Chebotarev Density Theorem

 In this post I try to organise the patchwork of things I have learned/heard about the important Chebotarev Density Theorem. This is probably not a good place to learn about this if you haven't seen any of this before. The proof is a combination of the algebraic approach here and Chebotarev's original argument as exposed here.

Today I want to discuss a fundamental theorem of algebraic number theory: Chebotarev's density theorem. It is a generalisation of two perhaps more famous results: Dirichlet's theorem and Frobenius' theorem, which we state below.

Dirichlet's theorem: Let $a$ and $b$ be positive integers such that $\mathrm{gcd}(a,b)=1$. Then there exist infinitely many primes in the arithmetic progression $an+b$.

In fact, one can quantify this. Recall the following notion:

Definition: If \( M \) is a set of primes of a number field \( K \), we define the limit
\[\lim_{s \to 1^+} \frac{\sum_{p \in M} \frac{1}{Np^s}}{\log \frac{1}{s-1}}\]
as the Dirichlet density of \( M \) (if it exists). 

 The prime number theorem for arithmetic progressions says that the density of primes in a given residue class $\mod{a}$ is $\frac{1}{\varphi(a)}$. The proof technique is quite similar to that of the prime number theorem given in a previous post. Thus, the primes behave randomly from the viewpoint of modular arithmetic: outside of the obvious congruence obstructions they don't favour any residue class anywhere.

We need a bit more background to state Frobenius' theorem. Let \( f \) be a polynomial with integer coefficients and with leading coefficient 1, and denote the degree of \( f \) by \( n \). Assume that the discriminant \( \Delta(f) \) of \( f \) does not vanish, so that \( f \) has \( n \) distinct zeroes \( \alpha_1, \alpha_2, \ldots, \alpha_n \) in a suitable extension field of the field \( \mathbb{Q} \) of rational numbers. Write \( K \) for the field generated by these zeroes: \( K = \mathbb{Q}(\alpha_1, \alpha_2, \ldots, \alpha_n) \). The Galois group \( G \) of \( f \) is the group of field automorphisms of \( K \). Each \( \sigma \in G \) permutes the zeroes \( \alpha_1, \alpha_2, \ldots, \alpha_n \) of \( f \), and is completely determined by the way in which it permutes these zeroes. Hence, we may consider \( G \) as a subgroup of the group \( S_n \) of permutations of \( n \) symbols. Writing an element \( \sigma \in G \) as a product of disjoint cycles (including cycles of length 1), and looking at the lengths of these cycles, we obtain the cycle pattern of \( \sigma \), which is a partition \( n_1, n_2, \ldots, n_t \) of \( n \).

If \( p \) is a prime number not dividing \( \Delta(f) \), then we can write \( f \) modulo \( p \) as a product of distinct irreducible factors over \( \mathbb{F}_p \). The degrees of these irreducible factors form the decomposition type of \( f \) modulo \( p \); this is also a partition of \( n \). Frobenius's theorem asserts, roughly speaking, that the ``number'' of primes with a given decomposition type is proportional to the number of \( \sigma \in G \) with the same cycle pattern.

Frobenius' theorem: The density of the set of primes \( p \) for which \( f \) has a given decomposition type \( n_1, n_2, \ldots, n_t \) exists, and it is equal to \( 1/\#G \) times the number of \( \sigma \in G \) with cycle pattern \( n_1, n_2, \ldots, n_t \).

One can see that both of these are results in the vein of 'primes behave randomly'. To state Chebotarev, we introduce the following:

Theorem 1: Let \( L/K \) be a Galois extension. If \( \mathfrak{p} \) is a prime unramified in \( K \), and \( \mathfrak{P} \) a prime above it in \( L \). Then there is a unique element of \( \operatorname{Gal}(L/K) \), denoted \( \operatorname{Frob}_{\mathfrak{P}} \), obeying
\[\operatorname{Frob}_{\mathfrak{P}}(\alpha) \equiv \alpha^{N\mathfrak{p}} \pmod{\mathfrak{P}} \quad \forall \alpha \in \mathcal{O}_L.\]

