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4.1 Lifting of Prime Ideals
Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite,
separable extension of K of degree n, and B is the integral closure of A in L. If A = Z,
then K = Q, L is a number field, and B is the ring of algebraic integers of L.
4.1.1 Definitions and Comments
Let P be a nonzero prime ideal of A. The lifting (also called the extension) of P to B is
the ideal PB. Although PB need not be a prime ideal of B, we can use the fact that B
is a Dedekind domain [see (3.1.3)] and the unique factorization theorem (3.3.1) to write
PB =
g
i=1
Pei
i
where the Pi are distinct prime ideals of B and the ei are positive integers [see (3.3.2)].
On the other hand, we can start with a nonzero prime ideal Q of B and form a prime
ideal of A via
P = Q ∩ A.
We say that Q lies over P, or that P is the contraction of Q to A.
Now suppose that we start with a nonzero prime ideal P of A and lift it to B. We
will show that the prime ideals P1, . . . , Pg that appear in the prime factorization of PB
are precisely the prime ideals of B that lie over P.
4.1.2 Proposition
Let Q be a nonzero prime ideal of B. Then Q appears in the prime factorization of PB
if and only if Q ∩ A = P.
Proof. If Q ∩ A = P, then P ⊆ Q, hence PB ⊆ Q because Q is an ideal. By (3.3.5), Q
divides PB. Conversely, assume that Q divides, hence contains, PB. Then
P = P ∩ A ⊆ PB ∩ A ⊆ Q ∩ A.
But in a Dedekind domain, every nonzero prime ideal is maximal, so P = Q ∩ A. ♣
4.1.3 Ramification and Relative Degree
If we lift P to B and factor PB as g
i=1 Pei
i , the positive integer ei is called the ramification
index of Pi over P (or over A). We say that P ramifies in B (or in L) if ei > 1 for at
least one i. We will prove in a moment that B/Pi is a finite extension of the field A/P.
The degree fi of this extension is called the relative degree (or the residue class degree, or
the inertial degree) of Pi over P (or over A).
4.1.4 Proposition
We can identify A/P with a subfield of B/Pi, and B/Pi is a finite extension of A/P.
Proof. The map from A/P to B/Pi given by a+P → a+Pi is well-defined and injective,
because P = Pi ∩ A, and it is a homomorphism by direct verification. By (3.1.2), B is a
finitely generated A-module, hence B/Pi is a finitely generated A/P-module, that is, a
finite-dimensional vector space over A/P. ♣
4.1.5 Remarks
The same argument, with Pi replaced by PB, shows that B/PB is a finitely generated
A/P-algebra, in particular, a finite-dimensional vector space over A/P. We will denote
the dimension of this vector space by [B/PB : A/P].
The numbers ei and fi are connected by an important identity, which does not seem
to have a name in the literature. We will therefore christen it as follows.
4.1.6 Ram-Rel Identity
g
i=1
eifi = [B/PB : A/P] = n.
Proof. To prove the first equality, consider the chain of ideals
B ⊇ P1 ⊇ P2
1
⊇ ·· · ⊇ Pe1
1
⊇ Pe1
1 P2 ⊇ Pe1
1 P2
2
⊇ ·· · ⊇ Pe1
1 Pe2
2
⊇ ·· · ⊇ Pe1
1
· · · Peg
g = PB.
By unique factorization, there can be no ideals between consecutive terms in the sequence.
(Any such ideal would contain, hence divide, PB.) Thus the quotient β/βPi of any two
consecutive terms is a one-dimensional vector space over B/Pi, as there are no nontrivial
proper subspaces. (It is a vector space over this field because it is annihilated by Pi.)
But, with notation as in (4.1.5), [B/Pi : A/P] = fi, so [β/βPi : A/P] = fi. For each i,
we have exactly ei consecutive quotients, each of dimension fi over A/P. Consequently,
[B/PB : A/P] = g
i=1 eifi, as claimed.
To prove the second equality, we first assume that B is a free A-module of rank n. By
(2.3.8), this covers the case where A is a PID, in particular, when L is a number field. If
x1, . . . , xn is a basis for B over A, we can reduce mod PB to produce a basis for B/PB
over A/P, and the result follows. Explicitly, suppose n
i=1(ai+P)(xi+PB) = 0 in B/PB.
Then n
i=1 aixi belongs to PB, hence can be written as j bjyj with bj ∈ B, yj ∈ P.
Since bj = k cjkxk with cjk ∈ A, we have ak = j cjkyj ∈ P for all k.
The general case is handled by localization. Let S = A\P, A = S−1A,B = S−1B. By
(1.2.6), (1.2.9), and the Dedekind property (every nonzero prime ideal of A is maximal),
it follows that A has exactly one nonzero prime ideal, namely P = PA . Moreover, P
is principal, so A is a discrete valuation ring, that is, a local PID that is not a field. [By
unique factorization, we can choose an element a ∈ P \(P )2, so (a) ⊆ P but (a) ⊆ (P )2.
Since the only nonzero ideals of A are powers of P (unique factorization again), we have
(a) = P .] Now B is the integral closure of A in L, so B is the integral closure of A in
S−1L = L. [The idea is that we can go back and forth between an equation of integral
dependence for b ∈ B and an equation of integral dependence for b/s ∈ B either by
introducing or clearing denominators.] We have now reduced to the PID case already
analyzed, and [B /PB : A /PA ] = n.
Now PB = g
i=1 Pei
i , and Pi is a nonzero prime ideal of B not meeting S. [If
y ∈ Pi ∩ S, then y ∈ Pi ∩ A = P by (4.1.2). Thus y ∈ P ∩ S, a contradiction.] By the
basic correspondence (1.2.6), we have the factorization PB = g
i=1(PiB )ei . By the PID
case,
n = [B
/PB
: A
/PA
] =
g
i=1
ei[B
/PiB
: A
/PA
].
