Download Finite Abelian Groups as Galois Groups

Finite Abelian Groups as Galois Groups
The purpose of this note is to record for my algebra class a proof of the Inverse Galois Problem
for finite abelian groups. Recall that the Inverse Galois Problem is stated as follows: Given a finite
group G, is there a Galois extension Q ⊆ K such Gal(K/Q) = G? The crucial point in the problem
is that the base field is Q, since given any finite group G, there is a Galois extension of fields F ⊆ K
with Galois group equal to G. The solution to the inverse problem is not known in general, but it
does have a positive answer in a number of different cases, including solvable groups and symmetric
groups. Moreover, it was shown in [FK] that an arbitrary finite group can be realized as the Galois
group of a finite extension of Q, but the extension in this construction is not always (or often)
Galois.
In proving the main result for abelian groups, we will use the fact that any finite abelian group
is the direct product of cyclic groups. Let’s first see the idea behind the case when A is cyclic,
say A = Zn . Using a special case of Dirichlet’s theorem on primes in an arithmetic progression,
one can find a prime number p such that n divides p − 1. We will prove this below. If we write
p − 1 = n · m, then any cyclic group of order p − 1 has a (unique) cyclic subgroup of order m, and
if we factor out that subgroup, the quotient group is then cyclic of order n. Let ϵ is a primitive
pth root of unity. We will also see that Gal(Q(ϵ)/Q) is isomorphic to Z∗p , a cyclic group of order
p − 1. Let H ⊆ Gal(Q(ϵ)/Q) be the subgroup of order m and K be the fixed field of H. Then,
since Gal(Q(ϵ)/Q) is abelian, H is normal and K is Galois over Q. Its Galois group over Q is the
factor group Gal(Q(ϵ)/Q)/H ∼
= Zn , which gives what we want. The general case will follow along
similar lines. Thus, we need to deal with the Galois group obtained by adjoining a root of unity
and also the statement about primes in a particular type of arithmetic progression.
Before taking the first step along our path, we’ll make some general comments. For any positive
integer n, Zn is a ring. As such, not every element has a multiplicative inverse. Recall that an
element in a commutative ring is said to be a unit precisely when it does have a multiplicative
inverse. The set of units in a ring forms a group under multiplication. Now, it is not hard to see
that the units in the ring Zn are precisely the residue classes of the positive integers that are less
than n and relatively prime to n. Let Z∗n denote the multiplicative group of units in the ring Zn
and set ϕ(n) := |Z∗n |. Then ϕ(n) is called the the Euler phi function. Thus, ϕ(n) is the number of
positive integers less than n and relatively prime to n. An important property of this function is
that ϕ(nm) = ϕ(n)ϕ(m), whenever n and m are relatively prime. One way to see this is as follows.
Since n and m are relatively prime, Znm ∼
= Zn × Zm . Thus the groups of units in these rings are
isomorphic. However, it is easy to see that an element in a product of rings is a unit if and only
if each coordinate is a unit in the corresponding factor. Thus, (Zn × Zm )∗ = Z∗n × Z∗m , so we have
Z∗nm ∼
= Z∗n × Z∗m . Counting elements in these groups gives ϕ(nm) = ϕ(n)ϕ(m).
In Theorem A below, we will use Gauss’s Lemma. The most common form of this lemma
asserts the following. If f (X), g(X) ∈ Z[X] are primitive polynomials, then f (X)g(X) is a primitive
polynomial. Recall that a polynomial with integer coefficients is said to be primitive if the greatest
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common divisor of its coefficients is 1. Now suppose h(X) ∈ Z[X] is a monic polynomial that factors
as h(X) = f (X)g(X), with f (X), g(X) monic polynomials in Q[X]. Then in fact, f (X) and g(X)
have integer coefficients. To see this, write f (X) = ab f0 (X) and g(X) = dc g0 (X), where, f0 (X)
and g0 (X) are primitive polynomials in Z[X]. Without loss of generality, assume the indicated
fractions are in lowest terms. Then (bd)h(X) = (ac)f0 (X)g0 (X). By Gauss’s Lemma, the product
f0 (X)g0 (X) is primitive. If p is a prime dividing b, then p does not divide f0 (X)g0 (X) in the ring
Z[X]. This is because an integer divides a polynomial in Z[X] precisely when it divides all of its
coefficients. Thus p must divide ac. Since a and b are relatively prime, p divides c. Thus, we can
cancel p from both sides of the equation (bd)h(X) = (ac)f0 (X)g0 (X). Continuing, we end up with
an equation h(X) = (a′ c′ )f0 (X)g0 (X). Since h(X) is monic, a′ c′ = 1. So h(X) = f0 (X)g0 (X). We
can now take both f0 (X), g0 (X) to be monic. But since these are just rational number multiples
of the original f (X) and g(X), which are monic, we must have f0 (X) = f (X) and g0 (X) = g(X),
which gives what we wanted.