Lemma 2: The Frobenius element \( \operatorname{Frob}_{\mathfrak{P}} \in \operatorname{Gal}(L/K) \) of an extension \( L/K \) has order equal to the inertial degree of \( \mathfrak{P} \), that is,
\[\operatorname{ord} \operatorname{Frob}_{\mathfrak{P}} = f(\mathfrak{P} \mid \mathfrak{p}).\]
In particular, \( \operatorname{Frob}_{\mathfrak{P}} = \operatorname{id} \) if and only if \( \mathfrak{p} \) splits completely in \( L/K \).

Proof: We want to understand the order of the map \( T \colon x \mapsto x^{N\mathfrak{p}} \) on the field \( \mathcal{O}_K/\mathfrak{P} \). But the latter is isomorphic to the splitting field of \( X^{N\mathfrak{p}} - X \) in \( \mathbb{F}_p \), by Galois theory of finite fields. Hence the order is \( \log_{N\mathfrak{p}}(N\mathfrak{P}) = f(\mathfrak{P} \mid \mathfrak{p}) \). $\blacksquare$


We are particularly interested in abelian extensions, (which just means that \( \operatorname{Gal}(L/K) \) is abelian) in which case the theory becomes extra nice: the conjugacy classes have size one.

Definition: Assume \( L/K \) is an abelian extension. Then for a given unramified prime \( \mathfrak{p} \) in \( K \), the element \( \operatorname{Frob}_{\mathfrak{P}} \) doesn't depend on the choice of \( \mathfrak{P} \). We denote the resulting \( \operatorname{Frob}_{\mathfrak{P}} \) by the Artin symbol:
\[\left( \frac{L/K}{\mathfrak{p}} \right).\]
Note that the Artin symbol looks a lot like the Legendre symbol. This is by design, since one of the main theorems of class field theory is Artin reciprocity, which is supposed to be the ultimate reciprocity law: other ones, such as cubic or quadratic reciprocity, follow from it. I have often felt that the proofs of class field theory are distinctly unenlightening, so I will forego discussion here and leave it to the very interested reader to flip through a text, e.g. Milne's class field theory notes. One could probably get away with reading this other blog post, which I consulted when writing the current blog post. Class field theory is very powerful, but I think its foundations are better left as a blackbox for those who just want to enjoy some nice maths.

Chebotarev Density Theorem: Let \( K/F \) be a Galois extension with group \( G \) and let \( C \subseteq G \) be a union of conjugacy classes. The set
\[S = \left\{ p \text{ prime of } \mathcal{O}_F \mid \left( \frac{K/F}{p} \right) \subseteq C \right\}\]
admits (Dirichlet) density, given by \( \operatorname{Dens}(S) = \frac{\#C}{\#G} \).

Again this can be thought of as a randomness result: prime ideals don't favour any element of the Galois group. We will do the abelian case and then reduce the general case to this.


Theorem 3: Let \( K/F \) be a Galois extension with abelian group \( G \) and let \( C \subseteq G \) be a union of conjugacy classes. The set
\[
S = \left\{ p \text{ prime of } \mathcal{O}_F \mid \left( \frac{K/F}{p} \right) \subseteq C \right\}
\]
admits (Dirichlet) density, given by \( \operatorname{Dens}(S) = \frac{\#C}{\#G} \).

The proof given here closely follows that given here, which is apparently quite close in spirit to Chebotarev's original proof. The authors comment that it seems that this proof and the technique in it is rarely presented. I am not qualified to comment so leave it to the interested reader to look in other sources. The article itself also contains an interesting brief account of Chebotarev's life.

Proof: Let \( K \) be abelian over \( F \), with group \( G \) and degree \( n \). Let \( m \) be any prime number not dividing the discriminant \( \Delta \) of \( K \) over \( \mathbb{Q} \), and denote by \( \zeta \) a primitive \( m \)th root of unity. Then the Galois group \( H \) of \( F(\zeta) \) over \( F \) is isomorphic to \( (\mathbb{Z}/m\mathbb{Z})^* \), and the Galois group of \( K(\zeta) \) over \( F \) may be identified with \( G \times H \). 

