Free hyperplane arrangements over arbitrary fields
Elisa Palezzato Michele Torielli





In this paper, we study the class of free hyperplane arrangements. Specifically, we investigate the relations between freeness over a field of finite characteristic
and freeness over ℚ.






Elisa Palezzato, Department of Mathematics, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo 060-0810, Japan.
palezzato@math.sci.hokudai.ac.jp

Michele Torielli, Department of Mathematics, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo 060-0810, Japan.
torielli@math.sci.hokudai.ac.jp




Free hyperplane arrangements over arbitrary fields
    Michele Torielli
    December 30, 2023
==================================================







§ INTRODUCTION

Let V be a vector space of dimension l over a field K. Fix a system of coordinate (x_1,…, x_l) of V^∗. 
We denote by S = S(V^∗) = K[x_1,…, x_l] the symmetric algebra. 
A hyperplane arrangement = {H_1, …, H_n} is a finite collection of hyperplanes in V.

Freeness of an arrangement is a key notion which connects arrangement theory with algebraic geometry and combinatorics. 
The study of free arrangements was started by Saito <cit.> and a remarkable factorization theorem was proved by Terao <cit.>. 
This theorem asserts that the characteristic polynomial of a free arrangement completely factors into linear polynomials over the integers. 
This imposes a necessary condition on the structure of the intersection lattice for an arrangement to be free. The Terao conjecture is the converse problem,
i.e. to understand if the structure of the intersection lattice characterize freeness of arrangements. A lot of work has done to solve this conjecture especially in the case
of characteristic 0 (see for example <cit.>, <cit.>, <cit.>, <cit.> and <cit.>). 
However, also the case of finite characteristic has been analyzed (see for example <cit.> and <cit.>).
Despite this effort, this question is still open.

The purpose of this paper is to study the connections between freeness over a field of characteristic zero and over a finite field, and to describe in which cases
 the two situations are related and how. 
 







§ PRELIMINARES ON HYPERPLANE ARRANGEMENTS


In this section, we recall the terminology, the basic notations and some fundamental results related to hyperplane arrangements.


Let K be a field. A finite set of affine hyperplanes ={H_1, …, H_n} in K^l 
is called a hyperplane arrangement. For each hyperplane H_i we fix a defining equation α_i∈ S= K[x_1,…, x_l] such that H_i = α_i^-1(0), 
and let Q()=∏_i=1^nα_i. An arrangement  is called central if each H_i contains the origin of K^l. 
In this case, the defining equation α_i∈ S is linear homogeneous, and hence Q() is homogeneous of degree n. 

Let L()={⋂_H∈ℬH |ℬ⊆} be the lattice of intersection of , 
ordered by reverse inclusion, i.e. X≤ Y if and only if Y⊆ X, for X,Y∈ L().


Define a rank function on L() by (X)=(X). 
L() plays a fundamental role in the study of hyperplane arrangements, in fact it determines the combinatorics of the arrangement.



Let μ L()→ℤ be the Möbius function of L() defined by

    μ(X)=
    
          1    for  X=K^l,
    
          -∑_Y<Xμ(Y)    if  X>K^l.

The characteristic polynomial of  is defined by 
    χ(,t) = ∑_X∈ L()μ(X)t^(X).















The importance of the characteristic polynomial in combinatorics is justified by the following result 
from <cit.>, <cit.> and <cit.>.

 
We have that


  * If  is an arrangement in 𝔽_p^l, then |𝔽^l_p∖⋃_H∈H|=χ(, p).

  * If  is an arrangement in ^l, then the topological i-th Betti number of the complement is b_i(^l  ∖⋃_H∈H)=b_i().

  * If  is an arrangement in ^l, then |χ(,-1)| is the number of chambers and |χ(, 1)| is the number of bounded chambers.






§ FREE HYPERPLANE ARRANGEMENTS

We recall the basic notions and properties of free hyperplane arrangements.