We are finished if we can show that B /PiB ∼=B/Pi and A /PA ∼=
A/P. The statement
of the appropriate lemma, and the proof in outline form, are given in the exercises. ♣
Problems For Section 4.1
We will fill in the gap at the end of the proof of the ram-rel identity. Let S be a multiplicative
subset of the integral domain A, and let M be a maximal ideal of A disjoint
from S. Consider the composite map A → S−1A → S−1A/MS−1A, where the first map
is given by a → a/1 and the second by a/s → (a/s) +MS−1A.
1. Show that the kernel of the map is M, so by the factor theorem, we have a monomorphism
h : A/M→S−1A/MS−1A.
2. Let a/s ∈ S−1A. Show that for some b ∈ A we have bs ≡1 mod M.
3. Show that (a/s) +MS−1A = h(ab), so h is surjective and therefore an isomorphism.
Consequently, S−1A/MS−1A
∼=
A/M, which is the result we need.
4.2 Norms of Ideals
4.2.1 Definitions and Comments
We are familiar with the norm of an element of a field, and we are going to extend the
idea to ideals. We assume the AKLB setup with A = Z, so that B is a number ring,
that is, the ring of algebraic integers of a number field L. If I is a nonzero ideal of B, we
define the norm of I by N(I) = |B/I|. We will show that the norm is finite, so if P is a
nonzero prime ideal of B, then B/P is a finite field. Also, N has a multiplicative property
analogous to the formula N(xy) = N(x)N(y) for elements. [See (2.1.3), equation (2).]
4.2.2 Proposition
Let b be any nonzero element of the ideal I of B, and let m = NL/Q(b) ∈ Z. Then m ∈ I
and |B/mB| = mn, where n = [L : Q].
Proof. By (2.1.6), m = bc where c is a product of conjugates of b. But a conjugate of an
algebraic integer is an algebraic integer. (If a monomorphism is applied to an equation
of integral dependence, the result is an equation of integral dependence.) Thus c ∈ B,
and since b ∈ I, we have m ∈ I. Now by (2.3.9), B is the direct sum of n copies of Z,
hence by the first isomorphism theorem, B/mB is the direct sum of n copies of Z/mZ.
Consequently, |B/mB| = mn. ♣
4.2.3 Corollary
If I is any nonzero ideal of B, then N(I) is finite. In fact, if m is as in (4.2.2), then N(I)
divides mn.
Proof. Observe that (m) ⊆ I, hence
B/(m)
B/I
∼=
I/(m). ♣
4.2.4 Corollary
Every nonzero ideal I of B is a free abelian group of rank n.
Proof. By the simultaneous basis theorem, we may represent B as the direct sum of n
copies of Z, and I as the direct sum of a1Z, . . . ,arZ, where r ≤ n and the ai are positive
integers such that ai divides ai+1 for all i. Thus B/I is the direct sum of r cyclic groups
(whose orders are a1, . . . , ar) and n − r copies of Z. If r < n, then at least one copy of Z
appears, and |B/I| cannot be finite. ♣
4.2.5 Computation of the Norm
Suppose that {x1, . . . , xn} is a Z-basis for B, and {z1, . . . , zn} is a basis for I. Each zi is
a linear combination of the xi with integer coefficients, in matrix form z = Cx. We claim
that the norm of I is the absolute value of the determinant of C. To verify this, first look
at the special case xi = yi and zi = aiyi, as in the proof of (4.2.4). Then C is a diagonalmatrix with entries ai, and the result follows. But the special case implies the general
result, because any matrix corresponding to a change of basis of B or I is unimodular, in
other words, has integer entries and determinant ±1. [See (2.3.9) and (2.3.10).]
Now with z = Cx as above, the discriminant of x is the field discriminant d, and the
discriminant of z is D(z) = (detC)2d by (2.3.2). We have just seen that N(I) = | detC|,
so we have the following formula for computing the norm of an ideal I. If z is a Z-basis
for I, then
N(I) =
D(z)
d
1/2
.
There is a natural relation between the norm of a principal ideal and the norm of the
corresponding element.
4.2.6 Proposition
If I = (a) with a = 0, then N(I) = |NL/Q(a)|.
Proof. If x is a Z-basis for B, then ax is a Z-basis for I. By (2.3.3), D(ax) is the square
of the determinant whose ij entry is σi(axj) = σi(a)σi(xj ). By (4.2.5), the norm of I is
|σ1(a) · · · σn(a)| = |NL/Q(a)|. ♣
In the proof of (4.2.6), we cannot invoke (2.3.2) to get D(ax1, . . . ,axn) = (an)2D(x1, . . . , xn),
because we need not have a ∈ Q.
We now establish the multiplicative property of ideal norms.
4.2.7 Theorem
If I and J are nonzero ideals of B, then N(IJ) = N(I)N(J).
Proof. By unique factorization, we may assume without loss of generality that J is a
prime ideal P. By the third isomorphism theorem, |B/IP| = |B/I| |I/IP|, so we must
show that |I/IP| is the norm of P, that is, |B/P|. But this has already been done in the
first part of the proof of (4.1.6). ♣
4.2.8 Corollary
Let I be a nonzero ideal of B. If N(I) is prime, then I is a prime ideal.
Proof. Suppose I is the product of two ideals I1 and I2. By (4.2.7), N(I) = N(I1)N(I2),
so by hypothesis, N(I1) = 1 or N(I2) = 1. Thus either I1 or I2 is the identity element
of the ideal group, namely B. Therefore, the prime factorization of I is I itself, in other
words, I is a prime ideal. ♣
4.2.9 Proposition
N(I) ∈ I, in other words, I divides N(I). [More precisely, I divides the principal ideal
generated by N(I).]
