Primary decomposition of the ideal of polynomials whose fixed
divisor is divisible by a prime power
J. Algebra 398 (2014), 227-242, http://dx.doi.org/10.1016/j.jalgebra.2013.09.016

arXiv:1304.7450v3 [] 7 May 2014

Giulio Peruginelli1
Institut für Analysis und Comput. Number Theory, Technische Universität, Steyrergasse 30, A-8010 Graz,
Austria.

Abstract
We characterize the fixed divisor of a polynomial f (X) in Z[X] by looking at the contraction
of the powers of the maximal ideals of the overring Int(Z) containing f (X). Given a prime
p and a positive integer n, we also obtain a complete description of the ideal of polynomials
in Z[X] whose fixed divisor is divisible by pn in terms of its primary components.
Keywords: Integer-valued polynomial, Image of a polynomial, Fixed divisor,
Factorization of integer-valued polynomials, Primary components, Primary
decomposition. MSC Classification codes: 13B25, 13F20.

to Sergio Paolini, whose teachings and memory I deeply preserve.

1. Introduction
In this work we investigate the image set of integer-valued polynomials over Z. The set
of these polynomials is a ring usually denoted by:
Int(Z) + {f ∈ Q[X] | f (Z) ⊂ Z}.
Since an integer-valued polynomial f (X) maps the integers in a subset of the integers,
it is natural to consider the subset of the integers formed by the values of f (X) over the
integers and the ideal generated by this subset. This ideal is usually called the fixed divisor
of f (X). Here is the classical definition.

1

Email address: peruginelli@math.tugraz.at (Giulio Peruginelli)
Old affiliation, from June 2014: Department of Mathematics, University of Padova, Italy.

1

January 22, 2018

Definition 1.1. Let f ∈ Int(Z). The fixed divisor of f (X) is the ideal of Z generated
by the values of f (n), as n ranges in Z:
d(f ) = d(f, Z) = (f (n)|n ∈ Z).
We say that a polynomial f ∈ Int(Z) is image primitive if d(f ) = Z.
It is well-known that for every integer n ≥ 1 we have
d(X(X − 1) . . . (X − (n − 1))) = n!
so that the so-called binomial polynomials Bn (X) + X(X − 1) . . . (X − (n − 1))/n! are
integer-valued (indeed, they form a free basis of Int(Z) as a Z-module; see [4]).
Notice that, given two integer-valued polynomials f and g, we have d(f g) ⊂ d(f )d(g)
and we may not have an equality. For instance, consider f (X) = X and g(X) = X −1; then
we have d(f ) = d(g) = Z and d(f g) = 2Z. If f ∈ Int(Z) and n ∈ Z, then directly from the
definition we have d(nf ) = nd(f ). If cont(F ) denotes the content of a polynomial F ∈ Z[X],
that is, the greatest common divisor of the coefficients of F , we have F (X) =cont(F )G(X),
where G ∈ Z[X] is a primitive polynomial (that is, cont(G)=1). We have the relation:
d(F ) = cont(F )d(G).
In particular, the fixed divisor is contained in the ideal generated by the content. Hence,
given a polynomial with integer coefficients, we can assume it to be primitive. In the same
way, if we have an integer-valued polynomial f (X) = F (X)/N , with f ∈ Z[X] and N ∈ N,
we can assume that (cont(F ), N ) = 1 and F (X) to be primitive.
The next lemma gives a well-known characterization of a generator of the above ideal
(see [1, Lemma 2.7]).
Lemma 1.1. Let f ∈ Int(Z) be of degree d and set
1) d1 = sup{n ∈ Z |

f (X)
n

∈ Int(Z)}

2) d2 = GCD{f (n) | n ∈ Z}
3) d3 = GCD{f (0), . . . , f (d)}
then d1 = d2 = d3 .
Let f ∈ Int(Z). We remark that the value d1 of Lemma 1.1 is plainly equal to:
d1 = sup{n ∈ Z |f ∈ nInt(Z)}.
Moreover, given an integer n, we have this equivalence that we will use throughout the
paper, a sort of ideal-theoretic characterization of the arithmetical property that all the
values attained by f (X) are divisible by n:
f (Z) ⊂ nZ ⇐⇒ f ∈ nInt(Z)
2

(nInt(Z) is the principal ideal of Int(Z) generated by n). From 1) of Lemma 1.1 we see
immediately that if f (X) = F (X)/N is an integer-valued polynomial, where F ∈ Z[X] and
N ∈ N coprime with the content of F (X), then d(f ) = d(F )/N , so we can just focus our
attention on the fixed divisor of a primitive polynomial in Z[X].
We want to give another interpretation of the fixed divisor of a polynomial f ∈ Z[X] by
considering the maximal ideals of Int(Z) containing f (X) and looking at their contraction
to Z[X]. We recall first the definition of unitary ideal given in [12].
Definition 1.2. An ideal I ⊆ Int(Z) is unitary if I ∩ Z 6= 0.
That is, an ideal I of Int(Z) is unitary if it contains a non-zero integer, or, equivalently,
IQ[X] = Q[X] (where IQ[X] denotes the extension ideal in Q[X]). The whole ring Int(Z)
is clearly a principal unitary ideal generated by 1.
The next results are probably well-known, but for the ease of the reader we report
them. The first lemma says that a principal unitary ideal I is generated by a non-zero
integer, which generates the contraction of I to Z. In particular, this lemma establishes a
bijective correspondence between the nonzero ideals of Z and the set of principal unitary
ideals of Int(Z).
Lemma 1.2. Let I ⊆ Int(Z) be a principal unitary ideal. If I ∩ Z = nZ with n 6= 0
then I = nInt(Z). In particular, nInt(Z) ∩ Z = nZ. Moreover, n1 Int(Z) = n2 Int(Z) with
n1 , n2 ∈ Z if and only if n1 = ±n2 .
Proof : If I = (f ) for some f ∈ Int(Z) then deg(f ) = 0 since a non-zero integer n is in
I. Since f (X) is integer-valued it must be equal to an integer and so it is contained in
I ∩ Z = nZ. Hence we get the first statement of the lemma. If n1 Int(Z) = n2 Int(Z) then
n1 = n2 f with f ∈ Int(Z); this forces f to be a non-zero integer, so that n1 divides n2 .
Similarly, we get that n2 divides n1 . 
Lemma 1.3. Let I1 , I2 ⊆ Int(Z) be principal unitary ideals. Then I1 ∩ I2 is a principal
unitary ideal too.
Proof : Suppose Ii = ni Int(Z), where ni ∈ Z, ni Z = Ii ∩ Z, for i = 1, 2. We have
n1 Z ∩ n2 Z = nZ, where n = lcm{n1 , n2 }. The ideal I1 ∩ I2 is unitary since n ∈ I1 ∩ I2 .
In particular, we have I1 ∩ I2 ⊇ nInt(Z). We have to prove that I1 ∩ I2 ⊆ nInt(Z). Let
f ∈ I1 ∩ I2 . Then f (Z) ⊂ n1 Z ∩ n2 Z = nZ, so that f (X)
n ∈ Int(Z). 
The previous lemma implies the following decomposition
Q for a principal unitary ideal
generated by an integer n, with prime factorization n = i pai i . We have
\ a
Y a
nInt(Z) =
pi i Int(Z) =
pi i Int(Z)
i