We denote by _K^l ={∑_i=1^l f_i∂_x_i | f_i∈ S} the S-module of polynomial vector fields on K^l (or S-derivations). 
Let δ =  ∑_i=1^l f_i∂_x_i∈_K^l. Then δ is  said to be homogeneous of polynomial degree d if f_1, …, f_l are homogeneous polynomials of degree d in S. 
In this case, we write (δ) = d.

 
Let  be a central arrangement in K^l. Define the module of vector fields logarithmic tangent to  (or logarithmic vector fields) by

    D() = {δ∈_K^l | δ(α_i) ∈ (α_i) S, ∀ i}.



The module D() is obviously a graded S-module and we have that D()= {δ∈_K^l | δ(Q()) ∈ (Q()) S}. 
In particular, since the arrangement  is central, then the Euler vector field δ_E=∑_i=1^lx_i∂_x_i belongs to D(), in fact
δ_E(Q())=nQ(). 
In this case, if the characteristic of K does not divide n, we can write D()≅ S·δ_E⊕ D_0(), where 
    D_0()={δ∈_K^l | δ(Q())=0}.


 
A central arrangement  in K^l is said to be free with exponents (e_1,…,e_l) 
if and only if D() is a free S-module and there exists a basis δ_1,…,δ_l ∈ D() 
such that (δ_i) = e_i, or equivalently D()≅⊕_i=1^lS(-e_i).


In general the exponents of an arrangement depend on the characteristic of K. In fact, we have the following examples.
[<cit.>, Example 4.35] Consider the arrangement  in K^3 with defining equation
Q()=xyz(x-y)(x+z)(y+z)(x+y+z). Then  is free for any K, but its exponents depend on the characteristic of K.

If (K)2, then  is free with exponents (1,3,3), in fact we can take as basis of D() the following vector fields δ_E, 
δ_2=x(x+z)(x+y+z)∂_x+y(y+z)(x+y+z)∂_y and δ_3=x(x+z)(2y+z)∂_x+y(y+z)(2x+z)∂_y.

If (K)=2, then  is free with exponents (1,2,4), in fact we can take as basis of D() the following vector fields δ_E, 
δ_2=x^2∂_x+y^2∂_y+z^2∂_z and δ_3=x^4∂_x+y^4∂_y+z^4∂_z.


 Consider  the arrangement in K^3 as the cone of ^[-2,2] the Shi-Catalan arrangement of type B , see for example <cit.>.
Then  is free for any K, but its exponents depend on the characteristic of K.

If (K)=5, then  is free with exponents [1,5,15]. If (K)=7, then  is free with exponents [1,7,13].
If (K)5,7, then  is free with exponents [1,9,11].



Assume that the characteristic of K does not divide n, then a central arrangement  is free if and only if D_0() is a free S-module.


In general, it is important to distinguish between the case that the characteristic of K does not divide n and the case that it does.
In the latter, there might not exist δ∈ D() such that δ(Q()) is a non-zero scalar multiple of Q().
 Consider the arrangement  in 𝔽_3^3 with defining equation Q()=xyz(y+z)(x-y)(x+z). In this case n=6. Then
 is free and we can take {δ_E,δ_2,δ_3} as basis of D(), where δ_2=(xy-y^2)∂_y+(xz+z^2)∂_z and 
δ_3=(xyz +xz^2 +yz^2 +z^3)∂_z. In this situation, δ_E(Q())=δ_2(Q())=0 and δ_3(Q())=xQ().
Hence, there does not exist δ∈ D() such that δ(Q()) is a non-zero scalar multiple of Q().







Let δ_1,…,δ_l ∈ D(). 
Then (δ_i(x_j))_i,j is divisible by Q().
One of the most famous characterization of freeness is due to Saito <cit.> and it uses  the determinant of the coefficient 
matrix of δ_1,…,δ_l to check if the arrangement  is free or not. Notice that the original statement is for
characteristic 0, but in <cit.> Terao showed that this statement holds true for any characteristic.


