Let f ∈ C(z) be a rational function. We denote by f n the iterates of f by composition, i.e., f 0 (z) := z, and for each n ≥ 1, f n := f •f n-1 . Then f and its iterates map P 1 (C) = C∪{∞} to itself. The (forward) orbit of a point x ∈ P 1 (C) is Orb + f (x) := {f n (x) : n ≥ 0}.
We say that x ∈ P 1 (C) is periodic (of period n) if there is an integer n ≥ 1 such that f n (x) = x; in that case, the smallest such integer is the exact period of x. More generally, we say x is preperiodic if there is some m ≥ 0 such that f m (x) is periodic. Equivalently, x is preperiodic if and only if the orbit of x is finite. In that case, the smallest m ≥ 0 such that f m (x) is periodic is the tail length of x. We say x is preperiodic of type (m, n) if x is preperiodic with tail length m, and n is the exact period of f m (x). That is, we have f m+n (x) = f m (x) for minimal integers m ≥ 0 and n ≥ 1.
The critical points of f are the ramification points of f in P 1 (C). We say that f is postcritically finite, or PCF, if all of the critical points of f are preperiodic. If f ∈ C[z] is a polynomial, then the critical points of f consist of the point at ∞ (which is fixed by f ) and all the roots of f ′ in C. Thus, a polynomial is PCF if and only if all the roots of its derivative are preperiodic.
In this paper, we consider the case of unicritical polynomials, i.e., polynomials with a single finite critical point (of high multiplicity). After a change of coordinates, we assume this critical point is 0. Thus, throughout the paper, we fix an integer d ≥ 2, and we define We may consider f as an element of the two-variable polynomial ring Z[c, z], but we usually consider c to be a parameter, and we iterate f in the variable z only. That is,
For each integer i ≥ 0, define the polynomial a i (c) ∈ Z[c] by a i (c) := f i (0). Thus, the sequence
gives the iterates of the critical point 0 under f . We are interested in the case that f is PCF, i.e., that this orbit is finite.
To this end, fix a d-th root of unity ζ that is not 1. For any integers m ≥ 2 and n ≥ 1, we define the
where µ denotes the Möbius µ-function. A priori, G ζ d,m,n is a rational function in Q(ζ)(c), but in fact it is a monic polynomial in Z[ζ][c], as we prove in Section 2. Its roots are parameters c 0 , called Misiurewicz parameters, for which f m+n c 0 (0) = f m c 0 (0) but no earlier iterates f i c 0 (0) coincide; we say f c 0 is PCF of exact type (m, n). The root of unity ζ further specifies that f m+n-1 c 0 (0)/f m-1 c 0 (0) = ζ. Milnor [18,Remark 3.5] conjectured that related polynomials over Q are irreducible, and we make the following corresponding conjecture over Q(ζ):
Recent progress in [5,11,12] has proven Conjecture 1.1 in the case that d = 2 and n ≤ 3, but otherwise, very little is currently known. Such arithmetic questions have dynamical consequences, as illustrated by the work of Buff, Epstein and Koch in [5], who applied these known instances of Conjecture 1.1 to prove the first cases of a different conjecture of Milnor [16,17] on the irreducibility of certain moduli curves arising in complex dynamics [5,Theorem 1,Theorem 4]. See also [13,Section 2] for a brief survey of known results on Misiurewicz parameters, and our companion paper [3] for further results in the study of their arithmetic properties.
More broadly, postcritically finite polynomials play a fundamental role in polynomial dynamics. On the complex dynamical side, Douady and Hubbard [7,Chapter 8] proved that Misiurewicz parameters are dense in the boundary of the Mandelbrot set, and Favre and Gauthier [9, Theorem 1] further proved that they are equidistributed in an appropriate sense. Ghioca, Krieger, Nguyen, and Ye [10,Theorem 3.1] generalized this equidistribution to PCF maps in arbitrary dynamical moduli spaces. Indeed, as proposed by Baker and DeMarco in [1], PCF maps should play a role in dynamical moduli spaces analogous to CM points on modular curves, and more generally to special points on Shimura varieties.
