THE GRUNWALD-WANG THEOREM: A LOCAL-GLOBAL PRINCIPLE
FOR POWER RESIDUES
ZEV ROSENGARTEN
Abstract
We prove the Grunwald-Wang Theorem, a local-global principle for power residues.
1. Introduction
The following simple question is often posed in introductory number theory courses:
Suppose that a ∈ Z is a square modulo every prime p. Does it follow that a is a square
in Z? The answer is yes, but this is often proven in a rather laborious and unenlightening
manner by using quadratic reciprocity. A more enlightening proof, albeit one that requires
more background√in algebraic number theory, goes as follows: Suppose a is as above. Then
the extension Q( a)/Q is √
a Galois extension√in which all but finitely many primes of Q
split completely. Hence Q( a) = Q; that is, a ∈ Q, so a is the square of an integer.
It is natural, of course, to generalize this question to arbitrary powers: If a ∈ Z is such
that a is an nth power modulo every prime p, does it follow that a is the nth power of
an integer? The answer to this question, unfortunately, is no. Indeed, 16 is an 8th power
modulo every prime, since x8 − 16 = (x2 − 2)(x2 + 2)(x2 + 2x + 2)(x2 − 2x + 2), and modulo
each odd prime p, either 2 is a square, −2 is a square, or −1 is a square, in which case the
last two quadratics factor modulo p. But this is, in a certain well-defined sense, the only
counterexample, as we shall see.
Our goal in this paper is to solve the general version of this question. First, a definition
to streamline the presentation: Let K be a number field. We say that something holds for
almost all primes of K (or holds almost everywhere locally if the field K is clear from the
context) if it holds for a set of primes of K of density one (Throughout this paper, density
may be taken to be either natural or Dirichlet.). Now once again let K be a number field,
α ∈ K, n ∈ Z+ , and suppose that α ∈ Kp n for almost all primes p of K. What can one
say about α? Note that this is indeed the natural generalization of the above question: By
Hensel’s Lemma, for all but finitely many primes p of K, α ∈ Kp n iff xn ≡ α (mod p) is
solvable.
Now that we have given the necessary introduction, we can state our main result, which
gives a beautiful answer to the above question (Note: Throughout this paper ζl denotes
e2πi/l .):
Theorem 1 (Grunwald-Wang): Let K be a number field, α ∈ K, n ∈ Z+ . Then α ∈ Kp n
for almost every prime p of K iff one of the following holds:
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ZEV ROSENGARTEN
(i) α ∈ K n .
(ii) 8 | n and letting 2m k n and ζ = ζ2m , we have K ∩ Q(ζ) ⊂ R and ζ + ζ −1 ∈
/ K.
l
−1
2
(Equivalently, K ∩ Q(ζ) ( Q(ζ + ζ ).). In this case, Gal(K(ζ)/K) =< −1 > × < 5 >,→
(Z/2m Z)∗ for some 0 ≤ l < m − 2, and α ∈ β n/2 K n 6= K n , where β ∈ K is such that
√
√
l
K( β) ⊂ K(ζ) is the extension of K with Gal(K(ζ)/K( β)) =< −52 >.
The following result, which answers the question over Q, is immediate from
√ Theorem 1,
since in this case the extension mentioned in part (ii) of Theorem 1 is Q( 2).
Corollary 1.1: Let α ∈ Q, n ∈ Z+ . Then α is an nth power almost everywhere locally iff
one of the following holds:
(i) α ∈ Qn .
(ii) 8 | n and α ∈ 2n/2 Qn .
This corollary explains what we meant above when we said that 16 is essentially the
only counterexample over Q: If 8 | n, then x8 − 16 | xn − 2n/2 , so all counterexamples are
essentially generated from 16.
It is perhaps also worth noting that Theorem 1 tells us that the Hasse Principle for
power residues holds without exception for certain number fields; for example, it follows
from Theorem 1 that if i ∈ K, then α ∈ K is an nth power almost everywhere locally iff α
is an nth power globally.
Our proof of Theorem 1 will be very illuminating; we will be able to see very explicitly
why it is true, as well as why the counterexamples exist. The argument is essentially a
generalization of that given above for the case K = Q, n = 2, but the details are more
involved because K may not contain the relevant (that is, nth) roots of unity.
2. Proof For 8 - n
In this section we prove Theorem 1 in the case 8 - n, i.e. that if 8 - n, α ∈ K, and
α ∈ Kp n for almost all primes p of K, then α ∈ K n . We only assume, however, that 8 - n
when we say so explicitly. The following is the key result, and is the suitable generalization
of the argument for n = 2:
Lemma 2.1: If α ∈ Kp n almost everywhere locally, then α1/n ∈ K(ζn ).