i

3

where the last equality holds because the ideals pai i Z are coprime in Z, hence they are
coprime in Int(Z).
We are now ready to give the following definition.
Definition 1.3. Let f ∈ Int(Z). The extended fixed divisor of f (X) is the minimal
ideal of the set {nInt(Z) | n ∈ Z, f ∈ nInt(Z)}. We denote this ideal by D(f ).
Equivalently, in the above definition, we require that nInt(Z) contains the principal
ideal in Int(Z) generated by the polynomial f (X). Lemma 1.2 and 1.3 show that the
minimal ideal in the above definition does exist: it is equal to the intersection of all the
principal unitary ideals containing f (X). Notice that the extended fixed divisor is an ideal
of Int(Z), while the fixed divisor is an ideal of Z. The polynomial f (X) is image primitive
if and only if its extended fixed divisor is the whole ring Int(Z). In the next sections we
will study the extended fixed divisor by considering the p-part of it, namely the principal
unitary ideals of the form pn Int(Z), p ∈ Z being prime and n a positive integer.
The following proposition gives a link between the fixed divisor and the extended fixed
divisor: the latter is the extension of the former and conversely. So each of them gives
information about the other one.
Proposition 1.1. Let f ∈ Int(Z). Then we have:
a) D(f ) ∩ Z = d(f )
b) d(f )Int(Z) = D(f )
Proof : Let d, D ∈ Z be such that d(f ) = dZ and D(f ) = DInt(Z). Since d(f )Int(Z) =
dInt(Z) is a principal unitary ideal containing f (X), from the definition of extended fixed
divisor, we have D(f ) ⊆ dInt(Z). In particular, D ≥ d. We also have f (X)/D ∈ Int(Z)
and so d ≥ D, by characterization 1) of Lemma 1.1. Hence we get a). From that we deduce
that d(f ) ⊆ D(f ), so statement b) follows. 
As already remarked in [5], the rings Z and Int(Z) share the same units, namely {±1}.
Then [5, Proposition 2.1] can be restated as follows.
Proposition 1.2 (Cahen-Chabert). Let f ∈ Int(Z) be irreducible in Q[X]. Then f (X)
is irreducible in Int(Z) if and only if f (X) is not contained in any proper principal unitary
ideal of Int(Z).
The next lemma has been given in [6] and is analogous to the Gauss Lemma for polynomials
in Z[X] which are irreducible in Int(Z).
Lemma 1.4 (Chapman-McClain). Let f ∈ Z[X] be a primitive polynomial. Then f (X)
is irreducible in Int(Z) if and only if it is irreducible in Z[X] and image primitive.
4

For example, the polynomial f (X) = X 2 + X + 2 is irreducible in Q[X] and also in
Z[X] since it is primitive (because of Gauss Lemma). But it is reducible in Int(Z) since its
extended fixed divisor is not trivial, namely it is the ideal 2Int(Z). So in Int(Z) we have
the following factorization:
X2 + X + 2
f (X) = 2 ·
2
and indeed this is a factorization into irreducibles in Int(Z), since the latter polynomial is
image primitive and irreducible in Q[X], and by [5, Lemma 1.1], the irreducible elements in
Z remain irreducible in Int(Z). So the study of the extended fixed divisor of the elements
in Int(Z) is a first step toward studying the factorization of the elements in this ring (which
is not a unique factorization domain).
Here is an overview of the content of the paper. At the beginning of the next section we
recall the structure of the prime spectrum of Int(Z). Then, for a fixed prime p, we describe
the contractions to Z[X] of the maximal unitary ideals of Int(Z) containing p (Lemma 2.1).
In Theorem 2.1 we describe the ideal Ip of Z[X] of those polynomials whose fixed divisor
is divisible by p, namely the contraction to Z[X] of the principal unitary ideal pInt(Z),
which is the ideal of integer-valued polynomials whose extended fixed divisor is contained
in pInt(Z). It turns out that Ip is the intersection of the aforementioned contractions.
In the third section we generalize the result of the second section to prime powers, by
means of a structure theorem of Loper regarding unitary ideals of Int(Z). We consider the
contractions to Z[X] of the powers of the prime unitary ideals of Int(Z) (Lemma 3.1). In
Remark 2 we give a description of the structure of the set of these contractions; that allows
us to give the primary decomposition of the ideal Ipn = pn Int(Z) ∩ Z[X], made up of those
polynomials whose fixed divisor is divisible by a prime power pn . We shall see that we have
to distinguish two cases: p ≤ n and p > n (see also the examples in Remark 3). In Theorem
3.1 we describe Ipn in the case p ≤ n. This result was already known in a slightly different
context by Dickson (see [7, p. 22, Theorem 27]), but our different proof uses the primary
decomposition of Ipn and that gives an insight to generalize the result to the second case.
In Proposition 3.2 we give a set of generators for the primary components of Ipn , in the
case p > n. Finally in the last section, as an application, we explicitly compute the ideal
Ipp+1 .