Proof. Let N(I) = |B/I| = r. If x ∈ B, then r(x + I) is 0 in B/I, because the order of
any element of a group divides the order of the group. Thus rx ∈ I, and in particular we
may take x = 1 to conclude that r ∈ I. ♣
4.2.10 Corollary
If I is a nonzero prime ideal of B, then I divides (equivalently, contains) exactly one
rational prime p.
Proof. By (4.2.9), I divides N(I) = pm1
1
· · · pmt
t , so I divides some pi. But if I divides
two distinct primes p and q, then there exist integers u and v such that up+vq = 1. Thus
I divides 1, so I = B, a contradiction. Therefore I divides exactly one p. ♣
4.2.11 The Norm of a Prime Ideal
If we can compute the norm of every nonzero prime ideal P, then by multiplicativity, we
can calculate the norm of any nonzero ideal. Let p be the unique rational prime in P, and
recall from (4.1.3) that the relative degree of P over p is f(P) = [B/P : Z/pZ]. Therefore
N(P) = |B/P| = pf(P).
Note that by (4.2.6), the norm of the principal ideal (p) is |N(p)| = pn, so N(P) = pm
for some m ≤ n. This conclusion also follows from the above formula N(P) = pf(P) and
the ram-rel identity (4.1.6).
Here are two other useful finiteness results.
4.2.12 Proposition
A rational integer m can belong to only finitely many ideals of B.
Proof. We have m ∈ I iff I divides (m), and by unique factorization, (m) has only finitely
many divisors. ♣
4.2.13 Corollary
Only finitely many ideals can have a given norm.
Proof. If N(I) = m, then by (4.2.9), m ∈ I, and the result follows from (4.2.12). ♣
Problems For Section 4.2
This problem set will give the proof that a rational prime p ramifies in the number field
L if and only if p divides the field discriminant d = dL/Q.
1. Let (p) = pB have prime factorization i Pei
i . Show that p ramifies if and only if the
ring B/(p) has nonzero nilpotent elements.
Now as in (2.1.1), represent elements of B by matrices with respect to an integral basis
ω1, . . . , ωn of B. Reduction of the entries mod p gives matrices representing elements of
B/(p).
2. Show that a nilpotent element (or matrix) has zero trace.
Suppose that A(β), the matrix representing the element β, is nilpotent mod p. Then
A(βωi) will be nilpotent mod p for all i, because βωi is nilpotent mod p.
3. By expressing β in terms of the ωi and computing the trace of A(βωj ), show that if β
is nilpotent mod p and β /∈ (p), then d ≡0 mod p, hence p divides d.
Now assume that p does not ramify.
4. Show that B/(p) is isomorphic to a finite product of finite fields Fi of characteristic p.
Let Ï€i : B → B/(p) → Fi be the composition of the canonical map from B onto B/(p)
and the projection from B/(p) onto Fi.
5. Show that the trace form Ti(x, y) = TFi/Fp (Ï€i(x)Ï€i(y)) is nondegenerate, and conclude
that i Ti is also nondegenerate.
We have d = det T(ωiωj ), in other words, the determinant of the matrix of the bilinear
form T(x, y) on B, with respect to the basis {ω1, . . . , ωn}. Reducing the matrix entries
mod p, we get the matrix of the reduced bilinear form T0 on the Fp-vector space B/(p).
6. Show that T0 coincides with i Ti, hence T0 is nondegenerate. Therefore d ≡0 mod p,
so p does not divide d.
As a corollary, it follows that only finitely many primes can ramify in L.
4.3 A Practical Factorization Theorem
The following result, usually credited to Kummer but sometimes attributed to Dedekind,
allows, under certain conditions, an efficient factorization of a rational prime in a number
field.
4.3.1 Theorem
Let L be a number field of degree n over Q, and assume that the ring B of algebraic
integers of L is Z[θ] for some θ ∈ B. Thus 1, θ, θ2, . . . , θn−1 form an integral basis of B.
Let p be a rational prime, and let f be the minimal polynomial of θ over Q. Reduce the
coefficients of f modulo p to obtain f ∈ Z[X]. Suppose that the factorization of f into
irreducible polynomials over Fp is given by
f = he1
1
· · · her
r .
Let fi be any polynomial in Z[X] whose reduction mod p is hi. Then the ideal
Pi = (p, fi(θ))
is prime, and the prime factorization of (p) in B is
(p) = Pe1
1
· · · Per
r .
Proof. Adjoin a root θi of hi to produce the field Fp[θi] ∼=
Fp[X]/hi(X). The assignment
θ → θi extends by linearity (and reduction of coefficients mod p) to an epimorphism
λi : Z[θ] → Fp[θi]. Since Fp[θi] is a field, the kernel of λi is a maximal, hence prime,
ideal of Z[θ] = B. Since λi maps fi(θ) to hi(θi) = 0 and also maps p to 0, it follows that
Pi ⊆ ker λi. We claim that Pi = kerλi. To prove this, assume g(θ) ∈ ker λi. With asubscript 0 indicating reduction of coefficients mod p, we have g0(θi) = 0, hence hi, the
minimal polynomial of θi, divides g0. If g0 = q0hi, then g − qfi ≡0 mod p. Therefore
g(θ) = [g(θ) − q(θ)fi(θ)] + q(θ)fi(θ)
so g(θ) is the sum of an element of (p) and an element of (fi(θ)). Thus ker λi ⊆ Pi, so
Pi = kerλi, a prime ideal.
We now show that (p) divides Pe1
1
· · · Per
r . We use the identity (I+I1)(I+I2) ⊆ I+I1I2,
where I, I1 and I2 are ideals. We begin with P1 = (p) + (f1(θ)), and compute
P2
1
⊆ (p) + (f1(θ))2, . . . ,Pe1
1
· · · Per
r
⊆ (p) + (f1(θ))e1 · · · (fr(θ))er .