It should be noted that there is no ambiguity in the above statement, since if any nth
root of α lies in K(ζn ), then every nth root of α does (This may seem like a trivial point,
but this ambiguity will lie at the center of our proof; it is in some sense its main obstacle.).
Proof of Lemma 2.1: For all but finitely many primes p of K, p splits completely in
K(ζn , α1/n ) iff it splits completely in K(ζn ) and xn ≡ α (mod p) is solvable. This is easy
to see: If p splits completely in K(ζn , α1/n ), then it also does in K(ζn ) and in K(α1/n ),
which, for all but finitely many p, implies that xn ≡ α (mod p) is solvable. Conversely,
THE GRUNWALD-WANG THEOREM: A LOCAL-GLOBAL PRINCIPLE FOR POWER RESIDUES
3
if p splits completely in K(ζn ) and xn ≡ α (mod p) is solvable, then this implies, for all
but finitely many p, that ζn ∈ Kp , hence xn − α splits in Kp , so p splits completely in
K(α1/n ), hence in K(ζn , α1/n ). Thus, the density of the primes of K that split completely
in K(ζn ), but for which xn ≡ α (mod p) is not solvable is [K(ζ1n ):K] − [K(ζ ,α11/n ):K] . This is
n
zero by hypothesis, so [K(ζn ) : K] = [K(ζn , α1/n ) : K]; that is, α1/n ∈ K(ζn ), as desired. It should be noted that all we really needed to prove the lemma was to establish that
for all but finitely many primes p of K splitting completely in K(ζn ), α ∈ Kp n =⇒ p splits
completely in K(ζn , α1/n ). We established both implications, however, for completeness.
Now note that if Theorem 1 holds for m and n with (m, n) = 1, then it also holds for
mn, so it suffices to prove it when n is a power of a prime. Then letting n = 2, 4, or a
power of an odd prime in the following lemma completes the proof of Theorem 1 for 8 - n,
since in these cases G = Gal(K(ζn )/K) ,→ (Z/nZ)∗ is cyclic.
Lemma 2.2: Let n ∈ Z+ , α ∈ K, and suppose that α ∈ Kp n for almost all primes p of K.
If G = Gal(K(ζn )/K) is cyclic, then α ∈ K n .
Proof: Assume for the sake of contradiction that α 6∈ K n . By Lemma 2.1 α1/n ∈ K(ζn ).
Consider the extensions K(β)/K as β ranges over the nth roots of α. They are all Galois, since K(ζn )/K is abelian. Furthermore, by our assumption they are all nontrivial
extensions. Now since α ∈ Kp n for almost all p, almost every prime of K splits completely
in at least one of the K(β)’s. On the oher hand, by the Cebotarev Density Theorem
the set S of primes of K for which (p, K(ζn )/K) = σ, where (p, K(ζn )/K) denotes the
Frobenius element of p in K(ζn ) and G =< σ >, has positive density. But for p ∈ S,
(p, K(β)/K) = σ|K(β) 6= id, so p does not split completely in K(β) for any β, in violation
of what we stated earlier. Thus, we must have α ∈ K n , as claimed.
3. The Case 8 | n
In this section we handle the case 8 | n of Theorem 1. We may assume, of course, that
n is a power of 2, so let us fix some notation for the rest of this section: In what follows,
we shall assume that n = 2m , m ≥ 3. We also set ζ = ζ2m . Under these assumptions,
(Z/nZ)∗ =< −1 > × < 5 >. We need the following:
Lemma 3.1: Gal(K(ζ)/K is not cyclic iff K ∩ Q(ζ) ( Q(ζ + ζ −1 ).
Proof: We have G = Gal(K(ζ)/K) ,→ (Z/nZ)∗ = < −1 > × < 5 >, and the noncyclic
l
subgroups of the latter group are those of the form < −1 > × < 52 > with 0 ≤ l < m − 2.
l
Thus, G is noncyclic iff −1 ∈ G and 52 ∈ G for some such l. Now there exists a Kautomorphism of K(ζ) with ζ 7→ ζ −1 iff K ∩ Q(ζ) ⊂ Q(ζ + ζ −1 ), the subfield of Q(ζ)
fixed by ζ 7→ ζ −1 . (We can see this as follows: If −1 ∈ Gal(K(ζ)/K), then −1 fixes
K ∩ Q(ζ), so K ∩ Q(ζ) ⊂ Q(ζ + ζ −1 ). Conversely, suppose that K ∩ Q(ζ) ⊂ Q(ζ + ζ −1 ).