2. Fixed divisor via Spec(Int(Z))
The study of the prime spectrum of the ring Int(Z) began in [3]. We recall that the
prime ideals of Int(Z) are divided into two different categories, unitary and non-unitary.
Let P be a prime ideal of Int(Z). If it is unitary then its intersection with the ring of
5

integers is a principal ideal generated by a prime p.
Non-unitary prime ideals: P ∩ Z = {0}.
In this case P is a prime (non-maximal) ideal and it is of the form
Bq = qQ[X] ∩ Int(Z)
for some q ∈ Q[X] irreducible. By Gauss Lemma we may suppose that q ∈ Z[X] is
irreducible and primitive.
Unitary prime ideals: P ∩ Z = pZ.
In this case P is maximal and is of the form
Mp,α = {f ∈ Int(Z) | f (α) ∈ pZp }
for some p prime in Z and some α ∈ Zp , the ring of p-adic integers. We have Mp,α = Mq,β
if and only if (p, α) = (q, β). So if we fix the prime p, the elements of Zp are in bijection
with the unitary prime ideals of Int(Z) above the prime p. Moreover, Mp,α is height 1 if
and only if α is transcendental over Q. If α is algebraic over Q and q(X) is its minimal
polynomial then Mp,α ⊃ Bq . We have Bq ⊂ Mp,α if and only if q(α) = 0. Every prime
ideal of Int(Z) is not finitely generated.
For a detailed study of Spec(Int(Z)) see [4].
If we denote by d(f, Zp ) the fixed divisor of f ∈ Int(Z) viewed as a polynomial over
the ring of p-adic integers Zp (that is, d(f, Zp ) is the ideal (f (α) | α ∈ Zp )), Gunji and
McQuillan in [8] observed that
\
d(f, Zp )
d(f ) =
p

where the intersection is taken over the set of primes in Z. Moreover, d(f, Zp ) = d(f )Zp ⊂
Zp . Remember that given an ideal I ⊂ Z and a prime p we have IZp = Zp if and only
if I 6⊂ (p), so that in the previous equation we have a finite intersection. Since Zp is a
DVR we have d(f, Zp ) = pn Zp , for some integer n (which of course depends on p), so that
the exact power of p which divides f (Z) is the same as the power of p dividing f (Zp ).
Without loss of generality, we can restrict our attention to the p-part of the fixed divisor
of a polynomial f ∈ Z[X]. We begin our research by finding those polynomials in Z[X]
whose fixed divisor is divisible by a fixed prime p, namely the ideal pInt(Z) ∩ Z[X].
Lemma 2.1. Let p be a prime and α ∈ Zp . Then Mp,α ∩ Z[X] = (p, X − a), where
a ∈ Z is such that α ≡ a (mod p). Moreover, if β ∈ Zp is another p-adic integer, we have
Mp,α ∩ Z[X] = Mp,β ∩ Z[X] if and only if α ≡ β (mod p).
Proof : Let a be an integer as in the statement of the lemma; it exists since Z is dense in
Zp for the p-adic topology. We immediately see that p and X − a are in Mp,α . Then the
6

conclusion follows since (p, X −a) is a maximal ideal of Z[X] and Mp,α ∩Z[X] is not equal to
the whole ring Z[X]. The second statement follows from the fact that (p, X −a) = (p, X −b)
if and only if a ≡ b (mod p). 
We have just seen that the contraction of Mp,α to Z[X] depends only on the residue
class modulo p of α. So, if p is a fixed prime, the contractions of Mp,α to Z[X] as α ranges
through Zp are made up of p distinct maximal ideals, namely
{Mp,α ∩ Z[X] | α ∈ Zp } = {(p, X − j) | j ∈ {0, . . . , p − 1}}.
Conversely, the set of prime ideals of Int(Z) above a fixed maximal ideal of the form
(p, X − j) is { Mp,α | α ∈ Zp , α ≡ j (mod p)}, since Bq are non-unitary ideals and p is the
only prime integer in Mp,α .
For a prime p and an integer j ∈ {0, . . . , p − 1}, we set:
Mp,j = Mj + (p, X − j).
Whenever the notation Mp,j is used, it will be implicit that j ∈ {0, . . . , p − 1}.
The next lemma computes the intersection of the ideals Mp,j , for a fixed prime p, by
finding an ideal whose primary decomposition is given by this intersection (and its primary
components are precisely the p ideals Mp,j ). From now on we will omit the index p.
Lemma 2.2. Let p ∈ Z be a prime. Then we have

\
Y
Mj = p,
j=0,...,p−1



(X − j) .

j=0,...,p−1

Proof : Let J be the ideal on the right-hand side. If P is a prime minimal over J, then
we see immediately that P = Mj for some j ∈ {0, . . . , p − 1}, since Mj is a maximal
ideal. Conversely, every such a maximal ideal contains J and is minimal over it. Then the
minimal primary decomposition of J is of the form
\
J=
Qj
j=0,...,p−1

where Qj is an Mj -primary ideal. Since X − i 6∈ Mj for all i ∈ {0, . . . , p − 1} \ {j}, we
have (X − j) ∈ Qj , so indeed Qj = (p, X − j) for each j = 0, . . . , p − 1. 
The next proposition characterizes the principal unitary ideals in Int(Z) generated by
a prime p.

