Unknown
As usual, we will be working in the ring B of algebraic integers of a number field L. Two
factorizations of an element of B are regarded as essentially the same if one is obtained
from the other by multiplication by a unit. Our experience with the integers, where
the only units are ±1, and the Gaussian integers, where the only units are ±1 and ±i,
suggests that units are not very complicated, but this is misleading. The Dirichlet unit
theorem gives a complete description of the structure of the multiplicative group of units
in a number field.
6.1 Preliminary Results
6.1.1 Lemma
Let B∗ be the group of units of B. An element x ∈ B belongs to B∗ if and only if
N(x) = ±1.
Proof. If xx−1 = 1, then 1 = N(1) = N(xx−1) = N(x)N(x−1), so the integer N(x) must
be ±1. Conversely, if the norm of x is ±1, then the characteristic equation of x has the
form xn + an−1xn−1 + · · · + a1x ± 1 = 0, with the ai ∈ Z [see (2.1.3) and (2.2.2)]. Thus
x(xn−1 + an−1xn−2 + · · · + a2x + a1) = ∓ 1. ♣
6.1.2 The Logarithmic Embedding
Let σ : L → Rr1 × Cr2 = Rn be the canonical embedding defined in (5.3.1). The
logarithmic embedding is the mapping λ : L∗ → Rr1+r2 given by
λ(x) = (log |σ1(x)|, . . . , log |σr1+r2 (x)|).
Since the σi are monomorphisms, λ(xy) = λ(x)+λ(y), so λ is a homomorphism from the
multiplicative group of L∗ to the additive group of Rr1+r2 .
6.1.3Lemma
Let C be a bounded subset of Rr1+r2 , and let C = {x ∈ B∗ : λ(x) ∈ C}. Then C is a
finite set.
Proof. Since C is bounded, all the numbers |σi(x)|, x ∈ B∗, i = 1, . . . ,n, will be confined
to some interval [a−1, a] with a > 1. Thus the elementary symmetric functions of the
σi(x) will also lie in some interval of this type. But by (2.1.6), the elementary symmetric
functions are the coefficients of the characteristic polynomial of x, and by (2.2.2), these
coefficients are integers. Thus there are only finitely many possible characteristic polynomials
of elements x ∈ C , hence by (2.1.5), only finitely many possible roots of minimal
polynomials of elements x ∈ C . We conclude that x can belong to C for only finitely
many x. ♣
6.1.4 Corollary
The kernel G of the homomorphism λ restricted to B∗ is a finite group.
Proof. Take C = {0} in (6.1.3). ♣
The following result gives additional information about G.
6.1.5 Proposition
Let H be a finite subgroup of K∗, where K is an arbitrary field. Then H consists of roots
of unity and is cyclic.
Proof. Let z be an element of H whose order n is the exponent of H, that is, the least
common multiple of the orders of all the elements of H. Then yn = 1 for every y ∈ H, so
H consists of roots of unity. Since the polynomial Xn − 1 has at most n distinct roots,
we have |H| ≤ n. But 1, z, . . . , zn−1 are distinct elements of H, because z has order n.
Thus H is cyclic. ♣
For our group G, even more is true.
6.1.6 Proposition
The group G consists exactly of all the roots of unity in the field L.
Proof. By (6.1.5), every element of G is a root of unity. Conversely, suppose xm = 1.
Then x is an algebraic integer (it satisfies Xm − 1 = 0) and for every i,
|σi(x)|m = |σi(xm)| = |1| = 1.
Thus |σi(x)| = 1 for all i, so log|σi(x)| = 0 and x ∈ G. ♣
6.1.7 Proposition
B∗ is a finitely generated abelian group, isomorphic to G × Zs where s ≤ r1 + r2.
Proof. By (6.1.3), λ(B∗) is a discrete subgroup of Rr1+r2 . [“Discrete” means that any
bounded subset of Rr1+r2 contains only finitely many points of λ(B∗).] It follows that λ(B∗) is a lattice in Rs, hence a free Z-module of rank s, for some s ≤ r1 +r2. The proof
of this is outlined in the exercises. Now by the first isomorphism theorem, λ(B∗) ∼=
B∗/G,
with λ(x) corresponding to the coset xG. If x1G, . . . , xsG form a basis for B∗/G and
x ∈ B∗, then xG is a finite product of powers of the xiG, so x is an element of G times a
finite product of powers of the xi. Since the λ(xi) are linearly independent, so are the xi,
provided we translate the notion of linear independence to a multiplicative setting. The
result follows. ♣
We can improve the estimate of s.