[Saito's criterion]
Let  be a central arrangement in K^l and δ_1, …, δ_l ∈ D(). Then the following facts are equivalent


  * D() is free with basis δ_1, …, δ_l, i. e. D() = S·δ_1⊕⋯⊕ S ·δ_l.

  * (δ_i(x_j))_i,j=c Q(), where c∈ K∖{0}.

  * δ_1, …, δ_l are linearly independent over S and ∑_i=1^l(δ_i)=n.












Given an arrangement  in K^l, the Jacobian ideal of 
is the ideal of S generated by Q() and all its partial derivatives, and it is denoted by J().
This ideal has a central role in the study of free arrangements.
In fact, we can also characterize freeness by looking at the Jacobian ideal of .
This characterization by Terao <cit.> will be used in Section <ref>.
Notice that Terao described this result for characteristic 0, but the same proof also works for 
the case that the characteristic does not divide the cardinality of .

[Terao's criterion] 
A central arrangement  in K^l is free if and only if S/J() is 0 or (l-2)-dimensional Cohen-Macaulay.


By Remark <ref>, we need to prove that D_0() is a free S-module if and only if S/J() is 0 or Cohen-Macaulay. 

We have an exact sequence 
    0[r]   D_0()[r]^   α   S^l [r]^β   S [r]^γ       S/J()[r]    0,
 where 
    α(∑_i=1^l f_i ∂ / ∂ x_i)=(f_1, …, f_l)^t   for  ∑_i=1^l f_i ∂ / ∂ x_i ∈ D_0(),
 
    β((g_1, …, g_l)^t)= ∑_i=1^l g_i ∂ Q()/ ∂ x_i   for   (g_1, …, g_l)^t∈ S^l
 and γ is the natural projection.
Thus D_0() is free if and only if the homological dimension of S / J() is less than three.

Recall the Auslander-Buchsbaum equality 
    (S / J())+(S / J())=(S)=l,
 if S / J() 0. On the other hand, we have (S / J())≤ l-2, then 
    (S / J())≤ 2
 
    ⇔(S / J())≥ l-2 ≥(S / J())
 
    ⇔(S / J())= l-2 = (S / J())
 
    ⇔ S / J()  is Cohen-Macaulay of dimension  l-2,
 if S / J() 0. When S / J()=0, D_0() is obviously a free S-module. 






















Freeness has several consequences. For example we recall the following.

(<cit.>) 
Suppose that  is a free arrangement in K^l with exponents (e_1, … , e_l). 
Then 
    χ(,t)=∏_i=1^l(t-e_i).



The previous theorem imposes a necessary condition on the structure of  L() for the arrangement  to be free. 
The Terao conjecture is the problem to ask the converse, i.e if the structure of L() characterize freeness of . 
This conjecture is still unsolved.  In <cit.>, Yoshinaga gave an affirmative 
result when K=𝔽_p  and l = 3. Specifically, he proved the following result.
 Let  be a central arrangement in 𝔽_p^3. Then the following facts hold true.


  * When ||≥2p,  is free if and only if χ(,p)=0.

  * When ||=2p-1,  is free if and only if either χ(,p)=0 or χ(,t)=(t-1)(t-p+1)^2.

  * When ||=2p-2,  is free if and only if either χ(,p)=0 or χ(,t)=(t-1)(t-p+1)(t-p+2).


More in general, in <cit.>, Yoshinaga also proved the following.
 Let  be a central arrangement in 𝔽_p^l. If χ(,p^l-2)=0, then
 is free with exponents (1,p,…,p^l-2,||-1-p-⋯-p^l-2).




§ FROM CHARACTERISTIC 0 TO CHARACTERISTIC P


From now on we will assume that ={H_1,…, H_n} is a central arrangement in ℚ^l. After getting rid of the denominators, we can suppose that α_i ∈ℤ[x_1,…, x_l] for all i=1,…, n, and hence that Q()=∏_i=1^nα_i∈ℤ[x_1,…, x_l]. Moreover, we can also assume that there exists no prime number p that divides any α_i.