7

Proposition 2.1. Let p ∈ Z be a prime. Then the principal unitary ideal pInt(Z) is equal
to
\
Mp,α .
pInt(Z) =
α∈Zp

Proof : We trivially have that pInt(Z) is contained in the above intersection, since p
is in every ideal of the form Mp,α . On the other hand, this intersection is equal to
{f ∈ Int(Z)|f (Zp ) ⊂ pZp }. If f (X) is in this intersection, since f (X) is integer-valued
and pZp ∩ Z = pZ, we have f (Z) ⊂ pZ. This is equivalent to saying that f (X)/p ∈ Int(Z),
that is, f ∈ pInt(Z). 
In particular, the previous proposition implies that Int(Z) does not have the finite
character property (we recall that a ring has this property if every non-zero element is
contained in a finite number of maximal ideals).
From the above results we get the following theorem, which characterizes the ideal of
polynomials with integer coefficients whose fixed divisor is divisible by a prime p, that is,
the ideal pInt(Z) ∩ Z[X].
Theorem 2.1. Let p be a prime. Then


pInt(Z) ∩ Z[X] = p,

Y



(X − j) .

j=0,...,p−1

Notice that Lemma 2.2 gives the primary decomposition of pInt(Z) ∩ Z[X], so Mj for
j = 0, . . . , p − 1 are exactly the prime ideals belonging to it. As a consequence of this
theorem we get the following well-known result: if f ∈ Z[X] is primitive and p is a prime
such that Q
d(f ) ⊆ p then p ≤ deg(f ). This immediately follows from the theorem, since the
degree of j=0,...,p−1 (X − j) is p.
We remark that by Fermat’s little theorem the ideal on the right-hand side of the
statement of Theorem 2.1 is equal to (p, X p − X). This amounts to saying that the two
polynomials X · . . . · (X − (p − 1)) and X p − X induce the same polynomial function on
Z/pZ.

3. Contraction of primary ideals
We remark that Proposition 2.1 also follows from a general result contained in [11]:
every unitary ideal in Int(Z) is an intersection of powers of unitary prime ideals (namely
the maximal ideals Mp,α ). In particular, every Mp,α -primary ideal is a power of Mp,α itself,
8

since Mp,α is maximal. From the same result we also have the following characterization
of the powers of Mp,α , for any positive integer n:
Mnp,α = {f ∈ Int(Z) | f (α) ∈ pn Zp }.
This fact implies the following expression for the principal unitary ideal generated by pn :
\
pn Int(Z) =
Mnp,α .
(1)
α∈Zp

We remark again that the previous ideal is made up of those integer-valued polynomials
whose extended fixed divisor is contained in pn Int(Z). Similarly to the previous case n = 1
(see Theorem 2.1) we want to find the contraction of this ideal to Z[X], in order to find
the polynomials in Z[X] whose fixed divisor is divisible by pn . We set:
Ipn + pn Int(Z) ∩ Z[X].
(2)
T
Notice that by (1) we have Ipn = α∈Zp (Mnp,α ∩ Z[X]).
Like before, we begin by finding the contraction to Z[X] of Mnp,α , for each α ∈ Zp . The
next lemma is a generalization of Lemma 2.1.
Lemma 3.1. Let p be a prime, n a positive integer and α ∈ Zp . Then Mnp,α ∩ Z[X] =
(pn , X − a), where a ∈ Z is such that α ≡ a (mod pn ). The ideal Mnp,α ∩ Z[X] is Mp,j primary, where j ≡ α (mod p). Moreover, if β ∈ Zp is another p-adic integer, we have
Mnp,α ∩ Z[X] = Mnp,β ∩ Z[X] if and only if α ≡ β (mod pn ).
Proof : The case n = 1 has been done in Lemma 2.1. For the general case, let a ∈ Z
be such that a ≡ α (mod pn ) (again, such an integer exists since Z is dense in Zp for the
p-adic topology). We have (pn , X − a) ⊂ Mnp,α ∩ Z[X] (notice that if n > 1 then (pn , X − a)
is not a prime ideal). To prove the other inclusion let f ∈ Mnp,α ∩ Z[X]. By the Euclidean
algorithm in Z[X] (the leading coefficient of X − a is a unit) we have
f (X) = q(X)(X − a) + f (a)
Since f (α) ∈ pn Zp and pn |a − α we have pn |f (a). Hence, f ∈ (pn , X − a) as we wanted.
Since Mnp,α is an Mp,α -primary ideal in Int(Z) and the contraction of a primary ideal is
a primary ideal, by Lemma 2.1 we get the second statement. Finally, like in the proof
of Lemma 2.1, we immediately see that (pn , X − a) = (pn , X − b) if and only if a ≡ b
(mod pn ), which gives the last statement of the lemma. 

Remark 1. It is worth to write down the fact that we used in the above proof: given a
polynomial f ∈ Z[X], we have
f ∈ (pn , X − a) ⇐⇒ f (a) ≡ 0 (mod pn )
9

(3)

Remark 2. If p is a fixed prime and n is a positive integer, the Lemma 3.1 implies
Ip,n + {Mnp,α ∩ Z[X] | α ∈ Zp } = {(pn , X − i) | i = 0, . . . , pn − 1}.
Let us consider an ideal I = Mnp,α ∩ Z[X] = (pn , X − i) in Ip,n , with i ∈ Z, i ≡ α (mod pn ).
It is quite easy to see that I contains (Mp,α ∩ Z[X])n = Mnp,j = (p, X − j)n , where
j ∈ {0, . . . , p − 1}, j ≡ α (mod p) (notice that j ≡ i (mod p)). If n > 1 this containment is
strict, since X − i 6∈ (p, X − j)n . We can group the ideals of Ip,n according to their radical:
there are p radicals of these pn ideals, namely the maximal ideals Mp,j , j = 0, . . . , p − 1.
This amounts to making a partition of the residue classes modulo pn into p different sets
of elements congruent to j modulo p, for j = 0, . . . , p − 1; each of these sets has cardinality
pn−1 . Correspondingly we have:
[
Ip,n =
Ip,n,j
j=0,...,p−1

where Ip,n,j + {(pn , X − i) | i = 0, . . . , pn − 1, i ≡ j (mod p)}, for j = 0, . . . , p − 1. Every
ideal in Ip,n,j is Mp,j -primary and it contains the n-th power of its radical, namely Mnp,j .
Now we want to compute the intersection of the ideals in Ip,n , which is equal to the
ideal Ipn in Z[X] (see (1) and (2)). We can express this intersection as an intersection of
Mp,j -primary ideals as we have said above, in the following way (in the first equality we
make use of equation (1) and Lemma 3.1):
\
\
Qp,n,j
(4)
(pn , X − i) =
Ip n =
i=0,...,pn −1

where
Qp,n,j +

j=0,...,p−1

\

(pn , X − i)

i≡j(mod p)