If a prime \( p \) of \( F \) has Frobenius element \( (\sigma, \tau) \) in \( G \times H \), then it has Frobenius element \( \sigma \) in \( G \). Hence, writing \( d_{\inf} \) for lower density---defined as the density, but with \( \inf \) replaced by \( \liminf \)---we have \( d_{\inf}(K/F, \{\sigma\}) \geq \sum_{\tau \in H} d_{\inf}(K(\zeta)/F, \{(\sigma, \tau)\}) \). 

For what follows, we will need an intermediate claim:

Claim: the conclusion of theorem 3 holds for $K,F, C$ iff it holds for $K, E, \{\sigma\}$
Proof of claim: Let \( S_{K,\sigma} \) be the set of \( \mathfrak{P} \) in \( K \) such that \( \mathfrak{P} \mid \mathfrak{p} \) for \( \mathfrak{p} \in S \), and \( (\mathfrak{P}, K/k) = \sigma \). Let \( \mathfrak{P} \mid \mathfrak{q} \) for \( \mathfrak{q} \) in \( Z \). Then \( S_{K,\sigma} \) is in bijection with the set \( S_Z \) of \( \mathfrak{q} \) in \( Z \) which lie in a given class mod \( H \), and which divide \( \mathfrak{p} \) splitting completely in \( Z \). However, the density depends only on those primes of degree 1 over \( \mathbb{Q} \). $\blacksquare$

Now fix \( \sigma \in G \) and \( \tau \in H \), and suppose that \( n \) divides the order of \( \tau \). Then the subgroups \( \langle (\sigma, \tau) \rangle \) and \( G \times \{1\} \) of \( G \times H \) have trivial intersection. Therefore the field \( L \) of invariants of \( \langle (\sigma, \tau) \rangle \) satisfies \( L(\zeta) = K(\zeta) \), so that the extension \( L \subset K(\zeta) \) is cyclotomic. By what we proved in the cyclotomic case, the density \( d(K(\zeta)/L, \{(\sigma, \tau)\}) \) exists and has the correct value. 

This is, by the claim, then also true for \( d(K(\zeta)/F, \{(\sigma, \tau)\}) \), which consequently equals \( 1/(\#G \cdot \#H) \). Summing over \( \tau \), one obtains \( d_{\inf}(K/F, \{\sigma\}) \geq \#H_n/(\#G \cdot \#H) \), where \( H_n \) is the set of \( \tau \in H \) of order divisible by \( n \). Now it is easy to see that as \( m \) ranges over all prime numbers not dividing \( \Delta \), the fraction \( \#H_n/\#H \) gets arbitrarily close to 1 (use, for example, Dirichlet's theorem to choose \( m \equiv 1 \mod n^k \) for large \( k \)). Thus it follows that \( d_{\inf}(K/F, \{\sigma\}) \geq 1/\#G \). Applying this to all other elements of the group, one finds that the upper density \( d_{\sup}(K/F, \{\sigma\}) \) is at most \( 1/\#G \). Therefore the lower and the upper density coincide, and the density equals \( 1/\#G \). $\blacksquare$



Proof of Chebotarev Density Theorem: Choose an element \( g \in C \) and let \( E = K^{(g)} \) be the field fixed by the subgroup \( H = \langle g \rangle \). Consider the set
\[
T_g = \left\{ \mathfrak{q} \text{ prime of } E \mid \left( \frac{K/E}{\mathfrak{q}} \right) = g,~ N(\mathfrak{q}) \text{ is prime} \right\}.
\]
Suppose that \( \mathfrak{q} \) is in \( T_g \): we claim that \( \mathfrak{p} := \mathfrak{q} \cap \mathcal{O}_F \) is in \( S' \). Indeed, \( N(\mathfrak{p}) \) divides \( N(\mathfrak{q}) \), so \( N(\mathfrak{p}) = N(\mathfrak{q}) \) is a prime number. If \( \Omega \) is a prime of \( K \) lying over \( \mathfrak{q} \), this implies that \( \left( \frac{K/F}{\Omega} \right) = \left( \frac{K/E}{\Omega} \right) = g \in C \), hence \( \mathfrak{p} \in S' \). Moreover, we claim that \( \Omega \) is the unique prime of \( K \) lying over \( \mathfrak{q} \). To see this, notice that \( D(\Omega \mid \mathfrak{q}) \) is by definition generated by \( \left( \frac{K/E}{\Omega} \right) = g \), so \( D(\Omega \mid \mathfrak{q}) = H \): the whole Galois group of \( K \) over \( E \) sends \( \Omega \) to itself, and therefore \( \Omega \) is the only prime of \( K \) over \( \mathfrak{q} \).

