Let q = p^s with p an odd prime, and let P be a monic irreducible polynomial of degree d >= 2 in F_q[t]. The Wilson analogue F_d = -1 mod P holds for the product F_d of all nonzero polynomials of degree less than d, and Thakur proved that P is a Wilson prime, i.e. F_d = -1 mod P^2, exactly when all exponents of P are congruent to 0 or 1 mod p, in which case the congruence holds modulo P^(p-1). He asked whether it can hold modulo a higher power, and what the exact multiplicities are. Let r be the largest integer such that all exponents of P are congruent to 0 or 1 modulo p^r. We prove that the exact multiplicity m = v_P(F_d + 1) is 1 if r = 0, p^r - 1 if 1 <= r <= s, and q + 1 if r > s. In particular m <= q + 1, and the multiplicity of a Wilson prime in the numerator of the sum over i < d of 1/(t^(q^i) - t) is exactly m - 1. Examples over F_3, F_5 and F_9 show that all regimes occur. A script checks the theorem on all primes of small degree and on these examples.
Introduction
Let q=ps with p an odd prime, A=Fq[t], and [i]=tqi−t. For a monic irreducible ℘∈A of degree d let Fd be the product of all nonzero polynomials of degree less than d. Then Fd≡−1(mod℘), the analogue of Wilson’s theorem, and
Fd=(−1)dLdDd=(−1)di=1∏d−1[i]qd−i−1,(1)
where Dd=∏i=1d−1[i]qd−i and Ld=∏i=1d−1[i][2 p. 1842]. Following Thakur, ℘ is a Wilson prime if Fd≡−1(mod℘2). We write
m(℘)=v℘(Fd+1)
for the exact multiplicity of the Wilson congruence. For d=1 one has F1=−1, so we assume d≥2 throughout.
Thakur proved that ℘ is a Wilson prime if and only if d2℘/dt2=0, that is, all exponents of ℘ are ≡0 or 1(modp), and that in this case the congruence holds modulo ℘p−1[2 Thm. 2.9]; see also the remarks after [3 Thm. 5.1]. For the primes tp−t−a over Fp the exponent p−1 was already known to be exact [1]; see also [2 Rem. 2.12]. Thakur also reported numerical data of G. Todd suggesting that p−1 is exact even when q is not prime, and noted that it was unclear whether the power can be increased [2 Rem. 2.11]. In Question (1) of the closing list of questions of [3], he asked whether, for d>1 and p>2, the congruence ever (or infinitely often) holds modulo a higher power, and what the exact multiplicities are.
Definition. The depth r(℘) is the largest integer r≥0 such that every exponent n with a nonzero coefficient in ℘ satisfies n≡0 or 1(modpr). Since d≥2 is an exponent of ℘, the depth is finite. Equivalently, r(℘)≥ρ if and only if ℘=U(tpρ)+tV(tpρ) for some polynomials U,V.
Theorem 1. Let ℘ be monic irreducible of degree d≥2 in Fq[t], q=ps, p odd, and r=r(℘). Then
v℘(Fd+1)=⎩⎨⎧1pr−1q+1if r=0,if 1≤r≤s,if r>s.
Thus the only possible exact multiplicities are 1, p−1, p2−1, …, q−1 and q+1, and the Wilson congruence never holds modulo ℘q+2. Over Fp the only possible values are 1, p−1 and p+1. Since depth r forces d≥pr, the depth is at most 1 when d<p2, and then the value p−1 of [2] is exact, in agreement with the data of [2 Rem. 2.11]. The case r=1 of Lemma 4 below is [2 Thm. 2.9(iii)]. We assume p odd, as in the question; for p=2 see [2 Thm. 2.9(iv)]. The answer to the first part of the question is positive: the examples of Section 4 have multiplicity q+1 for q=3,5,9, and p2−1=8 for q=9. We do not address the infinitude of such primes.
Thakur also proved [3 Thm. 7.1] that the primes of degree d dividing the numerator of Hd=1/[1]+⋯+1/[d−1] are exactly the Wilson primes, with multiplicity at least p−2. Theorem 1 makes this exact.