(notice that the intersection is taken over the set {i ∈ {0, . . . , pn − 1} | i ≡ j (mod p)}).
The ideal Qp,n,j is an Mp,j -primary ideal, for j = 0, . . . , p − 1, since the intersection of
M -primary ideals is an M -primary ideal. We will omit the index p in Qp,n,j and in Mp,j if
that will be clear from the context. The Mp,j -primary ideal Qn,j is just the intersection of
the ideals in Ip,n,j , according to the partition we made. It is equal to the set of polynomials
in Z[X] which modulo pn are zero at the residue classes congruent to j modulo p (see (3) of
Remark 1). We remark that (4) is the minimal primary decomposition of Ipn . Notice that
there are no embedded components in this primary decomposition, since the prime ideals
belonging to it (the minimal primes containing Ipn ) are {Mj | j = 0, . . . , p − 1}, which are
maximal ideals.
We recall that if I and J are two coprime ideals in a ring R, that is I + J = R, then
IJ = I ∩ J (in general only the inclusion IJ ⊂ I ∩ J holds). The condition for two ideals I
and J to be coprime amounts to saying that I and J are not contained in a same maximal
10

ideal M , that is, I + J is not contained in any maximal ideal M . If M1 and M2 are
two distinct maximal ideals then they are coprime, and the same holds for any of their
respective powers. If R is Noetherian, then every primary ideal Q contains a power of its
radical and moreover if the radical of Q is maximal then also the converse holds (see [14]).
So if Qi is an Mi -primary ideal for i = 1, 2 and M1 , M2 are distinct maximal ideals, then
Q1 and Q2 are coprime.
Since {Mj }j=0,...,p−1 are p distinct maximal ideals, for what we have just said above
we have
\
Y
Qn,j =
Qn,j .
j=0,...,p−1

j=0,...,p−1

Now we want to describe the Mj -primary ideals Qn,j , for j = 0, . . . , p − 1. The next lemma
gives a relation of containment between these ideals and the n-th powers of their radicals.
Lemma 3.2. Let p be a fixed prime and n a positive integer. For each j = 0, . . . , p − 1,
we have
Qn,j ⊇ Mnj .
Proof : The statement follows from Remark 2. 
As a consequence of this lemma, we get the following result:
Corollary 3.1. Let p be a fixed prime and n a positive integer. Then we have:

n
Y
Ipn ⊇ p,
(X − j) .
j=0,...,p−1

Proof : By (4) and Lemma 3.2 we have
Y
Ip n =
Qn,j ⊇
j=0,...,p−1

Y

Mnj

j=0,...,p−1

where the last containment follows from Lemma 3.2. Finally, by Lemma 2.2, the product
of the ideals Mnj is equal to

n
Y
Y
Mnj = p,
(X − j)
j=0,...,p−1

j=0,...,p−1

Notice that the product of the Mj ’s is actually equal to their intersection, since they are
maximal coprime ideals. 
The last formula of the previous proof gives the primary decomposition of the ideal
Q
(p, j=0,...,p−1 (X − j))n .
11

Remark 3. In general, for a fixed j ∈ {0, . . . , p−1}, the reverse containment of Lemma 3.2
does not hold, that is, the n-th power of Mj can be strictly contained in the Mj -primary
ideal Qn,j . For example (again, we use (3) to prove the containment):


\
X(X − 2) ∈ 
(23 , X − 2k) \ (2, X)3
k=0,...,3

Because of that, in general we do not have an equality in Corollary 3.1. For example, let
p = 2 and n = 3. We have
X(X − 1)(X − 2)(X − 3) ∈ I23 \ (2, X(X − 1))3 .
It is also false that
\



(pn , X − i) = pn ,

i=0,...,pn −1

See for example: p = 2, n = 2: 2X(X − 1) ∈

T

Y



(X − i) .

i=0,...,pn −1

 Q

(4,
X
−
i)
\
4,
(X
−
i)
.
i=0,...,3
i=0,...,3

We want to study under which conditions the ideal Qn,j is equal to Mnj . Our aim is
to find a set of generators for Qn,j . For f ∈ Qn,j , we have f ∈ (pn , X − i) for each i ≡ j
(mod p), i ∈ {0, . . . , pn − 1}. By (3) that means pn |f (i) for each such an i. Equivalently,
such a polynomial has the property that modulo pn it is zero at the pn−1 residue classes
of Z/pn Z which are congruent to j modulo p.
Without loss of generality, T
we proceed by considering the case j = 0. We set M =
M0 = (p, X) and Qn = Qn,0 = i≡0 (mod p) (pn , X − i). Let f ∈ Qn , of degree m. We have
f (X) = q1 (X)X + f (0)

(5)

where q1 ∈ Z[X] has degree equal to m − 1. Since f ∈ (pn , X) we have pn |f (0).
Since f ∈ (pn , X − p), we have pn |f (p) = q1 (p)p + f (0), so pn−1 |q1 (p). By the Euclidean
algorithm,
q1 (X) = q2 (X)(X − p) + q1 (p)
(6)
for some polynomial q2 ∈ Z[X] of degree m − 2. So
f (X) = q2 (X)(X − p)X + q1 (p)X + f (0).
We set R1 (X) = q1 (p)X + f (0). Then R1 ∈ Mn , since pn−1 |q1 (p) and pn |f (0). Since
f ∈ (pn , X − 2p), we have pn |f (2p) = q2 (2p)2p2 + q1 (p)2p + f (0). If p > 2 then pn−2 |q2 (2p),