But the product of the fi(θ)ei coincides mod p with r
i=1 hi(θ) = f(θ) = 0. We conclude
that r
i=1 Pei
i
⊆ (p), as asserted.
We now know that (p) = Pk1
1
· · · Pkr
r with 0 ≤ ki ≤ ei. (Actually, ki > 0 since
p ∈ ker λi = Pi, so Pi divides (p). But we will not need this refinement.) By hypothesis,
B/Pi = Z[θ]/Pi, which is isomorphic to Fp[θi], as observed at the beginning of the proof.
Thus the norm of Pi is |Fp[θi]| = pdi , where di is the degree of hi. By (4.2.6), (4.2.7) and
equation (3) of (2.1.3),
pn = N((p)) =
r
i=1
N(Pi)ki =
r
i=1
pdiki
hence n = d1k1 +· · ·+drkr. But n is the degree of the monic polynomial f, which is the
same as deg f = d1e1 +· · ·+drer. Since ki ≤ ei for every i, we have ki = ei for all i, and
the result follows. ♣
4.3.2 Prime Factorization in Quadratic Fields
We consider L = Q(
√
m), where m is a square-free integer, and factor the ideal (p) in
the ring B of algebraic integers of L. By the ram-rel identity (4.1.6), there will be three
cases:
(1) g = 2, e1 = e2 = f1 = f2 = 1. Then (p) is the product of two distinct prime ideals P1
and P2, and we say that p splits in L.
(2) g = 1, e1 = 1, f1 = 2. Then (p) is a prime ideal of B, and we say that p remains prime
in L or that p is inert.
(3) g = 1, e1 = 2, f1 = 1. Then (p) = P2
1 for some prime ideal P1, and we say that p
ramifies in L.
We will examine all possibilities systematically.
(a) Assume p is an odd prime not dividing m. Then p does not divide the discriminant,
so p does not ramify.
(a1) If m is a quadratic residue mod p, then p splits. Say m ≡ n2 mod p. Then x2 − m
factors mod p as (x + n)(x − n), so (p) = (p, n +
√
m) (p, n −
√
m).
(a2) If m is not a quadratic residue mod p, then x2 − m cannot be the product of two
linear factors, hence x2 − m is irreducible mod p and p remains prime.
(b) Let p be any prime dividing m. Then p divides the discriminant, hence p ramifies.
Since x2 − m ≡ x2 = xx mod p, we have (p) = (p,
√
m)2.
This takes care of all odd primes, and also p = 2 with m even.
(c) Assume p = 2, m odd.
(c1) Let m ≡ 3 mod 4. Then 2 divides the discriminant D = 4m, so 2 ramifies. We have
x2 − m ≡ (x + 1)2 mod 2, so (2) = (2,1 +
√
m)2.
(c2) Let m ≡ 1 mod 8, hence m ≡ 1 mod 4. An integral basis is {1, (1 +
√
m)/2}, and
the discriminant is D = m. Thus 2 does not divide D, so 2 does not ramify. We claim
that (2) = (2, (1 +
√
m)/2) (2, (1 −
√
m)/2). To verify this note that the right side is
(2, 1 −
√
m,1 +
√
m, (1 − m)/4). This coincides with (2) because (1 − m)/4 is an even
integer and 1 −
√
m+1+
√
m = 2.
If m ≡ 3 or 7 mod 8, then m ≡ 3 mod 4, so there is only one remaining case.
(c3) Let m ≡ 5 mod 8, hence m ≡ 1 mod 4, so D = m and 2 does not ramify. Consider
f(x) = x2−x+(1−m)/4 over B/P, where P is any prime ideal lying over (2). The roots
of f are (1 ±
√
m)/2, so f has a root in B, hence in B/P. But there is no root in F2,
because (1−m)/4 ≡ 1 mod 2. Thus B/P and F2 cannot be isomorphic. If (2) factors as
Q1Q2, then the norm of (2) is 4, so Q1 and Q2 have norm 2, so the B/Qi are isomorphic
to F2, which contradicts the argument just given. Therefore 2 remains prime.
You probably noticed something suspicious in cases (a) and (b). In order to apply
(4.3.1), 1 and
√
m must form an integral basis, so m ≡ 1 mod 4, as in (2.3.11). But we
can repair the damage. In (a1), verify directly that the factorization of (p) is as given. The
key point is that the ideal (p, n+
√
m) (p, n−
√
m) contains p(n+
√
m+n−
√
m) = 2np,
and if p divides n, then p divides (m − n2) + n2 = m, contradicting the assumption of
case (a). Thus the greatest common divisor of p2 and 2np is p, so p belongs to the ideal.
Since every generator of the ideal is a multiple of p, the result follows. In (a2), suppose
(p) = Q1Q2. Since the norm of p is p2, each Qi has norm p, so B/Qi must be isomorphic
to Fp. But
√
m ∈ B, so m has a square root in B/Qi [see (4.1.4)]. But case (a2) assumes
that there is no square root of m in Fp, a contradiction. Finally, case (b) is similar to
case (a1). We have p|m, but p2 does not divide the square-free integer m, so the greatest
common divisor of p2 and m is p.
Problems For Section 4.3
1. In the exercises for Section 3.4, we factored (2) and (3) in the ring B of algebraic
integers of L = Q(
√
−5), using ad hoc techniques. Using the results of this section, derive
the results rigorously.
2. Continuing Problem 1, factor (5), (7) and (11).
3. Let L = Q( 3
√
2), and assume as known that the ring of algebraic integers is B = Z[ 3
√
2].
Find the prime factorization of (5).