Corollary 2. If ℘ is a Wilson prime of degree d and depth r, then v℘(Hd)=pr−2 if r≤s and v℘(Hd)=q if r>s. In all cases v℘(Hd)=m(℘)−1.
Proof. Since [i]′=−1 and qd−i−1≡−1(modp), logarithmic differentiation of (1) gives Fd′=FdHd. As ℘∤Fd, v℘(Hd)=v℘(Fd′)=v℘((Fd+1)′). Write Fd+1=℘mG with ℘∤G; then (Fd+1)′=℘m−1(m℘′G+℘G′), and ℘∤℘′ because ℘ is separable. Since none of the values pr−1, q+1 is divisible by p, the valuation is m−1. The primes ℘ of degree d do not divide Ld−1=[1]⋯[d−1], so v℘(Hd) is also the valuation of the numerator of Hd. □
Local expansion at a root
Fix ℘ as in Theorem 1 and a root θ of ℘, so that Fq(θ)=Fqd and θqd=θ. Put
βi=θqi−θ(1≤i≤d−1);
these are nonzero, and β1,…,βd−1 together with 0 are the roots of ℘(X+θ). For a polynomial G∈Fq[t], the Hasse derivatives G[k] are defined by G(t+u)=∑kG[k](t)uk. Since ℘ is separable, v℘(G) is the order of vanishing of G at θ, that is
v℘(G)=min{k≥0:G[k](θ)=0}.(2)
Lemma 3. With ei=qd−i−1,
Fd(θ+u)=(−1)di=1∏d−1(βi−u+uqi)eiin Fqd[u].
Proof. In characteristic p, [i](θ+u)=θqi+uqi−θ−u=βi−u+uqi; apply (1). □
Lemma 4. Let xi=βi−1 and let hk=hk(x1,…,xd−1) be the complete homogeneous symmetric polynomials, so ∑k≥0hkuk=∏i(1−xiu)−1. Then
k≥0∑hkuk=∑j≥1℘[j](θ)uj−1℘′(θ).
Moreover ℘[j](θ)=0 for $2\le j\le p^r-1$, and ℘[pr](θ)=0 if r≥1. Consequently h1=⋯=hpr−2=0 and hpr−1=0 when r≥1, and h1=0 when $r=0.
Proof. We have ℘(u+θ)=∑j℘[j](θ)uj=u∏i=1d−1(u−βi), so Q(u):=∑j≥1℘[j](θ)uj−1=∏i(u−βi) and Q(0)=℘′(θ). Hence ∏i(1−xiu)=Q(u)/Q(0), which is the first identity.
Write ℘=∑ncntn, so that ℘[j]=∑n(jn)cntn−j. For j≥1 this polynomial has degree less than d, so it vanishes at θ if and only if it is zero, that is, if and only if p∣(jn) for every n with cn=0. Let 2≤j≤pr−1 and cn=0. The lowest r base-p digits of n are all 0, or are 1,0,…,0; since j<pr has j≥2, some digit of j exceeds the corresponding digit of n, and Lucas’ theorem gives p∣(jn). If r≥1, by maximality of r some exponent n has nmodpr+1∈/{0,1}; as n≡0 or 1(modpr), its digit nr is nonzero, and Lucas’ theorem gives (prn)≡nr=0. Distinct n give distinct monomials tn−pr, so ℘[pr]=0. Finally ℘′(θ)=0 by separability, and if r=0 some exponent has n≡0,1(modp), so (2n)=n(n−1)/2=0 and ℘[2](θ)=0.
If r≥1, the denominator in the first identity is ℘′(θ)(1+cupr−1+O(upr)) with c=℘[pr](θ)/℘′(θ)=0, so the quotient is 1−cupr−1+O(upr). If r=0, the same computation gives h1=−℘[2](θ)/℘′(θ)=0.
Proof of Theorem 1
Let fk=Fd[k](θ), the coefficient of uk in Fd(θ+u). By (2), m=min{k≥1:fk=0} and f0=Fd(θ)=−1.