12

because pn |q1 (p)2p + f (0). If p = 2 then we can just say pn−3 |q2 (2p). By the Euclidean
algorithm again, we have
q2 (X) = q3 (X)(X − 2p) + q2 (2p)
for some q3 ∈ Z[X]. So we have
f (X) = q3 (X)(X − 2p)(X − p)X + q2 (2p)(X − p)X + q1 (p)X + f (0).
Like before, if we set R2 (X) = q2 (2p)(X − p)X + q1 (p)X + f (0), we have R2 ∈ Mn if p > 2,
or R2 ∈ Qn if p = 2.
We define now the following family of polynomials:
Definition 3.1. For each k ∈ N, k ≥ 1, we set
Gp,0,k (X) = Gk (X) +

Y

(X − hp).

h=0,...,k−1

We also set G0 (X) + 1.
From now on, we will omit the index p in the above notation.
Notice that the polynomials Gk (X), whose degree for each k is k, enjoy these properties:
i) For every t ∈ Z, Gk (tp) = pk t(t − 1) . . . (t − (k − 1)). Hence, the highest power of p
which divides all the integers in the set {Gk (tp) | t ∈ Z} is pk+vp (k!) . It is easy to see
that k + vp (k!) = vp ((pk)!).
ii) Gk (X) = (X − kp)Gk−1 (X).
iii) since for every integer h, X − hp ∈ M, we have Gk (X) ∈ Mk . We remark that k is
the maximal integer with this property, since deg(Gk ) = k and Gk (X) is primitive
(since monic).
Recall that, by Lemma 3.2, for every integer n we have Qn ⊇ Mn . By property iii)
above Gk ∈ Mn if and only if n ≤ k. By property i) we have Gk ∈ Qn if and only if
k + vp (k!) ≥ n. From these remarks, it is very easy to deduce that, in the case p ≥ n, if
Gk ∈ Qn then Gk ∈ Mn . In fact, if that is not the case, it follows from above that k < n.
Since n ≤ p we get k + vp (k!) = k. Since Gk ∈ Qn , we have n ≤ k, contradiction.
The next lemma gives a sort of division algorithm between an element of Qn and the
polynomials {Gk (X)}k∈N . In particular, we will deduce that Qn = Mn , if p ≥ n.
Lemma 3.3. Let p be a prime and n a positive integer. Let f ∈ Qp,n,0 = Qn be of degree
m. Then for each 1 ≤ k ≤ m there exists qk ∈ Z[X] of degree m − k such that
f (X) = qk (X)Gk (X) + Rk−1 (X)
P

where Rk−1 (X) + h=1,...,k−1 qh (hp)Gh (X) for k ≥ 2 and R0 (X) + f (0). We also have
qk (X) = qk+1 (X)(X − kp) + qk (kp) for k = 1, . . . , m − 1. Moreover, for each such a k the
following hold:
13

i) pn−vp ((pk)!) |qk (kp), if vp ((pk)!) < n.
ii) qk (kp)Gk (X) ∈ Qn and if k < p then qk (kp)Gk (X) ∈ Mn .
iii) If m ≤ p then Rk−1 ∈ Mn for k = 1, . . . , m.
If m > p then Rk−1 ∈ Mn for k = 1, . . . , p and Rk−1 ∈ Qn for k = p + 1, . . . , m.

Proof : We proceed by induction on k. The case k = 1 follows from (5), and by (6) we
have the last statement regarding the relation between q1 (X) and q2 (X). Suppose now the
statement is true for k − 1, so that
f (X) = qk−1 (X)Gk−1 (X) + Rk−2 (X)
with Rk−2 (X) +

P

h=1,...,k−2 qh (hp)Gh (X)

and

- pn−vp ((p(k−1))!) |qk−1 ((k − 1)p), if vp ((p(k − 1)!)) < n,
- qk−1 ((k − 1)p)Gk−1 (X) belongs to Qn and if k − 1 < p it belongs to Mn ,
- Rk−2 ∈ Qn and if k − 2 < p then Rk−2 ∈ Mn .
We divide qk−1 (X) by (X − (k − 1)p) and we get
qk−1 (X) = qk (X)(X − (k − 1)p) + qk−1 ((k − 1)p)
for some polynomial qk ∈ Z[X] of degree m − k. We substitute this expression of qk−1 (X)
in the equation of f (X) at the step k − 1 and we get:
f (X) = qk (X)(X − (k − 1)p)Gk−1 (X) + Rk−1 (X),

(7)

where Rk−1 (X) + qk−1 ((k − 1)p)Gk−1 (X) + Rk−2 (X). This is the expression of f (X) at
step k, since (X − (k − 1)p)Gk−1 (X) is equal to Gk (X). By the inductive assumption,
Rk−1 ∈ Qn and if k − 1 < p we also have Rk−1 ∈ Mn . We still have to verify i) and ii).
We evaluate the expression (7) in X = kp and we get
f (kp) = qk (kp)Gk (kp) + Rk−1 (kp) = qk (kp)pk k! + Rk−1 (kp).
Since pn divides both f (kp) and Rk−1 (kp) (by definition of Qn ), if vp ((pk)!) < n we
get that qk (kp) is divisible by pn−vp ((pk)!) , which is statement i) at the step k. Notice
that qk (kp)Gk (X) is zero modulo pn on every integer congruent to zero modulo p; hence,
qk (kp)Gk (X) ∈ Qn . Moreover, k < p ⇔ vp (k!) = 0, so in that case qk (kp)Gk (X) ∈ Mn .
So ii) follows. 