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4.1 Lifting of Prime Ideals
Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite,
separable extension of K of degree n, and B is the integral closure of A in L. If A = Z,
then K = Q, L is a number field, and B is the ring of algebraic integers of L.
4.1.1 Definitions and Comments
Let P be a nonzero prime ideal of A. The lifting (also called the extension) of P to B is
the ideal PB. Although PB need not be a prime ideal of B, we can use the fact that B
is a Dedekind domain [see (3.1.3)] and the unique factorization theorem (3.3.1) to write
PB =
g
i=1
Pei
i
where the Pi are distinct prime ideals of B and the ei are positive integers [see (3.3.2)].
On the other hand, we can start with a nonzero prime ideal Q of B and form a prime
ideal of A via
P = Q ∩ A.
We say that Q lies over P, or that P is the contraction of Q to A.
Now suppose that we start with a nonzero prime ideal P of A and lift it to B. We
will show that the prime ideals P1, . . . , Pg that appear in the prime factorization of PB
are precisely the prime ideals of B that lie over P.
4.1.2 Proposition
Let Q be a nonzero prime ideal of B. Then Q appears in the prime factorization of PB
if and only if Q ∩ A = P.
Proof. If Q ∩ A = P, then P ⊆ Q, hence PB ⊆ Q because Q is an ideal. By (3.3.5), Q
divides PB. Conversely, assume that Q divides, hence contains, PB. Then
P = P ∩ A ⊆ PB ∩ A ⊆ Q ∩ A.
But in a Dedekind domain, every nonzero prime ideal is maximal, so P = Q ∩ A. ♣
4.1.3 Ramification and Relative Degree
If we lift P to B and factor PB as g
i=1 Pei
i , the positive integer ei is called the ramification
index of Pi over P (or over A). We say that P ramifies in B (or in L) if ei > 1 for at
least one i. We will prove in a moment that B/Pi is a finite extension of the field A/P.
The degree fi of this extension is called the relative degree (or the residue class degree, or
the inertial degree) of Pi over P (or over A).
4.1.4 Proposition
We can identify A/P with a subfield of B/Pi, and B/Pi is a finite extension of A/P.
Proof. The map from A/P to B/Pi given by a+P → a+Pi is well-defined and injective,
because P = Pi ∩ A, and it is a homomorphism by direct verification. By (3.1.2), B is a
finitely generated A-module, hence B/Pi is a finitely generated A/P-module, that is, a
finite-dimensional vector space over A/P. ♣
4.1.5 Remarks
The same argument, with Pi replaced by PB, shows that B/PB is a finitely generated
A/P-algebra, in particular, a finite-dimensional vector space over A/P. We will denote
the dimension of this vector space by [B/PB : A/P].
The numbers ei and fi are connected by an important identity, which does not seem
to have a name in the literature. We will therefore christen it as follows.
4.1.6 Ram-Rel Identity
g
i=1
eifi = [B/PB : A/P] = n.
Proof. To prove the first equality, consider the chain of ideals
B ⊇ P1 ⊇ P2
1
⊇ ·· · ⊇ Pe1
1
⊇ Pe1
1 P2 ⊇ Pe1
1 P2
2
⊇ ·· · ⊇ Pe1
1 Pe2
2
⊇ ·· · ⊇ Pe1
1
· · · Peg
g = PB.
By unique factorization, there can be no ideals between consecutive terms in the sequence.
(Any such ideal would contain, hence divide, PB.) Thus the quotient β/βPi of any two
consecutive terms is a one-dimensional vector space over B/Pi, as there are no nontrivial
proper subspaces. (It is a vector space over this field because it is annihilated by Pi.)
But, with notation as in (4.1.5), [B/Pi : A/P] = fi, so [β/βPi : A/P] = fi. For each i,
we have exactly ei consecutive quotients, each of dimension fi over A/P. Consequently,
[B/PB : A/P] = g
i=1 eifi, as claimed.
To prove the second equality, we first assume that B is a free A-module of rank n. By
(2.3.8), this covers the case where A is a PID, in particular, when L is a number field. If
x1, . . . , xn is a basis for B over A, we can reduce mod PB to produce a basis for B/PB
over A/P, and the result follows. Explicitly, suppose n
i=1(ai+P)(xi+PB) = 0 in B/PB.
Then n
i=1 aixi belongs to PB, hence can be written as j bjyj with bj ∈ B, yj ∈ P.
Since bj = k cjkxk with cjk ∈ A, we have ak = j cjkyj ∈ P for all k.
The general case is handled by localization. Let S = A\P, A = S−1A,B = S−1B. By
(1.2.6), (1.2.9), and the Dedekind property (every nonzero prime ideal of A is maximal),
it follows that A has exactly one nonzero prime ideal, namely P = PA . Moreover, P
is principal, so A is a discrete valuation ring, that is, a local PID that is not a field. [By
unique factorization, we can choose an element a ∈ P \(P )2, so (a) ⊆ P but (a) ⊆ (P )2.
Since the only nonzero ideals of A are powers of P (unique factorization again), we have
(a) = P .] Now B is the integral closure of A in L, so B is the integral closure of A in
S−1L = L. [The idea is that we can go back and forth between an equation of integral
dependence for b ∈ B and an equation of integral dependence for b/s ∈ B either by
introducing or clearing denominators.] We have now reduced to the PID case already
analyzed, and [B /PB : A /PA ] = n.
Now PB = g
i=1 Pei
i , and Pi is a nonzero prime ideal of B not meeting S. [If
y ∈ Pi ∩ S, then y ∈ Pi ∩ A = P by (4.1.2). Thus y ∈ P ∩ S, a contradiction.] By the
basic correspondence (1.2.6), we have the factorization PB = g
i=1(PiB )ei . By the PID
case,
n = [B
/PB
: A
/PA
] =
g
i=1
ei[B
/PiB
: A
/PA
].