6.1.8 Proposition
In (6.1.7), we have s ≤ r1 + r2 − 1.
Proof. If x ∈ B∗, then by (6.1.1) and (2.1.6),
±1 = N(x) =
n
i=1
σi(x) =
r1
i=1
σi(x)
r1+r2
j=r1+1
σj(x)σj(x).
Take absolute values and apply the logarithmic embedding to conclude that λ(x) =
(y1, . . . , yr1+r2 ) lies in the hyperplane W whose equation is
r1
i=1
yi + 2
r1+r2
j=r1+1
yj = 0.
The hyperplane has dimension r1 + r2 − 1, so as in the proof of (6.1.7), λ(B∗) is a free
Z-module of rank s ≤ r1 + r2 − 1. ♣
In the next section, we will prove the Dirichlet unit theorem, which says that s actually
equals r1 + r2 − 1.
Problems For Section 6.1
We will show that if H is a discrete subgroup of Rn, in other words, for every bounded set
C ⊆ Rn, H ∩ C is finite, then H is a lattice in Rr for some r ≤ n. Choose e1, . . . , er ∈ H
such that the ei are linearly independent over R and r is as large as possible. Let T
be the closure of the fundamental domain determined by the ei, that is, the set of all
x = r
i=1 aiei, with 0 ≤ ai ≤ 1. Since H is discrete, H ∩ T is a finite set.
Now let x be any element of H. By choice of r we have x = r
i=1 biei with bi ∈ R.
1. If j is any integer, set xj = jx− r
i=1
jbi ei, where y is the maximum of all integers
z ≤ y. Show that xj ∈ H ∩ T.
2. By examining the above formula for xj with j = 1, show that H is a finitely generated
Z-module.
3. Show that the bi are rational numbers.
4. Show that for some nonzero integer d, dH is a free Z-module of rank at most r.
5. Show that H is a lattice in Rr.
view of (6.1.4)-(6.1.8), it suffices to prove that s ≥ r1 + r2 − 1. Equivalently,
by the proof of (6.1.7), the real vector space V = λ(B∗) contains r1 + r2 − 1 linearly
independent vectors. Now by the proof of (6.1.8), V is a subspace of the (r1 + r2 − 1)-
dimensional hyperplane W, so we must prove that V = W. To put it another way, every
linear form f that vanishes on V must vanish on W. This is equivalent to saying that if
f does not vanish on W, then it cannot vanish on V , that is, for some unit u ∈ B∗ we
have f(λ(u)) = 0.
Step 1. We apply Minkowski’s convex body theorem (5.1.3b) to the set
S = {(y1, . . . , yr1, z1, . . . , zr2 ) ∈ Rr1 × Cr2 : |yi| ≤ ai, |zj| ≤ ar1+j}
where i ranges from 1 to r1 and j from 1 to r2. We specify the ai as follows. Fix
the positive real number b ≥ 2n−r1(1/2Ï€)r2 |d|1/2. Given arbitrary positive real numbers
a1, . . . ,ar, where r = r1 + r2 − 1, we choose the positive real number ar+1 such that
r1
i=1
ai
r1+r2
j=r1+1
a2j
= b.
The set S is compact, convex, and symmetric about the origin, and its volume is
r1
i=1
2ai
r1+r2
j=r1+1
Ï€a2j
= 2r1Ï€r2b ≥ 2n−r2 |d|1/2.
We apply (5.1.3b) with S as above and H = σ(B) [see (5.3.3)], to get S ∩ (H \ {0}) = ∅.
Thus there is a nonzero algebraic integer x = xa, a = (a1, . . . , ar), such that σ(xa) ∈ S,
and consequently,
|σi(xa)| ≤ ai, i = 1, . . . , n,
where we set aj+r2 = aj, j = r1 + 1, . . . , r1 + r2.
Step 2. We will show that the norms of the xa are bounded by b in absolute value, and
0 ≤ log ai − log |σi(xa)| ≤ log b.
Using step 1, along with (2.1.6) and the fact that the norm of an algebraic integer is a
rational integer [see (2.2.2)], we find
1 ≤ |N(xa)| =
n
i=1
|σi(xa)| ≤
r1
i=1
ai
r1+r2
j=r1+1
a2j
= b.