Step 1:fk=−hk for 0≤k≤q−1. Modulo uq all terms uqi in Lemma 3 vanish. Moreover
Step 2: the case r≤s. If r=0, Step 1 and Lemma 4 give f1=−h1=0, so m=1. If 1≤r≤s, then pr−1≤q−1, and Step 1 with Lemma 4 gives f1=⋯=fpr−2=0 and fpr−1=0, so m=pr−1.
Step 3: the case r>s. Then pr−1≥pq−1≥q+2, so Lemma 4 gives h1=⋯=hq+1=0, and by Step 1 f1=⋯=fq−1=0. We compute modulo uq+2. For i≥2 the term uqi vanishes, since q2≥q+2. As in Step 1, (βi−u+δi,1uq)ei=Ni/Di with Di=βi−u+δi,1uq and Ni=Diqd−i. Modulo uq+2, Ni=βiqd−i for i≤d−2, since qd−i≥q2, while Nd−1=βd−1q−uq; here d≥pr≥pq≥9, so the factors i=1 and i=d−1 are distinct. Hence, with y=βd−1−q,
Fd(θ+u)≡−(1−yuq)1−x1u+x1uq1i=2∏d−11−xiu1.
Now (1−x1u+x1uq)−1=(1−x1u)−1−x1uq(1−x1u)−2+O(u2q), so
Since θqd=θ, we have βd−1q=θqd−θq=−β1, so y=−x1. With hq=hq+1=h1=0 we get fq=0 and fq+1=x12=β1−2=0. Hence m=q+1.
Remark 5. In Step 3 the vanishing of fq depends on the identity βd−1q=−β1, which links the factors i=1 and i=d−1 of Lemma 3: the term −x1 coming from i=1 cancels the term −y coming from i=d−1.
Examples and verification
We write F9=F3[a]/(a2−a−1). The following polynomials are irreducible, with the stated depth and multiplicity:
q
℘
r
regime
m
3
t3−t−1
1
r≤s
2
3
t18−t10+1
2
r>s
4
5
t5−t−1
1
r≤s
4
5
t51−2t50−t26+1
2
r>s
6
9
t3−(a+1)t−(a+1)
1
r≤s
2
9
t10+(a+1)t9+t+1
2
r=s
8
9
t54−at28+at27+at+1
3
r>s
10
Table 1.
The ancillary file verify_wilson_multiplicity.py (Python 3, no dependency, exact arithmetic in Fq[t], about fifteen seconds) checks:
the product formula (1) against the definition of Fd over F3 for d=2,3;
for every monic irreducible ℘ of degree 2≤d≤7 over F3, 2≤d≤4 over F5 and 2≤d≤3 over F9 (981 primes in all): the valuation v℘(Fd+1), computed in Fq[t] modulo a power of ℘ from (1), equals the value of Theorem 1, and v℘(Hd) equals m−1 (Corollary 2);
for the examples above: irreducibility (Rabin’s test), the depth, and the same two valuations computed directly in Fq[t] for the five examples of degree at most 18;
for all the examples, the coefficients fk of Fd(θ+u) computed in the residue field Fq[t]/℘ from Lemma 3: the first nonzero one is fm; fk=−hk for k<q, with hk computed from the first identity of Lemma 4; the vanishing of ℘[j](θ) for 2≤j≤min(pr−1,q+3), and ℘[pr]=0; βd−1q=−β1; and, when r>s, fq=0 and fq+1=β1−2.
In small degree only the depths 0 and 1 occur, and the multiplicity p−1 of [2] is then exact. These computations check the statements; the proofs do not depend on them.
Acknowledgement
Acknowledgement. Technical assistance: Claude (Anthropic).
ORCID 0009-0009-1526-1178
References
[1][SSTT] J. Sauerberg, L. Shu, D. S. Thakur and G. Todd, Infinitude of Wilson primes for F_q[t], Acta Arith. 157 (2013), 91–100.DOI
[2][Th13] D. S. Thakur, Differential characterization of Wilson primes for F_q[t], Algebra Number Theory 7 (2013), 1841–1848.
[3][Th24] D. S. Thakur, Fermat–Wilson supercongruences, arithmetic derivatives and strange factorizations, J. Théor. Nombres Bordeaux 36 (2024), no. 2, 481–491, doi:10.5802/jtnb.1285; arXiv:2211.01076.DOI