We are finished if we can show that B /PiB ∼=B/Pi and A /PA ∼=
A/P. The statement
of the appropriate lemma, and the proof in outline form, are given in the exercises. ♣
Problems For Section 4.1
We will fill in the gap at the end of the proof of the ram-rel identity. Let S be a multiplicative
subset of the integral domain A, and let M be a maximal ideal of A disjoint
from S. Consider the composite map A → S−1A → S−1A/MS−1A, where the first map
is given by a → a/1 and the second by a/s → (a/s) +MS−1A.
1. Show that the kernel of the map is M, so by the factor theorem, we have a monomorphism
h : A/M→S−1A/MS−1A.
2. Let a/s ∈ S−1A. Show that for some b ∈ A we have bs ≡1 mod M.
3. Show that (a/s) +MS−1A = h(ab), so h is surjective and therefore an isomorphism.
Consequently, S−1A/MS−1A
∼=
A/M, which is the result we need.
4.2 Norms of Ideals
4.2.1 Definitions and Comments
We are familiar with the norm of an element of a field, and we are going to extend the
idea to ideals. We assume the AKLB setup with A = Z, so that B is a number ring,
that is, the ring of algebraic integers of a number field L. If I is a nonzero ideal of B, we
define the norm of I by N(I) = |B/I|. We will show that the norm is finite, so if P is a
nonzero prime ideal of B, then B/P is a finite field. Also, N has a multiplicative property
analogous to the formula N(xy) = N(x)N(y) for elements. [See (2.1.3), equation (2).]
4.2.2 Proposition
Let b be any nonzero element of the ideal I of B, and let m = NL/Q(b) ∈ Z. Then m ∈ I
and |B/mB| = mn, where n = [L : Q].
Proof. By (2.1.6), m = bc where c is a product of conjugates of b. But a conjugate of an
algebraic integer is an algebraic integer. (If a monomorphism is applied to an equation
of integral dependence, the result is an equation of integral dependence.) Thus c ∈ B,
and since b ∈ I, we have m ∈ I. Now by (2.3.9), B is the direct sum of n copies of Z,
hence by the first isomorphism theorem, B/mB is the direct sum of n copies of Z/mZ.
Consequently, |B/mB| = mn. ♣
4.2.3 Corollary
If I is any nonzero ideal of B, then N(I) is finite. In fact, if m is as in (4.2.2), then N(I)
divides mn.
Proof. Observe that (m) ⊆ I, hence
B/(m)
B/I
∼=
I/(m). ♣
4.2.4 Corollary
Every nonzero ideal I of B is a free abelian group of rank n.
Proof. By the simultaneous basis theorem, we may represent B as the direct sum of n
copies of Z, and I as the direct sum of a1Z, . . . ,arZ, where r ≤ n and the ai are positive
integers such that ai divides ai+1 for all i. Thus B/I is the direct sum of r cyclic groups
(whose orders are a1, . . . , ar) and n − r copies of Z. If r < n, then at least one copy of Z
appears, and |B/I| cannot be finite. ♣
4.2.5 Computation of the Norm
Suppose that {x1, . . . , xn} is a Z-basis for B, and {z1, . . . , zn} is a basis for I. Each zi is
a linear combination of the xi with integer coefficients, in matrix form z = Cx. We claim
that the norm of I is the absolute value of the determinant of C. To verify this, first look
at the special case xi = yi and zi = aiyi, as in the proof of (4.2.4). Then C is a diagonalmatrix with entries ai, and the result follows. But the special case implies the general
result, because any matrix corresponding to a change of basis of B or I is unimodular, in
other words, has integer entries and determinant ±1. [See (2.3.9) and (2.3.10).]
Now with z = Cx as above, the discriminant of x is the field discriminant d, and the
discriminant of z is D(z) = (detC)2d by (2.3.2). We have just seen that N(I) = | detC|,
so we have the following formula for computing the norm of an ideal I. If z is a Z-basis
for I, then
N(I) =
D(z)
d
1/2
.
There is a natural relation between the norm of a principal ideal and the norm of the
corresponding element.
4.2.6 Proposition
If I = (a) with a = 0, then N(I) = |NL/Q(a)|.
Proof. If x is a Z-basis for B, then ax is a Z-basis for I. By (2.3.3), D(ax) is the square
of the determinant whose ij entry is σi(axj) = σi(a)σi(xj ). By (4.2.5), the norm of I is
|σ1(a) · · · σn(a)| = |NL/Q(a)|. ♣
In the proof of (4.2.6), we cannot invoke (2.3.2) to get D(ax1, . . . ,axn) = (an)2D(x1, . . . , xn),
because we need not have a ∈ Q.
We now establish the multiplicative property of ideal norms.
4.2.7 Theorem
If I and J are nonzero ideals of B, then N(IJ) = N(I)N(J).
Proof. By unique factorization, we may assume without loss of generality that J is a
prime ideal P. By the third isomorphism theorem, |B/IP| = |B/I| |I/IP|, so we must
show that |I/IP| is the norm of P, that is, |B/P|. But this has already been done in the
first part of the proof of (4.1.6). ♣
4.2.8 Corollary
Let I be a nonzero ideal of B. If N(I) is prime, then I is a prime ideal.
Proof. Suppose I is the product of two ideals I1 and I2. By (4.2.7), N(I) = N(I1)N(I2),
so by hypothesis, N(I1) = 1 or N(I2) = 1. Thus either I1 or I2 is the identity element
of the ideal group, namely B. Therefore, the prime factorization of I is I itself, in other
words, I is a prime ideal. ♣
4.2.9 Proposition
N(I) ∈ I, in other words, I divides N(I). [More precisely, I divides the principal ideal
generated by N(I).]