Returning to the unicritical family f d,c , while Misiurewicz parameters are ones for which the critical point is strictly preperiodic, those for which the critical point is periodic are roots of Gleason polynomials. Specifically, for n ≥ 1, the roots of the Gleason polynomial
are parameters c 0 for which the critical point 0 has exact period n under f c 0 . See, for example, [4,6,13].
Question 1.2. For which d ≥ 2, n ≥ 1, is the Gleason polynomial G d,0,n irreducible over Q?
As with Conjecture 1.1, very little is known about Question 1.2. Even for fixed degree d, there is no infinite family of Gleason polynomials which are known to be irreducible. When d = 2, calculations for small periods suggest that G 2,0,n is irreducible over Q for all n ≥ 1, but this conjecture remains wide open. Buff [4, Proposition 5] observed that the corresponding conjecture is false in general by showing that G d,0,3 has 2 irreducible factors if d ≡ 1 (mod 6).
Remark 1.3. The definitions of Misiurewicz and Gleason polynomials are not entirely consistent in the literature. For example, some authors use the family of maps az d + 1, as in [4,5,6], instead of z d + c. In addition, our choice of a root of unity ζ is another difference both from those authors and from previous work in [11,12,13].
In this paper, we consider various arithmetic properties of the orbits Orb + f (c 0 ), where c 0 is a Misiurewicz parameter. Theorem 1.4, which we prove using purely local methods, concerns the case that the degree d is a prime power. In particular, it generalizes [11,Theorem 3.1] from prime degrees to prime-power degrees, and it answers a question raised in [11]. Here and throughout the paper, when K is a number field, we denote by O K the ring of integers of K, and for any b ∈ O K , we write b for the principal ideal generated by b.
Theorem 1.4. Let d = p e be a prime power, and suppose
. Take a prime ideal p ⊆ O K which lies over p. By Theorem 1.4(b), the ramification index e(p|p) satisfies e(p|p) ≥ p r (p-1)(d m-1 -1), and hence
,n to be irreducible over Q(ζ), as desired. Let c 0 be a root of the Gleason polynomial G d,0,n , and set K = Q(c 0 ). The second author showed [11,Lemma 3.1] that G d,0,i (c 0 ), i.e., another Gleason polynomial evaluated at c 0 , is an algebraic unit in O K unless i = n. Buff, Epstein, and Koch studied the resultants of Misiurewicz polynomials with Gleason polynomials, and they proved that a Misiurewicz polynomial evaluated at a Gleason parameter is an algebraic unit unless the periods of these two polynomials are same [5,Lemma 26]. They have used these resultants to prove new irreducibility results for Misiurewicz polynomials. In this paper, we study the next natural question for Misiurewicz polynomials: Note that in the setting of Question 1.6, we have ζ ∈ K, because ζ = a m+n-1 (c 0 )/a m-1 (c 0 ). Question 1.6 is also motivated in part by analogy with the theory of cyclotomic polynomials. Specifically, the following classical result is well known and has several different proofs in the literature, the earliest of which is due to Emma T. Lehmer [14,Theorem 4]. Question 1.6 is also evocative of the study of dynamical units introduced by Morton and Silverman in [19]. However, whereas Morton and Silverman considered units arising from differences between periodic points of a single map f , the units and non-units we consider in this paper arise from parameters in a dynamical moduli space.
When d is a prime power, we are able to give the following answer to Question 1.6 in the case j = m. Theorem 1.8. Let d = p e , where p is a prime and e ≥ 1.
, where
The techniques needed to analyze the case j = m are very different from those used in the current paper for j = m. Therefore we discuss the above conjecture in greater detail in the sequel paper [3].
The structure of the paper is as follows. In Section 2, we prove that G ζ d,m,n is a polynomial. We prove Theorem 1.4 in Section 3, answering the question posed in [11] in the affirmative. We then consider the j = m case of Question 1.6, proving Theorem 1.8 for j < m in Section 4, and for j > m in Section 5.
The purpose of this section is to prove the following theorem. Our proof will require two auxiliary lemmas, as follows.
for some positive divisor k of n, and suppose that k is the smallest positive divisor of n for which this equality holds. Then for any integer ℓ|n, we have
. By definition of k, we have a m+k-1 (α) = ζa m-1 (α). Applying f k to both sides of this equality, we obtain a m+2k-1 (α) = a m+k-1 (α). Applying f k repeatedly, then, we have a m+ik-1 (α) = a m+k-1 (α) for any integer i ≥ 1.
