. Then f is irre-
ducible over Q.
Proof If f is not irreducible then f = gh where g, h ∈ Q[X] are monic
polynomials of degree less than that of f. Then 0 = f(α) = g(α)h(α) and
so either g(α) = 0 or h(α) = 0. We may assume g(α) = 0. Then f | g which
is impossible since deg(g) < deg(f). (cid:3)
Suppose the minimum polynomial f of α lies in Z[X]. Then, since f is
monic and f(α) = 0, α is an algebraic integer. In fact the converse holds: if
α ∈ B then its minimum polynomial lies in Z[X]. We need to study integer
polynomials in more detail to prove this.
A nonzero polynomial f ∈ Z[X] is primitive if the greatest common
divisor of its coefficients is 1. Equivalently f is primitive if there is no prime
number dividing all its coefficients.
Lemma 1.3 (Gauss’s Lemma) Let f, g ∈ Z[X] be primitive polynomials.
Then fg is also a primitive polynomial.
2
Proof Write
f(X) = a +a X +a X2 +···+a Xm
0 1 2 m
and
g(X) = b +b X +b X2 +···+b Xn.
0 1 2 n
We show that there is no prime p dividing all the coefficients of fg. Take a
prime p. As f is primitive there is a coefficient of f not divisible by p; let
a be the first such. Similarly let b be the first coefficient of g not divisible
r s
by p. Then p | a for i < r and p | b for j < s. The coefficient of Xr+s in fg
i j
is
X
c = a b .
r+s i j
i+j=r+s
This sum contains the term a b , which is not divisible by p. Its other terms
r s
a b are all divisible by p, since they either have i < r or j < s. Hence c
i j r+s
is not divisible by p.
As there is no prime dividing all the coefficients of fg, the polynomial fg
is primitive. (cid:3)
If f ∈ Z[X] is nonzero, let a be the greatest common divisor of the
coefficients of f. Then f = af where f is primitive. We call a the content
1 1
of f and denote it by c(f). More generally, let g be a nonzero element of
Q[X]. Then bg ∈ Z[X] where the positive integer b is the product of the
denominators of the coefficients of f. Then bg = cg where c is the content
1
of bg and g is primitive. Hence g = (c/b)g where q is primitive polynomial
1 1 1
in Z[X] and c/b is a positive rational. We can write any nonzero g ∈ Z[X] as
g = rg with r ∈ Q, r > 0 and g ∈ Z[X] being primitive. It’s an instructive
1 1
exercise to show that this r is uniquely determined; hence it makes sense
to call r the content of g. Putting s = 1/r we see that there is a positive
rational s with sg a primitive element of Z[X].
WenowshowthatifamonicpolynomialinZ[X]factorsovertherationals
then it factors over the integers.
Proposition 1.2 Let f and g be monic polynomials with f ∈ Z[X] and
g ∈ Q[X]. If g | f then g ∈ Z[X].
Proof Suppose that g | f. Then f = gh where h ∈ Q[X]. Then h is
monic, as both f and g are. There are positive rationals r and s with rg and
sh primitive elements of Z[X]. The leading coefficients of rg and sh are r
and s respectively, so that r, s ∈ Z. By Gauss’s lemma, (rg)(sh) = (rs)f
is primitive. But since f ∈ Z[X] all coefficients of (rs)f are divisible by rs.
3
Hence rs = 1 (as rs is a positive integer) and so r = s = 1 (as r and s are
positive integers). Thus g = rg ∈ Z[X] as required. (cid:3)
An immediate corollary is this important characterization of algebraic
integers.
Theorem 1.1 Let α ∈ A have minimum polynomial f. Then α ∈ B if and
only if f ∈ Z[X].
Proof Suppose f ∈ Z[X]. Since f is monic and f(α) = 0 then α ∈ B.
Converselysupposethatα ∈ B. Thereisamonicg ∈ Z[X]withg(α) = 0.
Then f | g, since f is the minimum polynomial of α. By Proposition 1.2,
f ∈ Z[X]. (cid:3)
Another corollary is this useful criterion for irreducibility.
Proposition 1.3 (Eisenstein’s criterion) Let p be a prime number and
let
n−1
X
f(X) = Xn + a Xj ∈ Z[X].
j
j=0
If
• p | a when 0 ≤ j < n, and
j
• p2 - a
0
then f is irreducible over Q.