Proof. Let N(I) = |B/I| = r. If x ∈ B, then r(x + I) is 0 in B/I, because the order of
any element of a group divides the order of the group. Thus rx ∈ I, and in particular we
may take x = 1 to conclude that r ∈ I. ♣
4.2.10 Corollary
If I is a nonzero prime ideal of B, then I divides (equivalently, contains) exactly one
rational prime p.
Proof. By (4.2.9), I divides N(I) = pm1
1
· · · pmt
t , so I divides some pi. But if I divides
two distinct primes p and q, then there exist integers u and v such that up+vq = 1. Thus
I divides 1, so I = B, a contradiction. Therefore I divides exactly one p. ♣
4.2.11 The Norm of a Prime Ideal
If we can compute the norm of every nonzero prime ideal P, then by multiplicativity, we
can calculate the norm of any nonzero ideal. Let p be the unique rational prime in P, and
recall from (4.1.3) that the relative degree of P over p is f(P) = [B/P : Z/pZ]. Therefore
N(P) = |B/P| = pf(P).
Note that by (4.2.6), the norm of the principal ideal (p) is |N(p)| = pn, so N(P) = pm
for some m ≤ n. This conclusion also follows from the above formula N(P) = pf(P) and
the ram-rel identity (4.1.6).
Here are two other useful finiteness results.
4.2.12 Proposition
A rational integer m can belong to only finitely many ideals of B.
Proof. We have m ∈ I iff I divides (m), and by unique factorization, (m) has only finitely
many divisors. ♣
4.2.13 Corollary
Only finitely many ideals can have a given norm.
Proof. If N(I) = m, then by (4.2.9), m ∈ I, and the result follows from (4.2.12). ♣
Problems For Section 4.2
This problem set will give the proof that a rational prime p ramifies in the number field
L if and only if p divides the field discriminant d = dL/Q.
1. Let (p) = pB have prime factorization i Pei
i . Show that p ramifies if and only if the
ring B/(p) has nonzero nilpotent elements.
Now as in (2.1.1), represent elements of B by matrices with respect to an integral basis
ω1, . . . , ωn of B. Reduction of the entries mod p gives matrices representing elements of
B/(p).
2. Show that a nilpotent element (or matrix) has zero trace.
Suppose that A(β), the matrix representing the element β, is nilpotent mod p. Then
A(βωi) will be nilpotent mod p for all i, because βωi is nilpotent mod p.
3. By expressing β in terms of the ωi and computing the trace of A(βωj ), show that if β
is nilpotent mod p and β /∈ (p), then d ≡0 mod p, hence p divides d.
Now assume that p does not ramify.
4. Show that B/(p) is isomorphic to a finite product of finite fields Fi of characteristic p.
Let Ï€i : B → B/(p) → Fi be the composition of the canonical map from B onto B/(p)
and the projection from B/(p) onto Fi.
5. Show that the trace form Ti(x, y) = TFi/Fp (Ï€i(x)Ï€i(y)) is nondegenerate, and conclude
that i Ti is also nondegenerate.
We have d = det T(ωiωj ), in other words, the determinant of the matrix of the bilinear
form T(x, y) on B, with respect to the basis {ω1, . . . , ωn}. Reducing the matrix entries
mod p, we get the matrix of the reduced bilinear form T0 on the Fp-vector space B/(p).
6. Show that T0 coincides with i Ti, hence T0 is nondegenerate. Therefore d ≡0 mod p,
so p does not divide d.
As a corollary, it follows that only finitely many primes can ramify in L.
4.3 A Practical Factorization Theorem
The following result, usually credited to Kummer but sometimes attributed to Dedekind,
allows, under certain conditions, an efficient factorization of a rational prime in a number
field.
4.3.1 Theorem
Let L be a number field of degree n over Q, and assume that the ring B of algebraic
integers of L is Z[θ] for some θ ∈ B. Thus 1, θ, θ2, . . . , θn−1 form an integral basis of B.
Let p be a rational prime, and let f be the minimal polynomial of θ over Q. Reduce the
coefficients of f modulo p to obtain f ∈ Z[X]. Suppose that the factorization of f into
irreducible polynomials over Fp is given by
f = he1
1
· · · her
r .
Let fi be any polynomial in Z[X] whose reduction mod p is hi. Then the ideal
Pi = (p, fi(θ))
is prime, and the prime factorization of (p) in B is
(p) = Pe1
1
· · · Per
r .
Proof. Adjoin a root θi of hi to produce the field Fp[θi] ∼=
Fp[X]/hi(X). The assignment
θ → θi extends by linearity (and reduction of coefficients mod p) to an epimorphism
λi : Z[θ] → Fp[θi]. Since Fp[θi] is a field, the kernel of λi is a maximal, hence prime,
ideal of Z[θ] = B. Since λi maps fi(θ) to hi(θi) = 0 and also maps p to 0, it follows that
Pi ⊆ ker λi. We claim that Pi = kerλi. To prove this, assume g(θ) ∈ ker λi. With asubscript 0 indicating reduction of coefficients mod p, we have g0(θi) = 0, hence hi, the
minimal polynomial of θi, divides g0. If g0 = q0hi, then g − qfi ≡0 mod p. Therefore
g(θ) = [g(θ) − q(θ)fi(θ)] + q(θ)fi(θ)
so g(θ) is the sum of an element of (p) and an element of (fi(θ)). Thus ker λi ⊆ Pi, so
Pi = kerλi, a prime ideal.
We now show that (p) divides Pe1
1
· · · Per
r . We use the identity (I+I1)(I+I2) ⊆ I+I1I2,
where I, I1 and I2 are ideals. We begin with P1 = (p) + (f1(θ)), and compute
P2
1
⊆ (p) + (f1(θ))2, . . . ,Pe1
1
· · · Per
r
⊆ (p) + (f1(θ))e1 · · · (fr(θ))er .