Armed with this fact, we can now prove the equivalence. For the reverse implication, i.e., assuming k|ℓ, we have ℓ = ik for some i ≥ 1, and hence
For the forward implication, we assume a m+ℓ-1 (α) = ζa m-1 (α). There exist positive integers i, j, t ≥ 1 such that ik + jℓ = tk + gcd(k, ℓ). As we saw at the start of this proof, we have a m+ik-1 (α) = ζa m-1 (α); applying f jℓ yields
On the other hand, by our choice of i, j, t, we have
In particular, in the polynomial ring Z[c], we have a n |a nt and G d,0,n |a n . Thus, we have
To prove the desired equivalence, we begin with the reverse implication, i.e., we suppose that i = n. Because we have n|m -1 and hence also n|m + i -1, it follows that a m+i-1 (α) = 0 = ζa m-1 (α), as desired.
Conversely, suppose a m+i-1 (α) = ζa m-1 (α). Because n|m -1, we have a m-1 (α) = 0, and hence a m+i-1 (α) = 0 as well. Therefore, we have
or equivalently, a i (α) = 0. However, α was a root of G d,0,n , and G d,0,n is known to be relatively prime to a i for 1 ≤ i < n. (See, for instance, [5,Lemma 30], which shows that the resultant of two different Gleason polynomials is ±1, and hence they share no roots. Since a i is a product of Gleason polynomials, it is indeed relatively prime to G d,0,n .) Thus, we must have i = n, as desired.
Proof of Theorem 2.1. Case 1. Suppose that n ∤ m -1. By definition, we have
Let α be a root of a m+k-1 -ζa m-1 for some minimal positive integer k|n. By Lemma 2.2, for any positive divisor ℓ of n, we have that α is a root of a m+ℓ-1 -ζa m-1 if and only if k|ℓ. In that case, as shown in the proof of Theorem A.1 of [8], the order of vanishing of a m+ℓ-1 -ζa m-1 at α is 1. Thus, the order of vanishing of
where we have applied the well-known identity ( 5)
Thus, the rational function G ζ d,m,n has order of vanishing either 0 or 1 at every point of
and it has only simple roots. Finally, because all of the multiplicands in equation ( 4) are monic polynomials in
Case 2. Suppose that n|m -1. By definition, we have
.
As we saw in Case 1, the numerator is a monic polynomial in Z[ζ][c] with simple roots, so we only need to consider roots of G d,0,n = i|n a µ(n/i) i
, which is also known to have simple roots (see, for instance, [7,Lemma 19.1] or [8,Proposition A.1]).
For any root α of G d,0,n , Lemma 2.3 says that the only term of the numerator that has α as a root is when i = n, i.e., the term (a m+n-1 -ζa m-1 ) µ(n/n) = a m+n-1 -ζa m-1 , which has a (simple) root at α. Thus, G ζ d,m,n has order of vanishing zero at α. As before, then, it follows that
with only simple roots.
The results of this section generalize estimates proven by the second author in [11]. Throughout this section, fix integers d, m, n with d, m ≥ 2 and n ≥ 1. Let c 0 ∈ Q be a Misiurewicz parameter of exact type (m, n), write f := f d,c 0 , and define K := Q(c 0 ).
For any finite place v of K, we define K v to be the v-adic completion of K, and C v to be the completion of an algebraic closure of K v . For any x ∈ C v and r > 0, we denote by
We begin with the following modest strengthening of [11,Lemma 2.4]. Proposition 3.1. If f is PCF of exact type (m, n), then for every finite place v of K, either:
• v(a i (c 0 )) = 0 for all i ≥ 1, or
• v(a n (c 0 )) > 0, and for all i ≥ 1, we have v(a i (c
Applying Proposition 3.1 at every finite place v of K, we immediately obtain:
Proof of Proposition 3.1. We already know v(a i (c 0 )) ≥ 0 for all i ≥ 1. If v(a i (c 0 )) = 0 for all i, then we are in the first case, and we are done. So we assume for the remainder of the proof that v(a ℓ (c 0 )) > 0 for some minimal ℓ ≥ 1. Thus, f ℓ maps D(0, 1) onto itself multiply-to-1, and hence by Theorem 4.18(b) of [2], the disk D(0, 1) contains a unique periodic point b of f , which is v-adically attracting and of exact period ℓ. Because f m (0) is a periodic point of exact period n lying in f m (D(0, 1)), it must be in the same cycle as b, and hence ℓ = n.