Let p be a prime number. Consider the image of Q() under the canonical homomorphism π_pℤ[x_1,…, x_l]→𝔽_p[x_1,…, x_l]. 
 Let  be a central arrangement in ℚ^l. We will call a prime number p good for  if π_p(Q()) is reduced.


 There is a finite number of primes p that are not good for .

 We have that π_p(Q()) is not reduced if and only if π_p(α_i)=π_p(α_j) for some i j and this can happens only for a finite number of primes.


Let now p be a good prime for , and consider _p the arrangement in 𝔽_p^l defined by π_p(Q()). Hence, by construction, Q(_p)=π_p(Q())0 and it is reduced.

 If  is free in ℚ^l with exponents (e_1,…, e_l), then _p is free in 𝔽_p^l with exponents (e_1,…, e_l), 
for all good primes except possibly a finite number of them.

 Let Δ={δ_1,…, δ_l} be a basis of D() with (δ_i)=e_i for all i=1,…, l. Since  has a defining equation with only integer coefficients, 
we can assume that every polynomial that appear in each δ∈Δ is in ℤ[x_1,…, x_l]. Hence we can consider δ̅_1,…, δ̅_l∈_𝔽_p^l 
 the image of δ_1,…, δ_l. 
We can assume that p∤δ for all δ∈Δ, and hence δ̅0 for all δ∈Δ. This implies that (δ̅_i)=(δ_i)=e_i for all i=1,…, l.

By the definition of D(), for each δ∈Δ there exists h∈ℤ[x_1,…, x_l] such that δ(Q())=hQ(). If we apply π_p to this expression we obtain that δ̅(Q(_p))=π_p(δ(Q()))=π_p(hQ())=π_p(h)Q(A_p), and hence δ̅∈ D(_p).

By Theorem <ref>, since in each δ∈Δ  every polynomial that appear have only integer coefficients, there exists c∈ℤ∖{0} such that (δ_i(x_j))_i,j=c Q(). 
If we apply π_p to the previous equality we obtain that (δ̅_i(x_j))_i,j=π_p((δ_i(x_j))_i,j)=π_p(c Q())=π_p(c)Q(_p). Hence if p does not divide c, 
we have that π_p(c)∈𝔽_p∖{0} and hence again by Theorem <ref>, we have that δ̅_1,…, δ̅_l are a basis of D(_p). 
This proves that _p is free with exponents (e_1,…, e_l).


By Lemma <ref>, the number of non good primes is finite. Hence we have the following.
 Let  be a central arrangement in ℚ^l and p a large prime number. If  is free in ℚ^l with exponents (e_1,…, e_l), then _p is free in 𝔽_p^l with exponents (e_1,…, e_l).


From Example <ref>, we have the following.
 Consider the arrangement  of Example <ref> with K=ℚ. Then the determinant of the coefficient matrix of the basis of D() is equal to
2Q(). A direct computation shows that if we take another basis of D() with only integer coefficients, then the determinant of the coefficient matrix is equal to cQ() with c∈ 2ℤ∖{0}.
This is why over 𝔽_2 the exponents of  change.


Similarly to the previous example, we have the following

 Consider  the cone of ^[-2,2] the Shi-Catalan arrangement of type B  as in Example <ref> with K=ℚ. 
Then the determinant of the coefficient matrix of any basis of D() with only integer coefficients is equal to
c Q(), with c∈ 35ℤ∖{0}. This is why over 𝔽_5 and 𝔽_7 the exponents of  change.

 
[<cit.>] Let _all be the arrangement of all hyperplanes of 𝔽_3^3 and L a line in 𝔽_3^3. Consider the arrangement
={H∈_all | H⊉ L} in 𝔽_3^3. Now  is not free. However, if we consider K a field such that (K)3 and ' an arrangement in K^3 satisfying L()≅ L('),
then ' is free with exponents (1,4,4). Notice that the corresponding matroid is not representable over  or ℚ.