But the product of the fi(θ)ei coincides mod p with r
i=1 hi(θ) = f(θ) = 0. We conclude
that r
i=1 Pei
i
⊆ (p), as asserted.
We now know that (p) = Pk1
1
· · · Pkr
r with 0 ≤ ki ≤ ei. (Actually, ki > 0 since
p ∈ ker λi = Pi, so Pi divides (p). But we will not need this refinement.) By hypothesis,
B/Pi = Z[θ]/Pi, which is isomorphic to Fp[θi], as observed at the beginning of the proof.
Thus the norm of Pi is |Fp[θi]| = pdi , where di is the degree of hi. By (4.2.6), (4.2.7) and
equation (3) of (2.1.3),
pn = N((p)) =
r
i=1
N(Pi)ki =
r
i=1
pdiki
hence n = d1k1 +· · ·+drkr. But n is the degree of the monic polynomial f, which is the
same as deg f = d1e1 +· · ·+drer. Since ki ≤ ei for every i, we have ki = ei for all i, and
the result follows. ♣
4.3.2 Prime Factorization in Quadratic Fields
We consider L = Q(
√
m), where m is a square-free integer, and factor the ideal (p) in
the ring B of algebraic integers of L. By the ram-rel identity (4.1.6), there will be three
cases:
(1) g = 2, e1 = e2 = f1 = f2 = 1. Then (p) is the product of two distinct prime ideals P1
and P2, and we say that p splits in L.
(2) g = 1, e1 = 1, f1 = 2. Then (p) is a prime ideal of B, and we say that p remains prime
in L or that p is inert.
(3) g = 1, e1 = 2, f1 = 1. Then (p) = P2
1 for some prime ideal P1, and we say that p
ramifies in L.
We will examine all possibilities systematically.
(a) Assume p is an odd prime not dividing m. Then p does not divide the discriminant,
so p does not ramify.
(a1) If m is a quadratic residue mod p, then p splits. Say m ≡ n2 mod p. Then x2 − m
factors mod p as (x + n)(x − n), so (p) = (p, n +
√
m) (p, n −
√
m).
(a2) If m is not a quadratic residue mod p, then x2 − m cannot be the product of two
linear factors, hence x2 − m is irreducible mod p and p remains prime.
(b) Let p be any prime dividing m. Then p divides the discriminant, hence p ramifies.
Since x2 − m ≡ x2 = xx mod p, we have (p) = (p,
√
m)2.
This takes care of all odd primes, and also p = 2 with m even.
(c) Assume p = 2, m odd.
(c1) Let m ≡ 3 mod 4. Then 2 divides the discriminant D = 4m, so 2 ramifies. We have
x2 − m ≡ (x + 1)2 mod 2, so (2) = (2,1 +
√
m)2.
(c2) Let m ≡ 1 mod 8, hence m ≡ 1 mod 4. An integral basis is {1, (1 +
√
m)/2}, and
the discriminant is D = m. Thus 2 does not divide D, so 2 does not ramify. We claim
that (2) = (2, (1 +
√
m)/2) (2, (1 −
√
m)/2). To verify this note that the right side is
(2, 1 −
√
m,1 +
√
m, (1 − m)/4). This coincides with (2) because (1 − m)/4 is an even
integer and 1 −
√
m+1+
√
m = 2.
If m ≡ 3 or 7 mod 8, then m ≡ 3 mod 4, so there is only one remaining case.
(c3) Let m ≡ 5 mod 8, hence m ≡ 1 mod 4, so D = m and 2 does not ramify. Consider
f(x) = x2−x+(1−m)/4 over B/P, where P is any prime ideal lying over (2). The roots
of f are (1 ±
√
m)/2, so f has a root in B, hence in B/P. But there is no root in F2,
because (1−m)/4 ≡ 1 mod 2. Thus B/P and F2 cannot be isomorphic. If (2) factors as
Q1Q2, then the norm of (2) is 4, so Q1 and Q2 have norm 2, so the B/Qi are isomorphic
to F2, which contradicts the argument just given. Therefore 2 remains prime.
You probably noticed something suspicious in cases (a) and (b). In order to apply
(4.3.1), 1 and
√
m must form an integral basis, so m ≡ 1 mod 4, as in (2.3.11). But we
can repair the damage. In (a1), verify directly that the factorization of (p) is as given. The
key point is that the ideal (p, n+
√
m) (p, n−
√
m) contains p(n+
√
m+n−
√
m) = 2np,
and if p divides n, then p divides (m − n2) + n2 = m, contradicting the assumption of
case (a). Thus the greatest common divisor of p2 and 2np is p, so p belongs to the ideal.
Since every generator of the ideal is a multiple of p, the result follows. In (a2), suppose
(p) = Q1Q2. Since the norm of p is p2, each Qi has norm p, so B/Qi must be isomorphic
to Fp. But
√
m ∈ B, so m has a square root in B/Qi [see (4.1.4)]. But case (a2) assumes
that there is no square root of m in Fp, a contradiction. Finally, case (b) is similar to
case (a1). We have p|m, but p2 does not divide the square-free integer m, so the greatest
common divisor of p2 and m is p.
Problems For Section 4.3
1. In the exercises for Section 3.4, we factored (2) and (3) in the ring B of algebraic
integers of L = Q(
√
−5), using ad hoc techniques. Using the results of this section, derive
the results rigorously.
2. Continuing Problem 1, factor (5), (7) and (11).
3. Let L = Q( 3
√
2), and assume as known that the ring of algebraic integers is B = Z[ 3
√
2].
Find the prime factorization of (5).
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