Conversely, given \( \mathfrak{p} \in S' \), by definition there exists a prime \( \Omega \) of \( K \) lying over \( \mathfrak{p} \) with \( \left( \frac{K/F}{\Omega} \right) \in C \). Replacing \( \Omega \) by a conjugate if necessary, we can assume that \( \left( \frac{K/F}{\Omega} \right) = g \). If we define \( \mathfrak{q} = \Omega \cap E \), then \( \mathfrak{q} \) lies over \( \mathfrak{p} \) (obvious), and we claim that it is in \( T_g \). To see this, notice that again we have
\[
E = K^H = K^{\langle g \rangle} = K^{D(\Omega \mid \mathfrak{p})},
\]
so \( E \) is the decomposition field of \( \mathfrak{p} \). This means that \( f(\mathfrak{q} \mid \mathfrak{p}) = 1 \) and \( f(\Omega \mid \mathfrak{q}) = |H| = \operatorname{ord}(g) \). In particular, \( N(\mathfrak{q}) = N(\mathfrak{p})^{f(\mathfrak{q} \mid \mathfrak{p})} \) is prime, and as above it follows that \( \left( \frac{K/F}{\Omega} \right) = \left( \frac{K/E}{\Omega} \right) = g \in C \). Hence \( \mathfrak{q} \) is in \( T_g \) as claimed.

Summarising, there is a bijection between the primes in \( T_g \) lying over \( \mathfrak{p} \in S' \) and the primes \( \Omega \) of \( K \) that divide \( \mathfrak{p} \) and satisfy \( \left( \frac{K/F}{\Omega} \right) = g \).

Now, let \( Z_{G}(g) \) be the centralizer of \( g \) in \( G \). By the orbit-stabilizer lemma, \( [C] = \frac{|G|}{|Z_G(g)|} \). We count the primes in \( T_g \) lying over each prime \( \mathfrak{p} \in S' \). By the previous paragraph, it suffices to count the primes \( \Omega \) of \( K \) lying over \( \mathfrak{p} \) with \( \left( \frac{K/F}{\Omega} \right) = g \). We have already shown that there is at least one such prime, call it \( \Omega_1 \). Any other prime \( \Omega' \) of \( K \) lying over \( \mathfrak{p} \) is conjugate to \( \Omega_1 \) by an element \( \sigma \in G \), say \( \Omega' = \sigma \Omega_1 \). Then, the Artin symbol \( \left( \frac{K/F}{\Omega'} \right) \) is given by
\[\left( \frac{K/F}{\Omega'} \right) = \sigma \left( \frac{K/F}{\Omega_1} \right) \sigma^{-1} = \sigma g \sigma^{-1}.\]
Hence, \( \left( \frac{K/F}{\Omega'} \right) = g \) if and only if \( \sigma \in Z_G(g) \). By the orbit-stabilizer lemma again, the number of distinct primes \( \Omega' \) with \( \left( \frac{K/F}{\Omega'} \right) = g \) is
\[\frac{|Z_G(g)|}{|\mathrm{Stab}(\Omega_1) \cap Z_G(g)|} = \frac{|Z_G(g)|}{|D(\Omega_1 \mid \mathfrak{p}) \cap Z_G(g)|} = \frac{|Z_G(g)|}{|D(\Omega_1 \mid \mathfrak{p})|},\]
where we have used both the definition of \( D(\Omega_1 \mid \mathfrak{p}) \) and the fact that \( D(\Omega_1 \mid \mathfrak{p}) = \langle g \rangle \subseteq Z_G(g) \). In conclusion, the number of primes of \( K \) lying over \( \mathfrak{p} \) and having Artin symbol \( g \) is \( \frac{|Z_G(g)|}{|D(\Omega_1 \mid \mathfrak{p})|} = \frac{|G|}{|C| f} \), where \( f = |D(\Omega_1 \mid \mathfrak{p})| = \operatorname{ord}(g) = |H| \). By what we already argued above, this is also the number of primes in \( T_g \) lying over \( \mathfrak{p} \).