We’re now ready for our first major step.
Theorem A. Let ϵ be a primitive nth root of unity. Then Q ⊆ Q(ϵ) is a Galois extension with
Galois group isomorphic to Z∗n . In particular, [Q(ϵ) : Q] = ϕ(n).
Proof. Since Q(ϵ) is the splitting field for X n −1 over Q and the extension is separable, the extension
is Galois. We first show that [Q(ϵ) : Q] = ϕ(n). To see this, set G := Gal(Q(ϵ)/Q) and let σ ∈ G.
Then σ is determined by its value σ(ϵ). On the one hand, its easy to check that σ(ϵ) is also a
primitive nth root of unity. Since the number of primitive nth roots of unity equals ϕ(n), it follows
that ϕ(n) ≥ |G| = [Q(ϵ) : ϵ]. On the other hand, let f (X) ∈ Q(X) be the minimal polynomial for ϵ
over Q. We claim that if p is a prime less than n and relatively prime to n, then f (ϵp ) = 0. Suppose
the claim holds. Then, since ϵp is also a primitive nth root of unity, (ϵp )q = ϵpq is also a root of
f (X) for any prime q less than n and relatively prime to n. By iterating, it follows than for any r
less than n and relatively prime to n, ϵr is a root of f (X). Thus [Q(ϵ) : Q] = degree(f (X)) ≥ ϕ(n),
which gives what we want.
To prove the claim, write X n − 1 = f (X)g(X), with g(X) ∈ Q[X]. By the discussion above,
both f (X) and g(X) have integer coefficients. Since ϵp is also an nth root of unity, it is a root of
X n − 1, and thus either a root of f (X) or g(X). Suppose g(ϵp ) = 0. Then ϵ is a root of g(X p ).
Thus, f (X) divides g(X p ), and again by the discussion above, we can write g(X p ) = f (X)h(X),
where h(X) has integer coefficients. If we reduce the coefficents mod pZ, then we have an equation
g(X p ) ≡ f (X)h(X) holding in Zp [X]. Now, for any integer a, ap ≡ a, modulo pZ, so it follows that
in Zp [X], g(X p ) ≡ g(X)p . Thus, g(X)p ≡ f (X)h(X) in Zp [X]. It follows from this that f (X) and
g(X) have a common factor modulo pZ. Since X n − 1 ≡ f (X)g(X) in Zp [X], this means that in the
algebraic closure of Zp , X n − 1 has a repeated root. On the other hand, X n − 1 and its derivative
are relatively prime as elements of Zp [X]. This is because p does not divide n, so nX n−1 is not the
zero polynomial in Zp [X]. Thus, X n − 1 has distinct roots, a contradiction. Therefore, g(ϵp ) ̸= 0,
so we must have f (ϵp ) = 0, which establishes the claim and thus the equality [Q(ϵ) : Q] = ϕ(n).
We now show that G is isomorphic to Z∗n . To see this, as above, we note that if σ ∈ G, then
σ(ϵ) is a primitive nth root of unity, so σ(ϵ) = ϵr , for a unique positive integer r less than n and
relatively prime to n. Define θ : G → Z∗n , by θ(σ) = r, the image of r in Z∗n . If τ ∈ G and τ (ϵ) = ϵs ,
with s less than n and relatively prime to n, then τ σ(ϵ) = (ϵr )s = ϵsr . Now, sr is relatively prime
to n, but may be larger than n, so we write sr = a · n + q, with q less than n and relatively prime
to n. Thus, θ(τ σ) = q = sr ∈ Z∗n . But sr = s · r = θ(τ )θ(σ), so θ is a group homomorphism.