Proof Suppose that f is reducible over Q. Then f = gh where g, h ∈ Q[X],
g and h are monic, and also 0 < r = deg(g) < n and deg(g) = s = n−r. By
Proposition 1.2, g, h ∈ Z[X]. Write
r s
X X
g(X) = b Xi and h(X) = c Xj.
i j
i=0 j=0
Note that b = 1 = c . Certainly p - b and p - c . Let u and v be the least
r s r s
nonnegative integers with p - b and p - c . Then u ≤ r and v ≤ s. I claim
u v
that u = r and v = s. Otherwise u+v < r+s = n and
X
a = b c .
u+v i j
i+j=u+v
This sum contains the term b c which which is not divisible by p. The
u v
remaining terms have the form b c with either i < u or j < v. In each case
i j
4
one of b and c is divisible by p. Hence a is the sum of a nonmultiple of
i j u+v
p with a collection of multiples of p and so p - a contrary to hypothesis.
u+v
Hence u = r and v = s. As r, s > 0 both b and c are divisible by p so that
0 0
a = b c is divisible by p2 again contrary to hypothesis. This contradiction
0 0 0
shows that f is irreducible over Q. (cid:3)
Example Let p be a prime number and let
p−1
X Xp −1
f(X) = 1+X +X2 +···+Xp−1 = Xj = .
X −1
j=0
We cannot apply Eisenstein to f directly, but if we set f (X) = f(X+1) we
1
get
(X +1)p −1 (X +1)p −1 Xp−1 (cid:18)p(cid:19)
f (X) = = = Xp−j−1.
1
(X +1)−1 X j
j=0
This is a monic polynomial, but its remaining coefficients have the form (cid:0)p(cid:1)
j
for0 < j < pandsoaredivisiblebyp. Thefinalcoefficientis(cid:0) p (cid:1) = pwhich
p−1
is not divisible by p2. By Eisenstein’s criterion, f is irreducible over Q. It
1
followsthatf isirreducibleoverQ, foriff(X) = g(X)h(X)wereanontrivial
factorization of f, then f (X) = g(X + 1)h(X + 1) would be a nontrivial
1
factorization of f .
1
We now show that A is a subfield of C and B is a subring of C.
Theorem 1.2 (i) Let α, β ∈ A. Then α+β, α−β, αβ ∈ A, and if α 6= 0
then α−1 ∈ A.
(ii) Let α, β ∈ B. Then α+β, α−β, αβ ∈ B.
Proof We first prove (ii) in detail, since the bulk of the proof of (i) follows
mutatis mutandis.
Let α and β have minimum polynomials f and g of degrees m and n
respectively. Write
m−1 n−1
X X
f(X) = Xm + a Xi and g(X) = Xn + b Xj.
i j
i=0 j=0
Then the a and b are integers and
i j
m−1 n−1
X X
αm = − a αi and βn = − b βj. (∗)
i j
i=0 j=0
5
Let v be the column vector of height mn given by
v> = (1 α α2 ··· αm−1 β αβ α2β ··· αm−1β β2 ··· αm−1βn−1).
In other words the entries of v are the numbers αiβj for 0 ≤ i < m and
0 ≤ j < n. I claim that there are (mn-by-mn) matrices A and B with
entries in Z such that Av = αv and Bv = βv. The typical entry in αv has
the form αiβj where 1 ≤ i ≤ m and 0 ≤ j < n. If i < m this already is an
entry of v while if i = m, (∗) gives
m−1
X
αmβj = − a αkβj.
k
k=0
In any case this entry αiβj of αv is a linear combination, with integer coeffi-
cients, of the entries of v. Putting these coefficients into a matrix A we get
αv = Av. Similarly there is a matrix B with integer entries with βv = Bv.
Now (A+B)v = (α+β)v, (A−B)v = (α−β)v and (AB)v = (αβ)v. As
v 6= 0 the numbers α+β, α−β and αβ are eigenvalues of the matrices A+B,
A − B and AB each of which has integer entries. But if the matrix C has
integer entries, its eigenvalues are algebraic integers, since the characteristic
polynomial of C is a monic polynomial with integer coefficients. It follows
that α+β, α−β and αβ are all algebraic integers.
If we assume instead that α, β ∈ A, the above argument shows (when
we replace ‘integer’ by ‘rational’ etc.) that α+β, α−β, αβ are all algebraic
numbers.
Finally suppose that α is a nonzero algebraic number with minimum
polynomial
n−1
X
f(X) = Xn + a Xi.
i
i=0
Then a 6= 0 (why?) and dividing the equation f(α) = 0 by a αn gives
0 0
n−1
X a 1
α−n + n−iα−i + = 0
a a
0 0
i=1
so that α−1 ∈ A. (cid:3)
√
Example Let us see what the matrices A and B are for say α = 2
√
and β = 1(1+ 5). The minimum polynomials of α and β are X2 −2 and
2
X2−X−1 respectively so that α2 = 2 and β2 = β+1. Let v> = (1 α β αβ).