 



 The following is an example of a free arrangements in ℚ^l that is not free in 𝔽_p^l, for some good prime p.
  Consider  the arrangement in ℚ^4 as the cone of ^[-2,2] the Shi-Catalan arrangement of type B. As described in <cit.>,
  is free with exponents (1,13,15,17). Moreover, the determinant of the coefficient matrix of a basis of D() with only integer coefficients is equal to
c Q(), with c∈ 56595ℤ∖{0}. Notice that 56595=3·5·7^3·11. Now, 3 is not a good prime but 5,7 and 11 are.
A direct computation shows that the arrangement _5 over 𝔽_5 is free with exponents (1,5,15,25). However, both _7 over 𝔽_7
and _11 over 𝔽_11 are not free.
 




§ HOMOGENEOUS IDEALS IN ℤ[X_1,…, X_L]

In this section, we study ideals in ℤ[x_1,…, x_l], their Betti numbers and their zero divisors. 
This will play an important role in Section <ref>.




 Given a sequence of numbers {c_i,j}, we obtain a new sequence by a cancellation as follows. 
Fix a j, and choose i and i' so that one of the numbers is odd and the other is even. Then replace c_i,j by c_i,j-1, 
and replace c_i',j by c_i',j-1. We have a consecutive cancellation when i' = i+1.


The following result is a generalization of Proposition 5.2 from <cit.> and it will play an important role in Theorem <ref>.
 Let I be a homogeneous ideal of R= ℤ[x_1,… ,x_l] and p a prime number that is a non-zero divisor in R/I. 
Then, the graded Betti numbers of (R/I) ⊗_ℤℚ can be obtained from the one of (R/I) ⊗_ℤ𝔽_p by a sequence of consecutive cancellations.

 Since p is a non-zero divisor in R/I, then (R/I) ⊗_ℤℤ_(p) is flat over ℤ_(p). 
Let 𝔽_∙ be a minimal graded free R⊗_ℤℤ_(p)-resolution of (R/I) ⊗_ℤℤ_(p). 
Note that 𝔽_∙⊗_ℤℤ_(p) is a free resolution of
(R/I)⊗_ℤℤ_(p) over R ⊗_ℤℤ_(p) = ℤ_(p)[x_1, …, x_l], and 
that the graded Betti numbers of 𝔽_∙ and 𝔽_∙⊗_ℤℤ_(p) are the same.
The flatness of (R/I) ⊗_ℤℤ_(p) over ℤ_(p) ensures that the complex 𝔽_∙⊗_ℤ𝔽_p =
𝔽_∙⊗_ℤℤ_(p)⊗_ℤ_(p)𝔽_p
is a free resolution of (R/I) ⊗_ℤ𝔽_p = (R/I) ⊗_ℤℤ_(p)⊗_ℤ_(p)𝔽_p
over R⊗_ℤ𝔽_p = 𝔽_p[x_1, …, x_l].
Now matrices giving the maps in this complex are in (x_1, …, x_l), i.e., it is a minimal resolution. This shows that the graded Betti number of 𝔽_∙ 
coincides with the one of  (R/I) ⊗_ℤ𝔽_p.

Now, 𝔽_∙⊗_ℤ_(p)ℚ is a graded free (R⊗_ℤℚ)-resolution of (R/I)⊗_ℤℚ. 
Therefore we can write 𝔽_∙⊗_ℤ_(p)ℚ=𝔾_∙⊕𝔾'_∙,
 where 𝔾_∙ is a minimal graded free (R⊗_ℤℚ)-resolution of (R/I)⊗_ℤℚ and 𝔾'_∙ is a 
 graded trivial complex of free (R⊗_ℤℚ)-modules <cit.>. Therefore, the graded Betti numbers of 
 (R/I) ⊗_ℤℚ can be obtained from the one of (R/I) ⊗_ℤ𝔽_p by a sequence of consecutive cancellations.


From the notion of consecutive cancellations, we directly obtain the relation between homological dimensions.