Since the disk D(0, 1) has exact period ℓ = n, we have v(a i (c 0 )) = 0 for all i ≥ 1 for which n ∤ i. It remains to consider i of the form i = nj for j ≥ 1.
If b = 0, then z = 0 itself is periodic, so m = 0, and we have a n (c 0 ) = a i (c 0 ) = 0, and we are done. Thus, we assume for the rest of the proof that b = 0.
Because the periodic point b = 0 is attracting, we have
and hence |a nj (c 0 )| v = |b| v for all j ≥ 1. In particular, writing i = nj, we have
as desired.
We have a m+n-1 (c 0 ) d = a m-1 (c 0 ) d but a m+n-1 (c 0 ) = a m-1 (c 0 ), and hence there is a d-th root of unity ζ = 1 such that a m+n-1 (c 0 ) = ζa m-1 (c 0 ). We also have a m-1 (c 0 ) = 0.
Applying Theorem 3.3 at every finite place v immediately yields Theorem 1.4.
To prove part (2) of Theorem 3.3, we will need the following two lemmas. We denote by C p the completion of an algebraic closure of the p-adic field Q p .
Then
, w is a root of the polynomial h. However, the Newton polygon of g has vertices at (p r , e -r) for r = 0, 1, . . . , e, and the hypotheses say that v(x) ≤ p/(p -1). Thus, the Newton polygon of h consists of a single segment of length d and slope -v(x)/d. Hence, the root w satisfies dv(w) = v(x), and therefore |w| d p = |x| p . Multiplying both sides of this equation by |b| d p yields the desired result. Lemma 3.5. Let p be a prime, let e ≥ 1 be an integer, let d = p e , let c 0 ∈ C p , and suppose that f (z
Step 1. We claim that for every 0 ≤ i ≤ m -1, we have ( 6)
Indeed, if inequality (6) fails for any 0 ≤ i ≤ m -1, then because |a j (c 0 )| p ≤ 1 for all j, we have
so that the inequality also fails for i + 1. By induction, then, it fails for m -1, meaning that
where we have used the fact that |a m-1 (c 0 )| p ≥ |a n (c 0 )| p by Proposition 3.1. However, both the map z → z d , and hence also f , are one-to-one on the open disk
But the distinct points a m+n-1 (c 0 ) and a m-1 (c 0 ) both lie in this disk, and they both map to a m+n (c 0 ) = a m (c 0 ) under f . This contradiction proves our claim.
Step 2. Note that
, yielding the desired equality for i = 0, because a 0 = 0. Moreover, combining equation (7) with inequality (6), we have |p| In particular, since |a j (c 0 )| p ≤ 1 for all j, we have
Therefore, we may apply Lemma 3.4 inductively, yielding the desired conclusion.
Proof of Theorem 3.3. Case (1). Suppose first that v(d) = 0. If v(a n (c 0 )) > 0, then again by Theorem 4.18(b) of [2], there is a unique periodic point b of f in D(0, 1), which is v-adically attracting and of exact period n. (And we must have b = f nk (0) for some k ≥ 0 with nk ≥ m.) But because v(d) = 0, we have that f (z) = z d + c 0 is one-to-one on each disk D(x, |x|) for x ∈ C × v . In particular, f is one-to-one on each disk D(a i (c 0 ), 1) for i = 1, . . . , n -1, and on the disk D(b, |b|). Thus, f n maps D(0, 1) d-to-1 onto D(0, 1), with D(b, |b|) mapping bijectively onto a (proper) subdisk of itself.