14

Notice that by formula (3) of Remark 1, under the assumptions of Lemma 3.3 we have
for each k ∈ {1, . . . , p − 1} that
qk ∈ (pn−k , X − kp)
(see i) of Lemma 3.3: in this case vp ((pk)!) = k). If k = m = deg(f ) then qk ∈ Z. Hence,
we get the following expression for a polynomial f ∈ Qn in the case p ≥ n > m (this
assumption is not restrictive, since X n ∈ Qn ):
X
f (X) = qm Gm (X) + Rm−1 (X) = qm Gm (X) +
qk (kp)Gk (X)
(8)
k=1,...,m−1

where qm ∈ Z is divisible by pn−m and Rm−1 (X) is in Mn .
The next proposition determines the primary components Qn,j of Ipn of (4) in the case
p ≥ n. It shows that in this case the containment of Lemma 3.2 is indeed an equality.
Proposition 3.1. Let p ∈ Z be a prime and n a positive integer such that p ≥ n. Then
for each j = 0, . . . , p − 1 we have
Qn,j = Mnj .
Proof : It is sufficient to prove the statement for j = 0: for the other cases we consider
the Z[X]-automorphisms πj (X) = X − j, for j = 1, . . . , p − 1, which permute the ideals
Qn,j and Mj . Let Qn = Qn,0 and M = M0 .
The inclusion (⊇) follows from Lemma 3.2. For the other inclusion (⊆), let f (X) be
in Qn . We can assume that the degree m of f (X) is less than n, since X n is the smallest
monic monomial in Qn . By equation (8) above, f (X) is in Mn , since pn−m divides qm ,
Gm ∈ Mm and Rm−1 ∈ Mn by Lemma 3.3 (notice that m − 1 < p). 
Remark 4. In the case p ≥ n, Lemma 3.3 implies that Qn is generated by {pn−m Gm (X)}0≤m≤n :
it is easy to verify that these polynomials are in Qn (using (3) again) and (8) implies that
every polynomial f ∈ Qn is a Z-linear combination of {pn−m Gm (X)}0≤m≤n , since qm (mp)
is divisible by pn−m , for each of the relevant m.
The following theorem gives a description of the ideal Ipn in the case p ≥ n. In this
case the containment of the Corollary 3.1 becomes an equality.
Theorem 3.1. Let p ∈ Z be a prime and n a positive integer such that p ≥ n. Then the
ideal in Z[X] of those polynomials whose fixed divisor is divisible by pn is equal to

n
Y
Ipn = p,
(X − i) .
i=0,...,p−1

15

Proof : By Proposition 3.1, for each j = 0, . . . , p − 1 the ideal Qn,j is equal to Mnj . So,
by the last formula of the proof of Corollary 3.1, we get the statement. 
As a consequence, we have the following remark. Let p be a prime and n a positive
n such that the content of f (X) is not divisible
integer less than or equal to p. Let f ∈ IpQ
by p. Then deg(f ) ≥ np, since np = deg( i=0,...,p−1 (X − i)n ). Another well-known result
in this context is the following: if we fix the degree
P d of such a polynomial f , then the
maximum n such that f ∈ Ipn is bounded by n ≤ k≥1 [d/pk ] = vp (d!).
If we drop the assumption p ≥ n, the ideal Qn,j may strictly contain Mnj , as we observed
in Remark 3. The next proposition shows that this is always the case, if p < n. This result
follows from Lemma 3.3 as Proposition 3.1 does, and it covers the remaining
case p < n. It
T
is stated for the case j = 0. Remember that M = (p, X) and Qn = i≡0 (mod p) (pn , X − i).
Proposition 3.2. Let p ∈ Z be a prime and n a positive integer such that p < n. Then
we have
Qn = Mn + (qn,p Gp (X), . . . , qn,n−1 Gn−1 (X))
where, for each k = p, . . . , n − 1, qn,k is an integer defined as follows:
 n−v ((pk)!)
p p
, if vp ((pk)!) < n
qn,k +
1 , otherwise
In particular, Mn is strictly contained in Qn .
Proof : We begin by proving the containment (⊇). Lemma 3.2 gives Mn ⊆ Qn . We have
to show that the polynomials qn,k Gk (X), for k ∈ {p, . . . , n − 1}, lie in Qn . This follows
from property i) of the polynomials Gk (X) and the definition of qn,m.
Now we prove the other containment (⊆). Let f ∈ Qn be of degree m. If m < p then
f ∈ Mn (see Lemma 3.3 and in particular (8)). So we suppose p ≤ m. By Lemma 3.3 we
have
X
f (X) =
qk (kp)Gk (X) + Rp−1 (X)
(9)
k=p,...,m

where Rp−1 (X) = k=1,...,p−1 qk (kp)Gk (X) ∈ Mn and qm ∈ Z, so that qm (mp) = qn,m .
Then, since qn,k = pn−vp ((pk)!) |qk (kp) if vp ((pk)!) < n, it follows that the first sum on the
right-hand side of the previous equation belongs to the ideal (qn,p Gp (X), . . . , qn,n−1 Gn−1 (X)).
For the last sentence of the proposition, we remark that the polynomials {qn,k Gk (X)}k=p,...,n−1
are not contained in Mn : in fact, for each k ∈ {p, . . . , n − 1}, by property iii) of the polynomials Gk (X) we have that the minimal integer N such that qn,k Gk (X) is contained in
MN is n − vp (k!) if vp ((pk)!) = k + vp (k!) < n and it is k otherwise. In both cases it is
strictly less than n (since vp (k!) ≥ 1, if k ≥ p). 
P

16

Remark 5. The following remark allows us to obtain another set of generators for Qn .
We set
m = m(n, p) + min{m ∈ N | vp ((pm)!) ≥ n}
(10)
Remember that vp ((pm)!) = m+vp (m!). If p ≥ n then m = n and if p < n then p ≤ m < n.
Suppose now p < n. Then for each m ∈ {m, . . . , n} we have vp ((pm)!) ≥ n, since the
function e(m) = m + vp (m!) is increasing. So for each such m we have qn,m = 1, hence
Gm ∈ (Gm (X)). So we have the equalities:
Qn = Mn + (qn,m Gm (X) | m = p, . . . , m)

(11)