6
Then
      
α α 0 1 0 0 1
 α2   2   2 0 0 0  α 
αv =   =   =   
 αβ   αβ   0 0 0 1  β 
α2β 2β 0 0 2 0 αβ
and
      
β β 0 0 1 0 1
 αβ   αβ   0 0 0 1  α 
βv =   =   =   .
 β2   1+β   1 0 1 0  β 
αββ2 α+αβ 0 1 0 1 αβ
We can take
   
0 1 0 0 0 0 1 0
 2 0 0 0   0 0 0 1 
A =   and B =  .
 0 0 0 1   1 0 1 0 
0 0 2 0 0 1 0 1
Then, for instance, αβ is an eigenvalue of
 
0 0 0 1
 0 0 2 0 
AB =  
 0 1 0 1 
2 0 2 0
so that h(αβ) = 0 where h is the characteristic polynomial of AB.
2 Number fields
The set A of algebraic numbers is too large to handle all at once. We restrict
our consideration to looking at smaller subfields of A which contain all the
algebraic numbers “generated” from a given one. For instance consider
K = Q(i) = {a+bi : a,b ∈ Q}
1
and
√ √ √
K = Q( 3 2) = {a+b 3 2+c 3 4 : a,b,c ∈ Q}.
2
It is apparent that both K and K are rings, being closed under addition,
1 2
subtraction and multiplication. It’s not hard to see that K is a field since if
1
a+bi is a nonzero element of K then
1
1 a b
= − i ∈ Q(i).
a+bi a2 +b2 a2 +b2
7
√ √
But it is not so obvious that 1/(a + b 3 2 + c 3 4) is an element of K . But
2
this is in fact so, and is an example of a general phenomenon.
Let α be an algebraic number of degree n. Define
Q(α) = {a +a α+a α2 +···+a αn−1 : a ,a ,...,a ∈ Q}.
0 1 2 n−1 0 1 n−1
Proposition 2.1 For each α ∈ A, Q(α) is a subfield of A.
Proof Since A is closed under addition and multiplication, and α ∈ A and
Q ⊆ A then it is apparent that Q(α) ⊆ A.
Let α have degree n and minimum polynomial f. Then by definition
Q(α) = {g(α) : g ∈ Q[X], and either g = 0 or deg(g) < n}.
I claim that in fact
Q(α) = {g(α) : g ∈ Q[X]}.
Certainly Q(α) ⊆ {g(α) : g ∈ Q[X]} so that to prove equality we need
to show that g(α) ∈ Q(α) whenever g ∈ Q[X]. By the division algorithm
(Proposition A.1) there is q ∈ Q[X] such that h = g −qf either vanishes or
has deg(h) < n. Then h(α) ∈ Q(α) but h(α) = g(α)−f(α)q(α) = g(α) as
f(α) = 0. This proves that Q(α) = {g(α) : g ∈ Q[X]}.
It is now clear that, since Q[X] is closed under addition, subtraction and
multiplication then so is Q(α). Hence Q(α) is a subring of A. (Alternatively,
one sees that the map g 7→ g(α) from Q[X] to A is a ring homomorphism
with image Q(α) which must therefore be a subring of A.)
TocompletetheproofthatQ(α)isafield, wemustshowthat1/β ∈ Q(α)
whenever β is a nonzero element of Q(α). Write β = g(α) with g ∈ Q[X]
and note that f - g since otherwise g(α) = 0. Let h be the greatest common
divisor of f and g. By Proposition A.2, there exist u, v ∈ Q[X] with h =
uf+vg. But f is irreducible, and so either h = 1 or h = f. But this latter is
impossible since h - f. Hence 1 = uf+vg and so 1 = u(α)f(α)+v(α)g(α) =
v(α)β. It follows that 1/β = v(α) ∈ Q(α) and so K is a field. (cid:3)
The numbers 1,α,α2,...,αn−1, where n is the degree of α, form a basis
of Q(α) as a vector space over Q. Thus the degree n is also the dimension
of Q(α) as a vector space over Q, and so we call n the degree of Q(α). In
general when we speak of a basis for a number field K = Q(α) me mean a
basis for K as a vector space over Q.
Given α ∈ A of degree n, its minimum polynomial f factors over C as
n
Y
f(X) = (X −α )
j
j=1
8
where α = α say. The numbers α ,...,α are the conjugates of α. They
1 1 n
are all algebraic numbers with minimal polynomial f. It is important to
note that the conjugates of α are all distinct. This follows from the following
lemma.
