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\newcommand{\id}{\mathop{\rm id}\nolimits} \)
LOCAL-GLOBAL PRINCIPLE FOR \(n^{th}\) POWERS
ISHAN LEVY
Given a number field \(F\), when is \(a \in F\) an \(n^{th}\) power modulo every prime? Here the question will be
answered for many \(F\) including \(\QQ \) (and the technique in principle probably works for any \(F\) with some
calculation).
Theorem 0.1. Suppose \(F\) is a field. If \(4|n\), then \(x^n-a\) is reducible iff if \(a\) is a \(p^{th}\) power for some \(p|n\) or if \(a\)
is of the form \(-4k^4\) and \(\ch F \neq 2\). If \(4\nmid n\), then \(x^n-a\) is irreducible iff \(a\) is not a \(p^{th}\) power mod every \(p|n\).
Proof. First reduce to the case of a prime power. Suppose \(x^n-a,x^m-a\) are irreducible and \((n,m) = 1\). Then \(F[a^{\frac 1 m}]\) is
degree \(n\) and \(F[a^{\frac 1 n}]\) is degree \(m\), so their compositum, \(F[a^{\frac 1 mn}]\) is degree \(mn\), so \(x^{nm}-a\) is irreducible. Conversely, if
either is reducible, \(x^{nm}-a\) is reducible.
The prime power case will be done by induction on the power. If \(x^{p^{m}-a}\) is irreducible, then \(x^{p^{m+1}}-a\) is
reducible iff \(a^{\frac 1{p^m}}\) is a \(p^{th}\) power in \(F[a^{\frac 1{p^m}}]\). But if this is true, then its norm down to \(F\) is also a \(p^{th}\) power. The
norm is \(a\) unless \(p = 2\), in which case it is \(-a\). Thus \(a\), or \(-a\) is a \(p^{th}\) power, a contradiction if \(p>2\) or \(m>1\). If \(p=2,m = 1, \ch F \neq 2\), then
\(F[\sqrt{a}] = F[i] \). Now calculation shows \(ki\) is a square iff \(k\) is of the form \((1+i)^2y^2\), some \(y \in F\), so \(a\) is of the form \(-4y^4\). In this
case, conversely \((-4y^4)^{1/4} = (1+i)y\) is degree \(2\) over \(F\), so the polynomial \(x^4+4y^4\) is reducible. If \(\ch F=2\), then \(-a = a\), so \(a\) is a square,
a contradiction.
In the case \(n\) is a prime \(p\), If \(x^p-a\) factors with some polynomial \(f=x^i+\dots \), where \(i<p\), then the norm of a root
\(\alpha \) of \(f\) from \(F[\alpha ]\) to \(F\) will be \(\alpha ^{i/p}\), which is an integer iff \(\alpha ^i\) is a \(p^{th}\) power iff \(\alpha \) is a \(p^{th}\) power.
□
Let \(F\) be a number field.
Lemma 0.2. Let \(a \in F\) be an \(n^{th}\) power modulo every prime. Then \(a^{\frac 1 n} \in F[\zeta _n]\).
Proof. Let \(K\) be the splitting field of \(x^n-a\). If a prime splits in \(K\), it splits completely in the subfield
\(F[\zeta _n]\). Conversely, if it splits completely in \(F[\zeta _n]\), the polynomial \(x^n-a\) has a root mod \(p\) so it factors into
linear factors as \(\zeta _n\) is there. Thus if the prime doesn’t divide the discriminant, then it splits
completely in \(K\). Up to a finite set, the same primes split in \(F[\zeta _n]\) and \(K\) so \(K=F\). □
Corollary 0.3. Let \(q\) be prime. If \(a \in F\) is a \(q^{th}\) power mod every prime, it is a \(q^{th}\) power.
Proof. \(a^{1/q} \in F[\zeta _q]\), but \([F[\zeta _q]:F] = q-1\) is relatively prime to \(q\) so \(a \in F\). □
Now suppose that the cyclotomic polynomials are irreducible over \(F\), i.e. for all \(n\), \(F \cap \QQ [\zeta _n] = \QQ \).
Lemma 0.4. Let \(q\) be either a prime or \(4\) and suppose \(x^q-a\) is irreducible over \(F\) and if \(q=4\) suppose \(-a\) is
not a square in \(F\). Then the Galois group of \(x^q-a\) is \(\Hol (\ZZ /q\ZZ ) = \ZZ /q\ZZ \rtimes \Aut (\ZZ /q\ZZ )\).
Proof. This problem amounts to showing that \(F[a^{\frac 1 q}]\) is linearly disjoint over \(F\) from \(F[\zeta _{q}]\). Indeed, if that
is true, then the degree of the splitting field will be the degree of the compositum which is \(\phi (q)q\).