= (qn,m Gm (X) | m = 0, . . . , m)
where qn,m = pn−m , for m = 0, . . . , p − 1, and for m = p, . . . , m is defined as in the
statement of Proposition 3.2. The containment (⊇) is just an easy verification using the
properties of the polynomials Gm (X); the other containment follows by (9).
We can now group together Proposition 3.1 and 3.2 into the following one:
Proposition 3.3. Let p ∈ Z be a prime and n a positive integer. Then we have
Qn = (qn,0 G0 (X), . . . , qn,m Gm (X))
where m = min{m ∈ N | vp ((pm)!) ≥ n} and for each m = 0, . . . , m, qn,m is an integer
defined as follows:
 n−v ((pm)!)
p p
,m < m
qn,m +
1 ,m = m
It is clear what the primary ideals Qj , for j = 1, . . . , p − 1, look like:
\
Qn,j =
(pn , X − i) = Mnj + (qn,p Gp (X − j), . . . , qn,m Gm (X − j))
i≡j(mod p)

= (qn,0 G0 (X − j), . . . , qn,m Gm (X − j))
In fact, for each j = 1, . . . , p − 1, it is sufficient to consider the automorphisms of Z[X]
given by πj (X) = X − j. It is straightforward to check that πj (Ipn ) = Ipn . Moreover,
π(Qn,0 ) = Qn,j and π(M0 ) = Mj for each such a j, so that πj permutes the primary
components of the ideal Ipn .
The ideal Ipn = pn Int(Z) ∩ Z[X] was studied in [2] in a slightly different context, as the
kernel of the natural map ϕn : Z[X] → Φn , where the latter is the set of functions from
Z/pn Z to itself. In that article a recursive formula is given for a set of generators of this
ideal. Our approach gives a new point of view to describe this ideal.
For other works about the ideal Ipn in a slightly different context, see [9], [10], [13].
This ideal is important in the study of the problem of the polynomial representation of a
function from Z/pn Z to itself.
17

4. Case Ipp+1
As a corollary we give an explicit expression for the ideal Ipn in the case n = p + 1. By
Proposition 3.2 the primary components of Ipp+1 are
Qp+1,j = Mp+1
+ (Gp (X − j))
j

(12)

for j = 0, . . . , p − 1.
Corollary 4.1.

Ipp+1 = p,
where H(X) =

Q

i=0,...,p2 −1 (X

Y

p+1
(X − i)
+ (H(X))

i=0,...,p−1

− i).

We want to stress that the polynomial H(X) is not contained in the first ideal of the
right-hand side of the statement. In [2] a similar result is stated with another polynomial
H2 (X) instead of our H(X). Indeed
the two polynomials, as already remarked in [2], are
Q
congruent modulo the ideal (p, i=0,...,p−1 (X − i))p+1 .
Proof : Like before, we set Qp,p+1,j = Qp+1,j . The containment
(⊇) follows from corollary
Q
3.1 and because the polynomial H(X) is equal to j=0,...,p−1 Gp (X − j) and for each
j = 0, . . . , p − 1 the polynomial Gp (X − j) is in Qp+1,j by Proposition 3.2. Since Qp+1,j , for
j = 0, . . . , p − 1, are exactly the primary components of Ipp+1 (see
T (4)), we get the claim.
Now we prove the other containment (⊆). Let f ∈ Ipp+1 = j=0,...,p−1 Qp+1,j . By (12)
we have:
)
f (X) ≡ Cp,j (X)Gp (X − j) (mod Mp+1
j

for some Cp,j ∈ Z[X], for j = 0, . . . , p − 1.
Since the ideals {Mp+1
= (p, X − j)p+1 | j = 0, . . . , p − 1} are pairwise coprime (bej
cause they are powers of distinct maximal ideals, respectively), by the Chinese Remainder
Theorem we have the following isomorphism:


p−1
Y p+1
p+1
p+1
Mj  ∼
(13)
Z[X]/ 
= Z[X]/M0 × . . . × Z[X]/Mp−1
j=0

We need now the following lemma, which tells us what is the residue of the polynomial
H(X) modulo each ideal Mp+1
:
j
Lemma 4.1. Let p be a prime and let H(X) =
k = 0, . . . , p − 1 we have

Q

j=0,...,p−1 Gp (X

H(X) ≡ −Gp (X − k) (mod Mp+1
k )
18

− j). Then for each

Proof : Let k ∈ {0, . . . , p − 1} and set Ik = {0, . . . , p − 1} \ {k}. For each j ∈ Ik we have
Gp (k − j) ≡ (k − j)p (mod p). We have
Y
Gp (X − j)]
H(X) + Gp (X − k) = Gp (X − k)[1 +
j∈Ik

Q
Since Gp (X − k) ∈ Mpk we have just to prove that Tk (X) = 1 + j∈Ik Gp (X − j) ∈ Mk .
By formula (3) in Remark 1 it is sufficient to prove that Tk (k) is divisible by p. We have
Y
Tk (k) ≡ 1 +
(k − j)p (mod p)
j∈Ik

≡1+(

Y

s)p

s=1,...,p−1
p

≡ 1 + (p − 1)!

(mod p)

(mod p)

p

≡ (1 + (p − 1)!)

(mod p)

which is congruent to zero by Wilson’s theorem. 
We finish now the proof of the corollary.
By the Chinese Remainder Theorem, there exists a polynomial P ∈ Z[X] such that
), for each j = 0, . . . , p − 1. Then by the previous lemma
P (X) ≡ −Cp,j (X) (mod Mp+1
j
P (X)H(X) ≡ Cp,j (X)Gp (X −j) (mod Mp+1
) and so again by the isomorphism (13) above
j
we have
Y
f (X) ≡ P (X)H(X) (mod
Mp+1
)
j
j=0,...,p−1

so we are done since
lary 3.1). 

Q

p+1
j=0,...,p−1 Mj

= (p,

Q

i=0,...,p−1 (X

− i))p+1 (see the proof of Corol-

Acknowledgments
I want to thank the referee for his/her suggestions. I wish also to thank Sophie Frisch
for the useful references. I deeply thank Elvira Carlotti for her continuous and strenuous
help. The author was supported by the Austrian Science Foundation (FWF), Project
Number P23245-N18.

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20