If
, the inverse image of b under f n includes 0 counted with multiplicity d, and b with multiplicity 1, for a total of (at least) d + 1, contradicting the fact that f n has degree d on D(0, 1).
On the other hand, if f n (0) = b, then because b is attracting, we have |f n (0)-b| v < |0-b| v , so that f n (0) ∈ D(b, |b|). But then, because f n : D(b, |b|) → D(b, |b|) is one-to-one with b fixed, the iterates f nj (0) are never equal to b for j ≥ 1, contradicting the fact that b = f nk (0) for some k ≥ 0. Thus, either way, we have a contradiction, and hence our original assumption that v(a n (c 0 )) > 0 is impossible. That is, v(a n (c 0 )) = 0. By Proposition 3.1, we have v(a i (c 0 )) = 0 for all i ≥ 1, proving statement (1).
Case (2). For the remainder of the proof, we may assume that d = p e is a prime power, and that v|p (i.e., v(d) > 0). The map f (z) = z d + c 0 is a bijection on the residue field, since it is a composition of Frobenius and a translation. Thus, f acts as a bijection on the (finite) set of open unit disks {D(x, 1) : x ∈ O K }. Every disk is therefore periodic (as opposed to just preperiodic) under this action. In particular, there is some ℓ > 0 such that f ℓ (0) ∈ D(0, 1). By Proposition 3.1, then, we have v(a n (c 0 )) > 0, and v(a i (c 0 )) = v(a n (c 0 )) if and only if n|i. (And if n ∤ i, then v(a i (c 0 )) = 0.) Thus, it suffices to show the desired formula in the case that i = n.
Let ζ := a m+n-1 (c 0 )/a m-1 (c 0 ), and let 0 ≤ r ≤ e -1 be the smallest nonnegative integer such that ζ p r+1 = 1, as in the statement of the theorem. Then (8)
where the second equality is by repeated application of Lemma 3.5.
If n ∤ (m -1), then |a m-1 (c 0 )| p = 1 by Proposition 3.1, whence
where we have used the well known fact that
Thus, we have the desired equality
On the other hand, if n|(m-1), then |a m-1 (c 0 )| p = |a n (c 0 )| p by Proposition 3.1, and therefore equation ( 8) becomes
and hence
as desired.
4. G ζ d,j,ℓ (c 0 ) WHEN j < m. In this section, we answer Question 1.6 for j < m by proving Theorem 1.8 in that case. We begin with the following lemma, which is an analogue of part (1) of Theorem 3.3 for the principal ideal a j+ℓ-1 (c 0 ) -ζa j-1 (c 0 ) when 2 ≤ j ≤ m -1. Proof. Applying f := f d,c 0 to both sides of a j+ℓ-1 (c 0 ) ≡ wa j-1 (c 0 ) (mod p) yields (10) a j+ℓ (c 0 ) ≡ a j (c 0 ) (mod p).
Repeatedly applying f to both sides of (10), we obtain (11) a k+tℓ (c 0 ) ≡ a k (c 0 ) (mod p)
for any k ≥ j and t ≥ 1.
In particular, using k = m -1 ≥ j and t = n in (11), we have
,n , we have a m-1+nℓ (c 0 ) = ζa m-1 (c 0 ). Substituting this in (12), it follows that (ζ -1)a m-1 (c 0 ) ≡ 0 (mod p), and hence (13) either p| ζ -1 or p| a m-1 (c 0 ) .
(Alternatively, if d is a prime power, these two facts are immediate from ( 9) and our Theorem 1.4, respectively.) The desired result follows immediately from these two facts and (13).
We also need the following analogue of part (2) of Theorem 3.3 for the same setting as in Lemma 4.1, provided d is a prime power.
Proof. Case (a). Let f = f d,c 0 , and write ( 14)
Expanding the expression f j-1 (a ℓ (c 0 )), there exists a polynomial F ∈ Z[x] such that
Thus, equation ( 14) becomes ( 15)
Suppose there were a prime ideal p ⊆ O L dividing a j+ℓ-1 (c 0 ) -wa j-1 (c 0 ) . Then
We have p ≡ 0 (mod p) by Lemma 4.1, and hence
By ( 9), we have 1 -w p r (p-1) = p as ideals in O L , where 0 ≤ r ≤ e -1 is the smallest integer such that w p r+1 = 1. Hence 1 -w ≡ 0 (mod p), which forces a ℓ (c 0 ) ≡ 0 (mod p). This contradicts Corollary 3.2, which says that a ℓ (c 0 ) is a unit in O L .