Morover, by definition, if θ takes γ in G to 1, then γ(ϵ) = ϵ, so γ is the identity element in G, i.e., θ
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is one-to-one. Since G and Z∗n both have order ϕ(n), θ must also be onto, so θ is an ismomorphism
and the proof of the theorem is complete.
Remark. Let n ≥ 1 and ϵ be a primitive nth root of unity. It follows from the proof above that
if f (X) is the minimal polynomial for ϵ over Q then the roots of f (X) are precisely the primitive
nth roots of unity, i.e., f (X) = (X − ϵ1 ) · · · (X − ϵϕ(n) ), where ϵ1 , . . . , ϵϕ(n) are the primitive nth
roots of unity. Set Φn (X) := f (X). Then Φn (X) is called the nth cyclotomic polynomial. It also
follows from the proof above that Φn (X) is an irreducible monic polynomial with integer coefficients.
Moreover, if we let δ be any nth root of unity, then for the least d such that δ d = 1, d divides n (since
the nth roots of unity are a group of order n) and δ is a primitive dth root of unity. Conversely, for
any d dividing n, a∏
primitive dth root of unity is an nth root of unity. It follows immediately from
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this that X −1 = d|n Φd (X). Using this formula and recursion, one can compute the polynomials
Φn (X). For example, Φ1 (X) = X − 1, Φ2 (X) = X + 1, Φ3 (X) = X 2 + X + 1, Φ4 (X) = X 2 + 1,
Φ6 (X) = X 3 + 1. If p is prime, then Φp (X) = X p−1 + · · · + X + 1.
Corollary B. Let n1 , . . . , nt be pairwise relatively prime positive integers. For each 1 ≤ i ≤ t, let
ϵi be a primitive ni th root of unity. Then ϵ := ϵ1 · · · ϵt is a primitive n1 · · · nt root of unity and
Gal(Q(ϵ)/Q) ∼
= Z∗n1 × · · · × Z∗nt .
Proof. We use the following fact, whose proof we leave to the reader. Namely, that in an abelian
group, if elements x1 , . . . , xt have orders n1 , . . . , nt that are pairwise relatively prime, then the
order of x1 · · · xt is n1 · · · nt . If we apply this to the group (C\{0}, ·) and the elements ϵ1 , . . . , ϵt , it
follows that ϵ1 · · · ϵt has order n1 · · · nt . In other words, ϵ is a primitive n1 · · · nt root of unity. From
Theorem A, Gal(Q(ϵ)/Q) ∼
= (Zn1 ···nt )∗ . But since this group is isomorphic to Z∗n1 × · · · × Z∗nt , the
corollary follows.
The following theorem is a special case of the celebrated theorem of Dirichlet which states
that if a and b are relatively prime positive integers, then there are infinitely many primes in the
arithmetic progression {at + b}t≥1 . Theorem C will show that that for fixed n, there are infinitely
many primes in the arithmetic progression {nt + 1}t≥1 , or equivalently, there are infinitely many
primes congruent to 1 modulo nZ.
Theorem C. For a fixed positive integer n, there are infinitely many prime numbers p such that
p ≡ 1 (mod nZ).
Proof. The proof will proceed in a number of steps. The first step is interesting in its own right.
It generalizes Euclid’s proof that there are infinitely many prime numbers.
Step 1. Let g(X) be a polynomial with integer coefficients. Then there are infinitely many prime
numbers among the prime divisors of the elements in the set {g(0), g(1), g(2), . . .}. To see this, we
first reduce to the case that the constant term of g(X) is 1. If the constant term of g(X) is 0, the
result is trivial. Suppose that a ∈ Z is the constant term of g(X). Then every coefficient of g(aX)
is divisible by a. We can then write g(aX) = a · h(X), where h(X) has constant term equal to 1. If
we knew the set of values {h(0), h(1), h(2), . . .} had infinitely many prime divisors, the same would
certainly hold for g(aX), and thus for g(X), since the values of g(aX) are among those of g(X).