But for any i,
|σi(xa)| = |N(xa)|
j =i
|σj(xa)|−1 ≥
j =i
a
−1
j = aib
−1.
Thus aib−1 ≤ |σi(xa)| ≤ ai for all i, so 1 ≤ ai/|σi(xa)| ≤ b. Take logarithms to obtain
the desired chain of inequalities.
Step 3. Completion of the proof. In the equation of the hyperplane W, y1, . . . , yr can be
specified arbitrarily and we can solve for yr+1. Thus if f is a nonzero linear form on W,
then f can be expressed as f(y1, . . . , yr+1) = c1y1 + · · · + cryr with not all ci’s zero. By
definition of the logarithmic embedding [see (6.1.2)], f(λ(xa)) = r
i=1 ci log |σi(xa)|, so if
we multiply the inequality of Step 2 by ci and sum over i, we get
|
r
i=1
ci log ai − f(λ(xa))| = |
r
i=1
ci(log ai − log |σi(xa)|)| ≤
r
i=1
|ci| log b.
Choose a positive real number t greater than the right side of this equation, and for every
positive integer h, choose positive real numbers aih, i = 1, . . . , r, such that r
i=1 ci log aih
coincides with 2th. (This is possible because not all ci’s are zero.) Let a(h) = (a1h, . . . , arh),
and let xh be the corresponding algebraic integer xa(h). Then by the displayed equation
above and the choice of t to exceed the right side, we have |f(λ(xh)) − 2th| < t, so
(2h − 1)t < f(λ(xh)) < (2h + 1)t.
Since the open intervals ((2h − 1)t, (2h + 1)t) are (pairwise) disjoint, it follows that the
f(λ(xh)), h = 1, 2, . . . , are all distinct. But by Step 2, the norms of the xh are all bounded
in absolute value by the same positive constant, and by (4.2.13), only finitely many ideals
can have a given norm. By (4.2.6), there are only finitely many distinct ideals of the
form Bxh, so there are distinct h and k such that Bxh = Bxk. But then xh and xk are
associates, hence for some unit u we have xh = uxk, hence λ(xh) = λ(u) + λ(xk). By
linearity of f and the fact that f(λ(xh)) = f(λ(xk)), we have f(λ(u)) = 0. ♣
6.2.2 Remarks
The unit theorem implies that there are r = r1 + r2 − 1 units u1, . . . , ur in B such that
every unit of B can be expressed uniquely as
u = z un1
1
· · · unr
r
where the ui are algebraic integers and z is a root of unity in L. We call {u1, . . . , ur} a
fundamental system of units for the number field L.
As an example, consider the cyclotomic extension L = Q(z), where z is a primitive
pth root of unity, p an odd prime. The degree of the extension is Ï•(p) = p − 1, and an
embedding σj maps z to zj, j = 1, . . . , p − 1. Since these zj ’s are never real, we have
r1 = 0 and 2r2 = p − 1. Therefore r = r1 + r2 − 1 = (p − 3)/2.
6.3Units in Quadratic Fields
6.3.1 Imaginary Quadratic Fields
First, we look at number fields L = Q(
√
m), where m is a square-free negative integer.
There are no real embeddings, so r1 = 0 and 2r2 = n = 2, hence r2 = 1. But then
r1 + r2 − 1 = 0, so the only units in B are the roots of unity in L. We will use (6.1.1) to
determine the units.
Case 1. Assume m ≡ 1 mod 4. By (2.3.11), an algebraic integer has the form x = a+b
√
m
for integers a and b. By (6.1.1) and (2.1.10), x is a unit iff N(x) = a2 − mb2 = ±1. Thus
if m ≤ −2, then b = 0 and a = ±1. If m = −1, we have the additional possibility
a = 0, b = ±1.
Case 2. Assume m ≡ 1 mod 4. By (2.3.11), x = a + b(1 +
√
m)/2, and by (2.1.10),
N(x) = (a + b/2)2 − mb2/4 = [(2a + b)2 − mb2]/4. Thus x is a unit if and only if
(2a + b)2 − mb2 = 4. We must examine m = −3,−7,−11,−15, . . .. If m ≤ −7, then
b = 0, a = ±1. If m = −3, we have the additional possibilities b = ±1, (2a±b)2 = 1, that
is, a = 0, b = ±1; a = 1, b = −1; a = −1, b = 1.