On the other hand, by Theorem 3 Chebotarev's theorem holds for the extension \( E \subset K \), hence we have

\[\operatorname{Dens}_E(T_g) = \frac{1}{|H|} = \frac{1}{f}.\]

Recall that we replaced \( S \) by its subset \( S' \) of primes whose residue field is a prime field. From the above discussion, noticing that by definition the norm of a prime \( \mathfrak{p} \in S' \) is the same as the norm of any prime \( \mathfrak{q} \in T_g \) lying over \( \mathfrak{p} \), we obtain
\[\frac{|G|}{f |C|} \sum_{\mathfrak{p} \in S'} N(\mathfrak{p})^{-s} = \sum_{\mathfrak{q} \in T_g} N(\mathfrak{q})^{-s}.\]

Dividing by \( \log\left( \frac{1}{s-1} \right) \) and passing to the limit for \( s \to 1^{+} \), we obtain
\[\frac{|G|}{f |C|} \lim_{s \to 1^{+}} \frac{\sum_{\mathfrak{p} \in S'} N(\mathfrak{p})^{-s}}{\log\left( \frac{1}{s-1} \right)} = \lim_{s \to 1^{+}} \frac{\sum_{\mathfrak{q} \in T_g} N(\mathfrak{q})^{-s}}{\log\left( \frac{1}{s-1} \right)} = \operatorname{Dens}_E(T_g) = \frac{1}{f}.\]
This shows as desired that \( \operatorname{Dens}_F(S') \) exists and equals \( \frac{|C|}{|G|} \).
$\blacksquare$

Remark: Number fields and function fields are known collectively as global fields because they behave similarly in many ways. The function field Chebotarev density theorem is proved here. 

Chebotarev's density theorem has many far-reaching consequences. Here are some of my favourite (easily described) ones:

1. Let \(F_{1}\), \(F_{2}\) be two finite extensions of the number field \(K\). Then \(F_{1}\), \(F_{2}\) have the same Galois closure over \(K\) if and only if the primes of \(K\) that split completely in \(F_{1}\), \(F_{2}\) are the same, or even the same up to a subset of density \(0\).
2. If \(f(x) \in \mathbb{Z}[x]\) is irreducible and has a root modulo almost every prime \(p\), then \(\deg f = 1\). Here is an important subcase: Let \(p\) be a prime number. An element \(a \in \mathbb{Q}^{\times}\) is a \(p\)-th power if and only if it is a \(p\)-th power modulo \(\ell\) for almost all primes \(\ell\). 
3. Let \(f(x) \in \mathbb{Z}[x]\) be any polynomial. There exist infinitely many primes \(p\) such that \(f(x) \bmod p\) splits completely (this can also be proven elementarily, without relying on the Chebotarev theorem).
4. Let \(K\) be a number field. The 'probability' that a prime of \(\mathcal{O}_{K}\) is principal (that is, the density of principal primes) is \(1/h_{K}\). This partially justifies the oft-repeated claim that '\(h_{K}\) measures the failure of unique factorisation'---recall that unique factorisation is equivalent to the ring of integers being a PID, which in turn is equivalent to every prime ideal being principal, so \(h_{K}\) really is a measure of 'how much' unique factorisation fails.
5. A sufficiently strong version of Chebotarev with error term (unfortunately, one \textit{so strong} that we can only prove it under the assumption of the Generalised Riemann Hypothesis for Dedekind \(\zeta\) functions) implies
   
   Artin's primitive root conjecture: Let \(a\) be a non-zero integer which is not a square and is different from \(-1\). There exist infinitely many primes \(p\) such that \(a\) is a primitive root modulo \(p\).