Case (b). Putting ζ in the role of w in the proof of part (a), we have 1 -ζ p r (p-1) = p , where r is the same integer as in Theorem 3.3. Let ( 16)
for some units u 1 , u 2 in O L . Clearly EM ≥ M > d j-1 , so all of the exponents in (18) are positive, and hence Q ∈ O L . Suppose there were a prime ideal p ⊆ O L such that Q ≡ 0 (mod p). Then by (17) we would also have a j+ℓ-1 (c 0 ) -ζa j-1 (c 0 ) ≡ 0 (mod p), so that Lemma 4.1 yields p| d , and hence p| a ℓ (c 0 ) , since a ℓ (c 0 ) eEM = d . Equation ( 18) therefore yields
as desired. Note that we used Theorem 1.4 in the last equality.
We need one more lemma before we can prove Theorem 1.8 for j < m.
Proof. If ℓ ≡ 0(mod n), then by Proposition 4.2.(a), we have
Therefore, we may assume for the rest of the proof that ℓ ≡ 0(mod n). Write ℓ = nt for some t ∈ N, and as usual, write f := f d,c 0 . We proceed via a local argument.
For any place v of L that does not divide d, we have , which are the maximum and minimum v-adic distances (respectively) between a nontrivial d-th root of unity and 1.
For any x ∈ C v with |x| v < ρ, expanding (1 + x) d shows that
Thus, for any b, c ∈ C × v with |b -c| v < ρ|b| v , we have ( 19)
We claim that for any d-th root of unity η, we have (20)
To prove the claim, suppose inequality (20) fails for some such η. Then by inequality (19) with b = ηa j-1 (c 0 ) and c = a j+nt-1 (c 0 ), we have
Applying m -j -1 ≥ 0 more iterations of f , and noting that f does not expand distances on D(0, 1), we have
However, by Theorem 3.3, we have
This contradiction proves the claim of inequality (20). We now use the claim to prove the lemma. Observe that (22)
where the first inequality is by definition of ρ, and the second is by the claim applied to η = w. Therefore,
where the first inequality is the non-archimedean triangle inequality, and the second is by inequality (22).
We have just shown that |a j+nt-1 (c 0 ) -w ′ a j-1 (c 0 )| v ≤ |a j+nt-1 (c 0 ) -wa j-1 (c 0 )| v . Applying the same argument with the roles of w and w ′ reversed, we similarly have
thus proving the lemma.
Proof of Theorem 1.8 for j < m. We will consider the cases ℓ ≡ 0 (mod n) and ℓ ≡ 0 (mod n) separately.
Case 1. Suppose that ℓ ≡ 0 (mod n). The result is immediate from part (a) of Proposition 4.2, because by the Möbius product definition of G ζ d,j,ℓ , we have
as ideals in O K .
Case 2. Suppose that ℓ ≡ 0 (mod n). Set ℓ = nt for some t ∈ N, and L := Q(c 0 , η) for some primitive d-th root of unity η. By [5, Lemma 27], there is a polynomial F ∈ Z[c] such that (23)
First consider the case t > 1. By equation ( 3), we know that G d,0,nt (c 0 ) divides ant(c 0 ) an(c 0 ) in O K . (See also Lemma 5.4 of [15].) By Corollary 3.2, it follows that u
Substituting this value in (23), we obtain (24)
+ pF (c 0 )
Since we have G w d,j,nt (c 0 ) a j+nt-1 (c 0 ) -wa j-1 (c 0 )
as ideals in O L , if there were a prime ideal p ⊆ O L such that G w d,j,nt (c 0 ) ≡ 0 (mod p), then Lemma 4.1 yields p ≡ 0 (mod p). This fact together with (24) implies u 1 ≡ 0 (mod p), a contradiction. Hence, there is no such a prime ideal p ⊆ O L , whence G w d,j,nt (c 0 ) = O L for each d-th root of unity w. In particular, we have G ζ d,j,ℓ (c 0 ) = O K , completing the proof of part (a) of Theorem 1.8 for j < m.