To summarize, if B is the ring of algebraic integers of an imaginary quadratic field,
then the group G of units of B is {1,−1}, except in the following two cases:
1. If L = Q(i), then G = {1, i,−1,−i}, the group of 4th roots of unity in L.
2. If L = Q(
√
−3), then G = {[(1 +
√
−3)/2]j, j = 0, 1, 2, 3, 4, 5}, the group of 6th roots
of unity in L. We may list the elements x = a + b/2 + b
√
−3/2 ∈ G as follows:
j = 0 ⇒ x = 1 (a = 1, b = 0)
j = 1 ⇒ x = (1+
√
−3)/2 (a = 0, b = 1)
j = 2 ⇒ x = (−1 +
√
−3)/2 (a = −1, b = 1)
j = 3 ⇒ x = −1 (a = −1, b = 0)
j = 4 ⇒ x = −(1 +
√
−3)/2 (a = 0, b = −1)
j = 5 ⇒ x = (1 −
√
−3)/2 (a = 1, b = −1).
6.3.2 Remarks
Note that G, a finite cyclic group, has a generator, necessarily a primitive root of unity.
Thus G will consist of all tth roots of unity for some t, and the field L will contain only
finitely many roots of unity. This is a general observation, not restricted to the quadratic
case.
6.3.3 Real Quadratic Fields
Now we examine L = Q(
√
m), where m is a square-free positive integer. Since the
Q-automorphisms of L are the identity and a + b
√
m → a − b
√
m, there are two real
embeddings and no complex embeddings. Thus r1 = 2, r2 = 0, and r1 + r2 − 1 = 1. The
only roots of unity in R are ±1, so by (6.2.1) or (6.2.2), the group of units in the ring of
algebraic integers is isomorphic to {−1, 1} × Z. If u is a unit and 0 < u < 1, then 1/u
is a unit and 1/u> 1. Thus the units greater than 1 are hn, n = 1, 2, . . . , where h, the
unique generator greater than 1, is called the fundamental unit of L.
Case 1. Assume m ≡ 1 mod 4. The algebraic integers are of the form x = a + b
√
m
with a, b ∈ Z. Thus we are looking for solutions for N(x) = a2 − mb2 = ±1. Note that
if x = a + b
√
m is a solution, then the four numbers ±a ± b
√
m are x,−x, x−1,−x−1 in
some order. Since a number and its inverse cannot both be greater than 1, and similarly
for a number and its negative, it follows that exactly one of the four numbers is greater
than 1, namely the number with a and b positive. The fundamental unit, which is the
smallest unit greater than 1, can be found as follows. Compute mb2 for b = 1, 2, 3, . . . ,
and stop at the first number mb21
that differs from a square a21
by ±1. Then a1 + b1
√
m
is the fundamental unit.
There is a more efficient computational technique using the continued fraction expansion
of
√
m. Details are given in many texts on elementary number theory.
Case 2. Assume m ≡ 1 mod 4. It follows from (2.2.6) that the algebraic integers are of
the form x = 1
2 (a + b
√
m), where a and b are integers of the same parity, both even or
both odd. Since the norm of x is 1
4 (a2 −mb2), x is a unit iff a2 −mb2 = ±4. Moreover, if
a and b are integers satisfying a2 − mb2 = ±4, then a and b must have the same parity,
hence 1
2 (a + b
√
m) is an algebraic integer and therefore a unit of B. To calculate the
fundamental unit, compute mb2, b = 1, 2, 3, . . . , and stop at the first number mb21
that
differs from a square a21
by ±4. The fundamental unit is 1
2 (a1 + b1
√
m).
Problems For Section 6.3
1. Calculate the fundamental unit of Q(
√
m) for m = 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17.
In Problems 2-5, we assume m ≡ 1 mod 4. Suppose that we look for solutions to
a2 − mb2 = ±1 (rather than a2 − mb2 = ±4). We get units belonging to a subring
B0 = Z[
√
m] of the ring B of algebraic integers, and the positive units of B0 form a
subgroup H of the positive units of B. Let u = 1
2 (a + b
√
m) be the fundamental unit of
the number field L.
2. If a and b are both even, for example when m = 17, show that H consists of the powers
of u, in other words, B∗
0 = B∗.
3. If a and b are both odd, show that u3 ∈ B0.
4. Continuing Problem 3, show that u2 /∈
B0, so H consists of the powers of u3.
5. Verify the conclusions of Problems 3 and 4 when m = 5 and m = 13.
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