There are \(\phi (q)\) conjugates of \(\zeta _q\) and \(q\) conjugates of \(a^{\frac 1 q}\), so the Galois group must act transitively on
the pairs of conjugates of \(\zeta _q\) and \(a^{\frac 1 q}\), giving an isomorphism with \(\Hol (\ZZ /q\ZZ )\). In the case that \(q\) is prime,
linearly disjointness follows from the degrees being relatively prime. In the case that \(q=4\), one
simply has to note that since \(a,-a\) are not squares, \(x^4-a\) is irreducible in \(F[i]\). □
Corollary 0.5. For \(a\) not a \(q^{th}\) power for \(q\) an odd prime, \(a^{1/q}\) doesn’t lie in any cyclotomic extension of \(F\).
If \(\pm a\) is not a square, then \(a^{\frac 1 4}\) doesn’t lie in a cyclotomic extension.
Proof. If it did, then the Galois group of \(x^q-a\) would be abelian by above, but it is not. □
Theorem 0.6. For an odd prime power \(q^n\), if \(a\) is a \((q^n)^{th}\) power mod every prime then \(a\) is a \((q^n)^{th}\) power.
Proof. Suppose the theorem holds for \(n-1\). Then \(a=b^{q^{n-1}}\) by hypothesis, and \(a^{\frac 1{q^n}}=b^{\frac 1 q} \in F[\zeta _{q^n}]\), so by applying the previous
corollary to \(b\), \(b\) is a \(q^{th}\) power in \(F\). □
Theorem 0.7. If \(a\) is a \((2^n)^{th}\) power mod every prime, either \(a\) is a \((2^n)^{th}\) power or \(n \geq 3\) and a is \(2^{2^{n-1}}\) times a \((2^{n})^{th}\)
power.
Proof. For \(n=2\), \(a = b^2\), and \(\sqrt b \in F[i] \implies b = \pm k^2\) for some \(k \in F\). Then \(k^4=a\). For \(n \geq 3\) assume the theorem holds for \(n-1\), \(a = b^{2^{n-1}}\) or \(2^{2^{n-2}}b^{2^{n-1}}\), and \(\sqrt{b}\) or \((2b^2)^{\frac 1 4} \in F[\zeta _{2^{n}}]\). If \(\sqrt{b} \in F[\zeta _{2^n}]\), \(\sqrt{b}\)
lies in a quadratic subfield of \(F[\zeta _{2^n}]\) which must be in \(F[\zeta _8]\) so \(b = \pm k^2\) or \(b = \pm 2k^2\), and \(a = k^{2^{n}},2^{2^{n-1}}k^{2^{n}}\). If \((2b^2)^{\frac 1 4} \in F[\zeta _{2^{n}}]\), note that by the corollary,
\(\pm 2b^2 = c^2\), which is impossible. Thus \(a = 2^{2^{n-1}}c^{2^n}\). □
Finally here is another problem of this sort:
Theorem 0.8. Let \(f\) be an irreducible polynomial of degree \(n\) with coefficient in a number
field \(F\). Let \(G\) be the Galois group of \(K\), the splitting field of \(f\), and let \(G \to S_n\) be the action of \(G\) on
the roots of \(f\). Then some element of \(G\) acts as an \(n\)-cycle iff \(f\) is irreducible modulo every
prime.
Proof. If \(f\) is not separable mod some \(p\), then it can’t be irreducible mod \(p\). So we can ignore
those primes. For the rest of the primes, note that how \(p\) splits in \(F[\alpha ]\), \(\alpha \) a root of \(f\), depends on
the cycle structure of Frobenius. In particular, it is an \(n\)-cycle iff \(p\) is inert. But if \(G\) has some
\(n\)-cycle, then by Chebotarev density theorem it is realized by some prime. Thus \(f\) is reducible
mod all \(p\) iff \(p\) is never inert iff there is no \(n\)-cycle. □
The action of \(G\) on \(S_n\) is transitive, and is its action on \(G/H\), where \(H\) is the subgroup of index \(n\)
corresponding to the subfield generated by one root of \(f\). Then some \(g \in G\) acts as an \(n\)-cycle iff \(\langle g \rangle H = G\). In
particular, suppose that adjoining one root of \(f\) gives a Galois extension, so that \(H\) is trivial. Then \(G\) is
not cyclic iff \(f\) is reducible modulo every prime. This lets us produce lots of examples, by
choosing primitive elements of Galois extensions with non-cyclic Galois groups. For
example, \(x^4+1\) has \(\zeta _8\) a root, which has non-cyclic Galois group, and so \(x^4+1\) is reducible modulo every
prime.
Theorem 0.9. Let \(f\) be a polynomial with \(\cO _K\) coefficients over a number field \(K\). Let \(G\) be the Galois
group of \(f\), let \(a_i\) be the roots of \(f\), and let \(G_i\) be the subgroup of \(G\) fixing \(K[a_i]\). Then \(f\) has a root modulo
(almost) every prime iff \(\cup G_i = G\).
Proof. By the Chebotarev density theorem, some prime has Frobenius not in \(\cup G_i\) iff \(\cup G_i \neq G\). An
unramified prime \(\pp \) lies in some \(G_i\) iff its Frobenius fixes some root \(a_i\) iff \(f\) has a root mod \(\pp \). □