Lemma 2.1 Let f ∈ Q[X] be a monic polynomial and suppose that f is
irreducible over Q. Then f(X) = 0 has n distinct roots in C.
Proof Suppose that α is a repeated root of f(X) = 0. Then f(X) =
(X −α)2g(X) where g(X) ∈ C[X]. Consequently f0(X) = (X −α)2g0(X)+
2(X − α)g(X) and so f(α) = f0(α) = 0. Let h be the greatest common
divisor of f and f0. Then h = uf + vf0 for some u, v ∈ Q[X]. Thus
h(α) = u(α)f(α) + v(α)f0(α) = 0. But as h | f and f is irreducible, then
h = 1 of h = f. Since h(α) = 0, h = f. But then f | f0 and as f0 has leading
term nXn−1 this is impossible. (cid:3)
The field Q(α) forms the set of numbers which can be expressed in terms
of rational numbers and α using the standard arithmetic operations. We
might instead consider what happens when we take two algebraic numbers
α and β and consider which numbers can be expressed in terms of both.
Suppose α and β have degrees m and n respectively, and define
( )
m−1n−1
XX
Q(α,β) = c αjβk : c ∈ Q .
jk jk
j=0 k=0
It is readily apparent that Q(α,β) is a ring; less apparent but nonetheless
true that it is a field. However this field can be expressed in terms of one
generator.
Theorem 2.1 (Primitive element) Let α, β ∈ A. Then there is γ ∈ A
with Q(α,β) = Q(γ).
Proof We show that for a suitable rational number c, γ = α+cβ suffices.
Let α and β have degrees m and n respectively, and let their minimum
polynomials be
m n
Y Y
f(X) = (X −α ) and g(X) = (X −β )
j k
j=1 k=1
respectively, with α = α and β = β . Suppose that 1 ≤ j ≤ m and
1 1
2 ≤ k ≤ n. The equation
α+xβ = α +xβ
j k
9
can be rewritten as
(β −β )x = α −α
1 k 1 j
and so has exactly one solution x = x as β 6= β . Choose c to be a nonzero
jk 1 k
rational which is not equal to any of the x . This is possible as Q is an
jk
infinite set. Then
α+cβ 6= α +cβ
j k
whenever k 6= 1, by the choice of c. Let γ = α + cβ. For convenience put
K = Q(γ). I claim that Q(α,β) = K.
Certainly γ ∈ Q(α,β) and as Q(α,β) is a ring, then K ⊆ Q(α,β). To
prove that K ⊇ Q(α,β) it suffices to show that α ∈ K and β ∈ K. Let
h(X) = f(γ − cX). Then h has degree m, as c 6= 0, and h ∈ K[X]. Also
h(β) = f(γ−cβ) = f(α) = 0. But of course g(β) = 0 so that g and h have β
as a common zero. Suppose that it had another one, so that g(δ) = h(δ) = 0.
Then δ = β for some k ≥ 2 as g(δ) = 0. But then 0 = h(β ) = f(γ −cβ )
k k k
so that γ − cβ = α for some j. Thus γ = α + cβ which is false by the
k j j k
choice of c.
The greatest common divisor of g(X) and h(X) must be X − β. As
g ∈ Q[X] ⊆ K[X] and h ∈ K[X] there exists u, v ∈ K[X] with u(X)g(X)+
h(X)v(X) = X −β. Thus β = −(u(0)g(0)+h(0)v(0)) ∈ K, and it follows
that α = γ −cβ ∈ K. This completes the proof. (cid:3)
More generally we can consider fields Q(β ,...,β ) generated by any
1 n
finite number of algebraic numbers. But by using the primitive element
theorem and induction we see that each such field still has the form Q(γ).
We call a field of the form Q(α) for α ∈ A an algebraic number field or
simply a number field.
Let α ,...,α be the conjugates of α. The fields Q(α ) are very similar
1 n j
to Q(α) each being generated by an element with minimum polynomial f. In
fact they are all isomorphic. We define an isomorphism σ : Q(α) → Q(α )
j j
by setting σ (g(α)) = g(α ) when g ∈ Q[X]. It is perhaps not immediately
j j
evident that σ is well-defined. But this follows since if g , g ∈ Q[X] and
j 1 2
g (α) = g (α)theng (α ) = g (α ). Thisisaconsequenceofα andα having
1 2 1 j 2 j j
the same minimum polynomial. Once σ is seen to be well-defined, then it
j
is straightforward to prove it is an isomorphism. As α = α then σ is the
1 1
identity map on Q(α).
Let β ∈ Q(α). We define the norm N(β) and trace T(β) of β as follows.
Let
n
Y
N(β) = σ (β)
j
j=1
10