 Let I be a homogeneous ideal of R= ℤ[x_1,… ,x_l] and p a prime number that is a non-zero divisor in R/I. 
Then, the Betti numbers of (R/I) ⊗_ℤℚ are smaller or equal to the one of (R/I) ⊗_ℤ𝔽_p, and hence
((R/I) ⊗_ℤℚ)≤((R/I) ⊗_ℤ𝔽_p).


In general, given I an ideal of R= ℤ[x_1,… ,x_l], the number of zero divisor in R/I is infinite.
However, if we restrict our attention to zero divisors that are prime numbers, we have the following.
 Let I be an ideal of R=ℤ[x_1,… ,x_l]. Then the number of distinct prime numbers
that are zero divisors in R/I is finite.

  By Theorem 14.4 of <cit.>, there exists a∈ℤ∖{0} such that (R/I)[a^-1] is a free ℤ[a^-1]-module. This implies that 
the set of distinct prime numbers that are zero divisors in R/I is included in the set of distinct prime numbers that divide a, that is finite by the unique factorization theorem.


Also if the number of prime numbers that are zero divisors in R/I is finite, it might be difficult to compute them.
The rest of this section is dedicated to describe a method to compute them.

Consider I an ideal of R = ℤ[x_1,… ,x_l] and σ a term ordering on R. As for the case with polynomial rings over fields, we can study the  theory of Gröbner basis
for ideals of R. We refer to Chapter 4.5 of <cit.> for details.
 Let I be an ideal of R=ℤ[x_1,… ,x_l], σ a term ordering on R and G={g_1,…, g_t} a set of non-zero polynomials in I.
Then we say that G is a minimal strong σ-Gröbner basis for I if the following conditions hold true


  * G forms a set of generators of I;

  * for each f∈ I, there exists i∈{1,…,t} such that _σ(g_i) divides _σ(f);

  * if i j, then _σ(g_i) does not divide _σ(g_j).



[c.f. <cit.>, Lemma 4.5.8] A minimal strong σ-Gröbner basis of an ideal I of ℤ[x_1,… ,x_l] is also a σ-Gröbner basis of I. 



[<cit.>, Exercise 4.5.9] Let I be a non-zero ideal of R=ℤ[x_1,… ,x_l] and σ a term ordering on R. 
Then there always exists a minimal strong σ-Gröbner basis of I

Notice that in general, a minimal strong σ-Gröbner basis is not unique. Consider the following example.
 Consider the ideal I=(x^2-y,3y) in ℤ[x,y] and σ=. Then both 
{x^2-y,3y} and {x^2+y,3y} are minimal strong σ-Gröbner basis for I.

 Let I be an ideal of R=ℤ[x_1,…,x_l], and σ a term ordering on R. Let G_1 and G_2 be two minimal strong 
σ-Gröbner bases of I. Then {_σ(g) | g∈ G_1} = {_σ(g) | g∈ G_2}. Consequently we have |G_1| = |G_2| 
and {_σ(g) | g∈ G_1} = {_σ(g) | g∈ G_2}.


Assume that there exists t∈_σ(G_1)∖_σ(G_2). Now, t∈_σ(I) and hence there exists f∈ I such that t=_σ(f).
Since G_2 is a minimal strong σ-Gröbner basis for I, there exists g∈ G_2 such that _σ(g)=t' divides _σ(f)=t. With a similar argument, we see that t', and hence t,
is a multiple of an element t” in _σ(G_1). The minimality of G_1 implies that t”=t. Hence, t=t'∈_σ(G_2), but this is a contradiction.
This shows that _σ(G_1)⊆_σ(G_2). 

With the same argument we can show that _σ(G_2)⊆_σ(G_2), and hence that _σ(G_1)=_σ(G_2).


 The previous lemma implies that {_σ(g) | g∈ G} generates the monomial ideal _σ(I).


By Lemma <ref>, we can introduce the following Definition. See <cit.>, for more details.

 Let I be an ideal of R=ℤ[x_1,…,x_l], and σ be a term ordering on R. 
If a prime number p does not divide the leading coefficient of any polynomial 
in a minimal strong σ-Gröbner basis for I, then we will call p σ-lucky for I.