Let L be the Galois closure of K/Q, and let F = LQ(ζ). Let H be the subgroup of
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ZEV ROSENGARTEN
Gal(Q(ζ)/Q) consisting of those σ ∈ Gal(Q(ζ)/Q) such that ∃ τ ∈ Gal(F/Q) such that
τ fixes K and τ|Q(ζ) = σ. Then H = Gal(Q(ζ)/Q(ζ)H ) = Gal(Q(ζ)/K ∩ Q(ζ)). Since
K ∩ Q(ζ) ⊂ Q(ζ + ζ −1 ), −1 ∈ H; that is, ∃ τ ∈ Gal(F/K) with τ (ζ) = ζ −1 . The restriction of τ to K(ζ) is then the desired automorphism.) Next, G 6= < −1 > iff ∃ σ ∈ G not
fixing ζ + ζ −1 , i.e. iff ζ + ζ −1 ∈
/ K. Putting everything together, we see that G is noncyclic
−1
iff K ∩ Q(ζ) ( Q(ζ + ζ ).
Now suppose that α ∈ K is almost everywhere locally an nth power, but α ∈
/ K n.
Combining Lemmas 2.2 and 3.1, we see that K is as in Theorem 1, part (ii). Now we shall
prove that we have α ∈ β n/2 K n , where β is as in Theorem 1, (ii).
Lemma 3.2: If α ∈ K − K n is almost everywhere locally an nth power, then α ∈ β n/2 K n ,
where β is as in Theorem 1, (ii).
Proof: We know that K must be as in Lemma 3.1. Let G = Gal(K(ζ)/K) =< −1 >
l
× < 52 >,→ (Z/nZ)∗ , 0 ≤ l < m − 2. By an argument similar to the one in the
l
proof of Lemma 2.1, −52 ∈ Gal(K(ζ)/K(γ)) for some nth root γ of α, where, by hyl
pothesis, γ ∈
/ K. (Otherwise, any prime p of K with (p, K(ζ)/K) = −52 doesn’t split
completely in K(γ) for any nth√root γ of α, so α√is not almost everywhere locally an
nth power.) Then K(γ) = K( β), so γ = a + b β for some a, b √∈ K. Then, since
√
γ’s conjugate over K must also be an nth root of α, we have a − b β = ζ j (a + b β)
√
l
for some j ∈ Z. Then ζ j ∈ K( β), so ζ j is fixed by −52 ∈ Gal(K(ζ)/K(β)); that is,
2l
l
l
2
ζ j = ζ −5 j =⇒ j(1 + 52 ) ≡√0 (mod n) =⇒
√ j ≡ 0 (mod n/2), since 1 + 5 ≡ 2 (mod 4).
j
But then ζ = ±1, so a + b β = ±(a − b β) =⇒√a = 0 or b = 0. If b = 0, then γ ∈ K,
contrary to hypothesis. If a = 0, then α = γ n = (b β)n ∈ β n/2 K n , as desired.
The following lemma completes the proof of Theorem 1.
Lemma 3.3: Let β, K, n be as in Theorem 1, part (ii). Then β n/2 is an nth power at all
but finitely many places of K, but β n/2 ∈
/ K n.
Proof:
in K(ζ), and let σ = (p, K(ζ)/K). If
√ p be a (finite) prime of K unramified
√
√ Let
σ( β) = β, then p splits completely in K( β). Otherwise, let l ∈ (Z/nZ)∗ be the image
∗
of σ under the canonical injection Gal(K(ζ)/K)
,→
√
√ (Z/nZ) r. Since
√ l 6= 1,
√∃ r ∈ Z with
r
rl
r β, so p splits
r(l − 1) ≡ n/2 (mod
n).
Then
σ(ζ
β)
=
(ζ
)(−
β)
=
(−ζ
)(−
β)
=
ζ
√
completely in K(ζ r β). Since p was an arbitrary (finite) prime of K unramified in K(ζ),
it follows that all but finitely many primes of K split completely in K(γ) for some nth root
γ of β n/2 , so β n/2 is an nth power at all but finitely many places of K, as claimed.
Finally, note that if β n/2 ∈ K n , then ζ r β ∈ K 2 for some r ∈ Z. In particular, ζ r ∈ K,
so ζ r = ±1, since K ∩ Q(ζ) ⊂ R. √
ζ r = 1 =⇒√β ∈ K 2 , contrary to our choice of
r
2
β. ζ = −1 =⇒ −β ∈ K =⇒ K( β) = K(i −β) = K(i), but i is not fixed by
THE GRUNWALD-WANG THEOREM: A LOCAL-GLOBAL PRINCIPLE FOR POWER RESIDUES
5
l
−52 ∈ Gal(K(ζ)/K), so this too contradicts our choice of β. Thus, β n/2 ∈
/ K n , which
proves the lemma.
Remark: It is also possible to give a proof of the first part of the above lemma in much
the same manner as one proves that x8 − 16 has a root modulo every rational prime,
m
m−1
namely by factoring the polynomial x2 − β 2
. We have given the above proof because
it is simpler and more conceptual. We invite the interested reader, however, to try to prove
the first part of Lemma 3.3 using this alternative approach.