It remains to consider the case that t = 1, i.e. ℓ = n. By [12, Lemma 2.2], there is a unit u 2 in O K such that G d,0,n (c 0 ) = u 2 a n (c 0 ). Substituting this value in (23), we obtain
for some unit u 3 in O K . Define E, M as in equations ( 16), and observe that EM > (d -1)N j,n . Recall from Theorem 3.3 that a n (c 0 ) EM = p . Hence, there exists a unit u 4 in O K such that (25)
For each w in the above product, G w d,j,n (c 0 ) divides a j+n-1 (c 0 ) -wa j-1 (c 0 ) (as ideals in O L ), by the Möbius product definition of G w d,j,n . By part (b) of Proposition 4.2 and by Lemma 4.3, then, any prime ideal of O L dividing G w d,j,n (c 0 ) must divide a n (c 0 ) . By equation (25), any prime ideal p ⊆ O L dividing u 3 + u 4 a n (c 0 ) EM -(d-1)N j,n F (c 0 ) must divide some G w d,j,n (c 0 ) and hence also divides a n (c 0 ) . Then
contradicting the fact that u 3 is a unit, and hence showing that no such p exists. Thus, u 3 + u 4 a n (c 0 ) EM -N j,n F (c 0 ) must be a unit in O L , and hence also in O K . Therefore, Proof. Case (a). Suppose for the sake of contradiction that there exists a prime ideal p ⊆ O K that satisfies a j+ℓ-1 (c 0 ) -ζa j-1 (c 0 ) ≡ 0 (mod p), i.e.
(26) a j+ℓ-1 (c 0 ) ≡ ζa j-1 (c 0 ) (mod p).
Applying n iterations of f := f d,c 0 to both sides of (26), we obtain a j+ℓ+n-1 (c 0 ) ≡ a j+n-1 (c 0 ) (mod p).
Since f has exact type (m, n) and j -1 ≥ m, it follows that (27) a j+ℓ-1 (c 0 ) ≡ a j-1 (c 0 ) (mod p).
Combining (26) and ( 27) yields Let t be a positive integer with nt ≥ j. Applying nt -j + 1 iterations of f to both sides of (27) yields (30) a nt+ℓ (c 0 ) ≡ a nt (c 0 ) (mod p).
However, because ℓ ≡ 0 (mod n), Theorem 1.4 implies that a nt+ℓ (c 0 ) is a unit in O K ; thus, equations ( 29) and (30) contradict one another. Hence, there is no such prime ideal p ⊆ O K . That is, a j+ℓ-1 (c 0 ) -ζa j-1 (c 0 ) = O K , as desired.
Case (b). If ℓ ≡ 0 (mod n), then because f has exact type (m, n) and j -1 ≥ m, we obtain a j+ℓ-1 (c 0 ) = a j-1 (c 0 ), which immediately implies the result.
Proof of Theorem 1.8 for j > m. We again consider the cases ℓ ≡ 0 (mod n) and ℓ ≡ 0 (mod n) separately. Case 2. Suppose that ℓ ≡ 0 (mod n). Write ℓ = nt for some t ∈ N. We first consider the case nt ∤ j -1. By definition, we have G ζ d,j,nt (c 0 ) = k|nt (a j+k-1 (c 0 ) -ζa j-1 (c 0 )) µ(nt/k) .
By Lemma 5.1, a j+k-1 (c 0 ) -ζa j-1 (c 0 ) is a unit in O K for each k ≡ 0 (mod n). Thus,
as desired. In particular, the third equality is by Lemma 5.1, the fourth is by the Möbius identity (5), and the fifth is by Theorem 1.4 together with the fact that n ∤ j -1.
It remains to consider the case nt | j -1. Using the first part of Case 2 and the definition of G ζ d,j,nt , we have
k|nt a k (c 0 ) -µ(nt/k) if t = 1 as ideals in O K . By Theorem 1.4, (31) immediately yields
as desired. Note that in the last equality, we used Theorem 1.4, equation (5), and the fact that n | j -1.