 Given I an ideal of R=ℤ[x_1,…,x_l] and σ be a term ordering on R, since a minimal strong σ-Gröbner basis is finite,
 then the number of primes that are not σ-lucky for I is finite.

Notice that the list of σ-lucky primes for an ideal depends on the choice of term ordering.
 Consider the ideal I=(3x-y) in ℤ[x,y]. Consider on ℤ[x,y] a term ordering σ
such that x>_σ y. Then {3x-y} is a minimal strong σ-Gröbner basis of I, and hence every prime p 3 is σ-lucky for I. 
However, if we consider on ℤ[x,y] a term ordering τ
such that y>_τ x, then {y-3x} is a minimal strong τ-Gröbner basis of I. Hence every prime p is τ-lucky for I, also p=3.


Notice that in general, we cannot determine the σ-lucky primes from the coefficients that appear in a set of generators.
 Consider the ideal I=(2x+3y,x-y) in ℤ[x,y] with σ=. In this case,
{5y,x-y} is a minimal strong σ-Gröbner basis of I, and hence p=5 is the only non σ-lucky prime of I.


We are now ready to describe the relation between lucky primes and primes that are non-zero divisors.
 Let I be an ideal of R=ℤ[x_1,…,x_l] and σ a term ordering on R. If p is a 
σ-lucky prime for I, then p is a non-zero divisor in R/I.

 Let G={g_1,…,g_t} be a minimal strong σ-Gröbner basis of I and p a σ-lucky prime for I. 
Suppose that p is a zero divisor in R/I. This implies that there exists f∈ R∖ I such that pf∈ I. By definition of minimal strong σ-Gröbner basis,
there exists i∈{1,…,t} such that _σ(g_i) divides _σ(pf). We can now consider h_1=pf-_σ(pf)/_σ(g_i)g_i∈ I. Since p σ-lucky,
p| _σ(pf_1)/_σ(g_j), and hence h_1=pf_1, for some f_1∈ R. 
We can now repeat the process with pf_1∈ I and obtain pf_2=pf_1-_σ(pf_1)/_σ(g_j)g_j∈ I, for some j. 
Since G is a σ-Gröbner basis this process will end after a finite number of steps, i.e. we will obtain that there exists a k≥1 such that pf_k∈ I with pf_k0 but pf_k+1=0.
Retrieving each step, we can write pf=∑_i=1^tpa_ig_i, for some a_i∈ R. This implies that f=∑_i=1^ta_ig_i∈ I, but this is impossible by assumption.































Notice that in general the set of distinct prime numbers that are non-zero divisors in R/I contains strictly the set of σ-lucky primes for I.
 Consider the ideal of Example <ref> with the term ordering σ. Then p=3 is a non-zero divisor in R/I, but
it is not a σ-lucky prime.







§ FROM CHARACTERISTIC P TO CHARACTERISTIC 0

As in Section <ref>, we will assume that ={H_1,…, H_n} is a central arrangement in ℚ^l and hence that α_i ∈ℤ[x_1,…, x_l] for all i=1,…, n.
Moreover, we can also assume that there exists no prime number p that divides any α_i.

In the rest of the paper, we will denote by J()_ℤ the ideal of ℤ[x_1,… ,x_l] generated by Q() and its partial derivatives.

 Let ={H_1,…, H_n} be a central arrangement in ℚ^l. Let p
be a good prime number for  that does not divide n and that is a non-zero divisor in ℤ[x_1,… ,x_l]/J()_ℤ. If _p is free in 𝔽_p^l with exponents (e_1,…, e_l), 
then  is free in ℚ^l with exponents (e_1,…, e_l).

 Denote S=ℚ[x_1,…, x_l], S_p=𝔽_p[x_1,…, x_l] and R=ℤ[x_1,… ,x_l]. 
Notice that  S_p/J(_p)≅ (R/J()_ℤ)⊗_ℤ𝔽_p and
S/J()≅ (R/J()_ℤ)⊗_ℤℚ.