We will use the first result to answer a fairly natural question: 

Question 4: Let $x$ and $y$ be nonzero integers and for a positive integer $w$ denote by $\mathrm{ord}_p(w)$ the multiplicative order of $w$ in $\mathbb{Z}/p\mathbb{Z}$. If $\mathrm{ord}_p(x)= \mathrm{ord}_p(y)$ for all primes not dividing $x$ or $y$, does this imply $x=y$? 

We are naturally lead to the study of polynomials of the form $t^k-a$. It turns out we will need the following:

Lemma 5: Let $k$ be an odd positive integer such that the rational polynomial $t^k-a$ has a root in $\mathbb{Q}(\zeta_k)$. Then $t^k-a$ has a rational root.

(I reproduce the answer given here)
After a reformulation, we are considering the kernel of the map $\mathbb Q^{\times}/(\mathbb Q^{\times})^k \to L^{\times}/(L^{\times})^k,$ where $L =\mathbb Q(\zeta_k)$, which we study for any $k$.

In general, for any field $K$ of char. prime to $k$, there is a natural isomorphism $K^{\times}/(K^{\times})^n \cong H^1(G_K,\mu_k)$.   (This is the content of Kummer theory, and follows from Hilbert's Thm. 90.)

Thus if $L$ is a Galois extension of $K$, the kernel of the map $K^{\times}/(K^{\times})^k \to L^{\times}/(L^{\times})^k$ is naturally identified with the kernel of the restriction map $H^1(G_K,\mu_k) \to H^1(G_L,\mu_k)$, which by the inflation-restriction exact sequence, is equal to  $H^1(Gal(L/K),\mu_k(L))$ (where $\mu_k(L)$ denotes the subgroup of $\mu_k$ consisting of element which lie in $L$).

If we apply this with $K = \mathbb Q$ and $L = \mathbb Q(\zeta_k)$, we find that the kernel we are interested in is identified with $H^1((\mathbb Z/k)^{\times},\mu_k)$, which is one can compute as follows: 
If we factor $k$ into a product of powers of distinct primes, say $k = \prod p^n,$ then $\mu_k = \oplus \mu_{p^n},$ and so we are reduced to computing $$H^1((\mathbb Z/m)^{\times} \times (\mathbb Z/p^n)^{\times}, \mu_{p^n})$$ for each $p$ (where, after having chosen a particular $p$, I have written $k = m p^n$, with $m$ coprime to $p$).  One can compute this lots of ways, e.g. via the Kunneth formula.

The key facts are that if $p$ is odd then (since the mod $p$ cyclotomic character is distinct from the trivial character) $H^i((\mathbb Z/p^n)^{\times},\mu_{p^n})$ vanishes for all $i$, while if $p = 2$ and $n \geq 1$ (resp. $n \geq 2$), then $H^0((\mathbb Z/2^n)^{\times},\mu_{2^n})$ (resp. $H^1((\mathbb Z/2^n)^{\times},\mu_{2^n})$) has order two.

From these, one deduces that

$H^1((\mathbb Z/k)^{\times},\mu_k)$ is trivial if $k$ is odd;

is a product of $l$ cyclic groups of order $2$ if $k$ is exactly divisible by $2$, and is divisible by $l$ distinct odd primes; 

and is a product of $l+1$ cyclic groups of order $2$ if $k$ is divisible by $4$, and by $l$ distinct odd primes.

Here are the concrete interpretations: 

1. If $k$ is odd, then any element of
$\mathbb Q^{\times}$ which becomes a $k$th power in $\mathbb Q(\zeta_k)$ was
already a $k$th power in $\mathbb Q$. This shows Lemma 5.