Therefore replacing h(X) by g(X), we may assume that the constant term of g(X) is 1. Now, by
way of contradiction, suppose the elements in the required set of values of g(X) had just finitely
many prime divisors, say p1 , . . . , pr . Consider the values of g(t · p1 · · · pr ) as t ∈ Z goes to infinity.
Eventually these values are larger in absolute value than any pi , yet are not divisble by any pi .
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Thus for t sufficiently large, g(t · p1 · · · pr ) must be divisible by a prime different from any of the pi ,
a contradiction. This proves Step 1.
Step 2. Let n be a positive integer and p be a prime number not dividing n. Then p divides Φn (c)
for some integer c if and only if the order of the element c in Z∗p equals n. To see this, suppose first
that p divides Φn (c). Then p divides cn − 1, so (c)n ≡ 1 (mod pZ). Suppose ∏
the order of c were
e
e
less than n. Then for some e dividing n, (c) ≡ 1 (mod pZ). Since X − 1 ≡ d|e Φe (X) modulo
pZ, it follows that c must also be a root modulo pZ of some Φd (X), with d|n and d < n. Since
Φd (X) and Φn (X) are factors of X n − 1 mod pZ, c is a multiple root of X n − 1 over Zp . But as in
the proof of Theorem A, X n − 1 and its derivative are relatively prime modulo pZ, so X n − 1 has
distinct roots modulo pZ. Therefore, the order of c in Z∗p is n. Conversely, if the order of c in Z∗p is
n, then p divides cn − 1. Therefore, p divides Φd (c), for some d dividing n. But if d were less than
n, then p would divide cd − 1, so the order of c in Z∗p would be less than n. Thus, p divides Φn (c),
as required. This finishes the proof of Step 2.
Step 3. For an integer n ≥ 1 and a prime p not dividing n, Z∗p has an element of order n if and
only if p ≡ 1 (mod nZ). To see this, just note that since Z∗p is cyclic, Z∗p has an element of order n
if and only if n divides |Z∗p | = p − 1, which happens if and only if p ≡ 1 (mod nZ).
We now finish the proof of the Theorem C. By Step 1, there are infinitely many primes p
dividing the values Φn (c) as c varies over the positive integers. Of course, infinitely many of these
primes do not divide n, so by Step 2, there are infinitely many primes p for which p does not divide
n and for which there exists c such that c has order n in Z∗p . By Step 3, there are infinitely many
primes congruent to 1 modulo nZ, which is what we wanted to show.
All of the pieces are now in place to prove the main result.
Theorem D. Let G be a finite abelian group. There there exists a finite Galois extension Q ⊆ K
such that Gal(K/Q) ∼
= G.
Proof. Since G is a finite abelian group, we may write it as a direct product of cylic groups, say
G = Zn1 × · · · × Znt . It follows from Theorem C that we can find t distinct primes p1 , . . . , pt such
that for each 1 ≤ i ≤ t, pi is congruent to 1 modulo ni Z. For each 1 ≤ i ≤ t, there is an integer mi
such that pi − 1 = ni · mi . Then for each 1 ≤ i ≤ t, there is a unique subgroup Hi ⊆ Z∗pi of order
mi and moreover, Z∗pi /Hi is isomorphic to Zni .
Now, for each 1 ≤ i ≤ t, let ϵi be a primitive pi th root of unity and write ϵ for the product
ϵ1 · · · ϵt . By Corollary B, Gal(Q(e)/Q) ∼
= Z∗p1 × · · · × Z∗pt . Let H ⊆ Gal(Q(ϵ)/Q) be the subgroup
corresponding to H1 × · · · × Ht and take K to be the fixed field of H. Then, since Gal(Q(ϵ)/Q) is
abelian, H is normal, so K is a Galois extension of Q. Moreover, Gal(K/Q) = Gal(Q(ϵ)/Q)/H is
isomorphic to G. Thus, G is the Galois group of a Galois extension of Q and the proof of Theorem
D is complete.
Bibliography
[FK] E. Fried and J. Kollar, Automorphism groups of algebraic number fields, Math Z. 165 (1978),
no. 2, 121-123.
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