Clearly S_p/J(_p)=0 if and only if S_/J()=0

Assume now that S_p/J(_p)0. By Theorem <ref>, S_p/J(_p) is Cohen-Macaulay of dimension
l-2.

By Auslander-Buchsbaum equality, we can write

    (S_p / J(_p))+(S_p / J(_p))=l=(S/ J())+(S/ J()).

By Corollary <ref>, (S_p/ J(_p))≥(S/ J()) and hence, by the previous equality, (S_p/ J(_p))≤(S/ J()).
Now S_p/J(_p) is Cohen-Macaulay, hence (S_p / J(_p))=(S_p / J(_p))=l-2=(S/ J()). This implies that (S/ J())≥ l-2=(S/ J()).
On the other hand, (S/ J())≤(S/ J()) and hence we have that (S/ J())=(S/ J()). This implies that S/J() is Cohen-Macaulay of dimension
l-2, and so  is free in ℚ^l.

This argument also shows that 2=(S_p/ J(_p))=(S/ J()), and hence, S/ J() have the same Betti numbers of S_p / J(_p). Since the exponents of  can be
computed uniquely from the Betti numbers of S/ J(), this implies that  has exponents (e_1,…, e_l).







In Theorem <ref>, the assumption that the prime p is a non-zero divisor in ℤ[x_1,… ,x_l]/J()_ℤ is fundamental. In fact we have the following.
 Consider the arrangement ⊆ℚ^3 with defining equation Q()=z(x+2y-4z)(y+4z)(x+3y-6z).  is non free and both 2 and 3 are zero divisors in ℤ[x_1,… ,x_l]/J()_ℤ.
In fact, we have that 3(y^2z^2+2yz^3-8z^4)∈ J()_ℤ but y^2z^2+2yz^3-8z^4∉ J()_ℤ, and similarly 
2(xyz^2+4y^2z^2+4xz^3+8yz^3-32z^4)∈ J()_ℤ but xyz^2+4y^2z^2+4xz^3+8yz^3-32z^4∉ J()_ℤ. However, both A_2 and A_3 are free with
exponents (1,1,2).


By Proposition <ref>, the number of prime numbers that are zero divisors in ℤ[x_1,… ,x_l]/J()_ℤ is finite. Hence, putting together Corollary <ref> 
and Theorem <ref>, we have the following.
 Let  be a central arrangement in ℚ^l and p a large prime number. _p is free in 𝔽_p^l with exponents (e_1,…, e_l) if and only if
 is free in ℚ^l with exponents (e_1,…, e_l).





§ APPLICATIONS

As in Sections <ref> and <ref>, we will assume that ={H_1,…, H_n} is a central arrangement in ℚ^l and hence that α_i ∈ℤ[x_1,…, x_l] for all i=1,…, n.
Moreover, we can also assume that there exists no prime number p that divides any α_i.

From Theorems <ref> and <ref>, we have the following.
 Let ={H_1,…, H_n} be a central arrangement in ℚ^l. Let p
be a good prime number for  that does not divide n and that is a non-zero divisor in ℤ[x_1,… ,x_l]/J()_ℤ. If χ(_p,p^l-2)=0, then
 is free with exponents (1,p,…,p^l-2,||-1-p-⋯-p^l-2).


As noted in <cit.>, if p is a large prime number, then  and _p have isomorphic intersection lattices, and hence have the same characteristic polynomial. 
This fact, together with the finiteness of good prime numbers that are zero divisors in ℤ[x_1,… ,x_l]/J()_ℤ, gives us the following.

 Let  be a central arrangement in ℚ^l and p
a large prime number. If χ(,p^l-2)=0, then  is free with exponents (1,p,…,p^l-2,||-1-p-⋯-p^l-2).










  
Acknowledgements The authors would like to thank G. Caviglia, M. Kummini, N. Nakashima and M. Yoshinaga for many helpful discussions.



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