2. If $k = 2m,$ where $m$ is odd, divisible by primes $p_1,\ldots,p_l$,
then any element of $\mathbb Q^{\times}$ which becomes a $k$th power
in $\mathbb Q(\zeta_k) = \mathbb Q(\zeta_m)$ is a product
of powers of $p_1^m,\ldots,p_l^m$.

3. If $k = 2^n m,$ where $m$ is odd, divisible by primes $p_1,\ldots,p_l$, and $n \geq 4,$ then any element
of $\mathbb Q^{\times}$ which becomes a $k$th power in $\mathbb Q(\zeta_k)$
is a product of powers of 
$p_1^{2^{n-1}m},\ldots,p_l^{2^{n-1}m}, (-4)^{2^{n-2}m}.$


The answer to question 5: The answer is yes, they must be equal. I reproduce the answer from this MSE question. The condition implies that the polynomials $t^k - x, t^k - y$ split at the same primes over $\mathbb{Q}$ for all $k$, hence by the first consequence of Chebotarev's density theorem we listed their splitting fields over $\mathbb{Q}$ are identical. Let $L_k$ denote this splitting field. Then $\mathbb{Q}(\zeta_k) \subset L_k$ where $\zeta_k$ is a primitive $n^{th}$ root of unity; moreover, $L_k$ is a Kummer extension of $\mathbb{Q}(\zeta_k)$ with Galois group $\text{Gal}(L_k/\mathbb{Q}(\zeta_k)) \cong C_k$ acting on a root of either $t^k - x$ or $t^k - y$ by multiplication by a primitive $k^{th}$ root of unity.

Let $a^k = x, b^k = y$. WLOG a generator of $C_k$ acts on $a$ by multiplication by $\zeta_k$ and acts on $b$ by multiplication by $\zeta_k^m$ for some $m \in (\mathbb{Z}/k\mathbb{Z})^{\ast}$. It follows that $\frac{a^m}{b}$ is fixed under the action of $C_k$, hence $\frac{a^m}{b} \in \mathbb{Q}(\zeta_k)$ and 

$$\left( \frac{a^m}{b} \right)^k = \frac{x^m}{y}.$$

When $k = 2$ we conclude that $x, y$ have the same squarefree parts; in particular, $x, y$ have the same sign. By Lemma 5, for all odd $k$ we have $\frac{a^m}{b} \in \mathbb{Q}$ for some choice of $a, b$. 

Given a prime $p$ let $\nu_p$ denote the function which, given an integer, returns the exponent of $p$ in the prime factorization of that integer. Given two primes $p_i, p_j$ and fixing $k$, the condition that $\frac{a^m}{b} \in \mathbb{Q}$ implies that

$$\nu_{p_i}(x) \nu_{p_j}(y) \equiv \nu_{p_j}(x) \nu_{p_i}(y) \bmod k.$$

If this is true for infinitely many $k$, it follows that $\nu_{p_i}(x) \nu_{p_j}(x) = \nu_{p_j}(x) \nu_{p_i}(x)$ for all $i, j$. Together with the fact that $x, y$ have the same sign, it follows that one of $x, y$ is a power of the other, say an $n^{th}$ power. If $n$ has a nontrivial odd factor $o$, then this is a contradiction by taking $k$ a sufficiently large power of $o$; otherwise, $n$ is a power of $2$, and by taking $k$ a sufficiently large power of $2$ we get a contradiction. $\blacksquare$

I think this is a very nice and natural result. Question 5 came up while I was a coordinator at IMO 2025 and one of the students quoted this result on the way to failing to solve P3, which prompted some debate among the coordinators about how well know this fact actually was and whether it should be quotable. It seems there is no proof that doesn't use Chebotarev. For the record, I think most known facts, including this, should be quotable, but I don't think that they should be awarded marks. Sadly, that isn't what happened in the end due to a particularly persistent and silver-tongued leader. Someone later suggested that weaker countries should receive advocates from the IMO so that their students aren't at a disadvantage because of their leader's incompetence, which is a suggestion I'm still considering whether I support.