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\title[Exact multiplicity of the Wilson supercongruence]{The exact multiplicity of the Wilson
supercongruence in \texorpdfstring{$\mathbb{F}_q[t]$}{Fq[t]}}
\author{David Niedbala Giraudin}
\address{ORCID 0009-0009-1526-1178}
\date{3 November 2026}

\begin{document}

\begin{abstract}
Let $q=p^s$ with $p$ an odd prime, and let $\wp$ be a monic irreducible polynomial of degree
$d\ge2$ in $\F_q[t]$. The Wilson analogue $F_d\equiv-1\pmod\wp$ holds for the product $F_d$ of
all nonzero polynomials of degree less than $d$, and Thakur proved that $\wp$ is a Wilson prime,
i.e.\ $F_d\equiv-1\pmod{\wp^2}$, exactly when all exponents of $\wp$ are $\equiv0$ or
$1\pmod p$, in which case the congruence holds modulo $\wp^{p-1}$. 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 $\wp$ are $\equiv0$ or $1$ modulo $p^r$. We prove that the exact
multiplicity $m=v_\wp(F_d+1)$ is $1$ if $r=0$, $p^r-1$ if $1\le r\le s$, and $q+1$ if $r>s$.
In particular $m\le q+1$, and the multiplicity of a Wilson prime in the numerator of
$\sum_{i<d}1/(t^{q^i}-t)$ 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.
\end{abstract}

\maketitle

\section{Introduction}

Let $q=p^s$ with $p$ an odd prime, $A=\Fq[t]$, and $[i]=t^{q^i}-t$. For a monic irreducible
$\wp\in A$ of degree $d$ let $F_d$ be the product of all nonzero polynomials of degree less than
$d$. Then $F_d\equiv-1\pmod\wp$, the analogue of Wilson's theorem, and
\begin{equation}\label{eq:Fd}
F_d=(-1)^d\frac{D_d}{L_d}=(-1)^d\prod_{i=1}^{d-1}[i]^{\,q^{d-i}-1},
\end{equation}
where $D_d=\prod_{i=1}^{d}[i]^{q^{d-i}}$ and $L_d=\prod_{i=1}^{d}[i]$ \cite[p.~1842]{Th13}. Following Thakur, $\wp$ is a \emph{Wilson prime} if $F_d\equiv-1\pmod{\wp^2}$. We
write
\[
m(\wp)=v_\wp(F_d+1)
\]
for the exact multiplicity of the Wilson congruence. For $d=1$ one has $F_1=-1$, so we assume
$d\ge2$ throughout.

Thakur proved that $\wp$ is a Wilson prime if and only if $d^2\wp/dt^2=0$, that is, all exponents
of $\wp$ are $\equiv0$ or $1\pmod p$, and that in this case the congruence holds modulo
$\wp^{p-1}$ \cite[Thm.~2.9]{Th13}; see also the remarks after \cite[Thm.~5.1]{Th24}. For the primes
$t^p-t-a$ over $\F_p$ the exponent $p-1$ was already known to be exact \cite{SSTT}; see also
\cite[Rem.~2.12]{Th13}. 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 \cite[Rem.~2.11]{Th13}. In Question~(1) of the closing list of questions of \cite{Th24}, 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.

\emph{Definition.} The \emph{depth} $r(\wp)$ is the largest integer $r\ge0$ such that every
exponent $n$ with a nonzero coefficient in $\wp$ satisfies $n\equiv0$ or $1\pmod{p^r}$.
Since $d\ge2$ is an exponent of $\wp$, the depth is finite. Equivalently, $r(\wp)\ge\rho$ if and
only if $\wp=U(t^{p^\rho})+t\,V(t^{p^\rho})$ for some polynomials $U,V$.

\begin{theorem}\label{thm:main}
Let $\wp$ be monic irreducible of degree $d\ge2$ in $\Fq[t]$, $q=p^s$, $p$ odd, and $r=r(\wp)$.
Then
\[
v_\wp(F_d+1)=\begin{cases}1&\text{if } r=0,\\ p^r-1&\text{if } 1\le r\le s,\\ q+1&\text{if } r>s.\end{cases}
\]
\end{theorem}

Thus the only possible exact multiplicities are $1,\ p-1,\ p^2-1,\ \dots,\ q-1$ and $q+1$, and the Wilson
congruence never holds modulo $\wp^{q+2}$. Over $\F_p$ the only possible values are $1$, $p-1$ and $p+1$. Since depth $r$ forces $d\ge p^r$,
the depth is at most $1$ when $d<p^2$, and then the value
$p-1$ of \cite{Th13} is exact, in agreement with the data of \cite[Rem.~2.11]{Th13}. The case $r=1$
of Lemma~\ref{lem:hasse} below is \cite[Thm.~2.9(iii)]{Th13}. We assume $p$ odd, as in the
question; for $p=2$ see \cite[Thm.~2.9(iv)]{Th13}. The answer to the first part of the
question is positive: the examples of Section~\ref{sec:ex} have multiplicity $q+1$ for $q=3,5,9$,
and $p^2-1=8$ for $q=9$. We do not address the infinitude of such primes.

Thakur also proved \cite[Thm.~7.1]{Th24} that the primes of degree $d$ dividing the
numerator of $H_d=1/[1]+\dots+1/[d-1]$ are exactly the Wilson primes, with multiplicity at least
$p-2$. Theorem~\ref{thm:main} makes this exact.

\begin{corollary}\label{cor:H}
If $\wp$ is a Wilson prime of degree $d$ and depth $r$, then $v_\wp(H_d)=p^r-2$ if $r\le s$ and
$v_\wp(H_d)=q$ if $r>s$. In all cases $v_\wp(H_d)=m(\wp)-1$.
\end{corollary}

\begin{proof}
Since $[i]'=-1$ and $q^{d-i}-1\equiv-1\pmod p$, logarithmic differentiation of \eqref{eq:Fd} gives
$F_d'=F_dH_d$. As $\wp\nmid F_d$, $v_\wp(H_d)=v_\wp(F_d')=v_\wp\bigl((F_d+1)'\bigr)$. Write
$F_d+1=\wp^mG$ with $\wp\nmid G$; then $(F_d+1)'=\wp^{m-1}(m\wp'G+\wp G')$, and $\wp\nmid\wp'$
because $\wp$ is separable. Since none of the values $p^r-1$, $q+1$ is divisible by $p$, the
valuation is $m-1$. The primes $\wp$ of degree $d$ do not divide $L_{d-1}=[1]\cdots[d-1]$, so
$v_\wp(H_d)$ is also the valuation of the numerator of $H_d$.
\end{proof}

\section{Local expansion at a root}

Fix $\wp$ as in Theorem~\ref{thm:main} and a root $\theta$ of $\wp$, so that $\Fq(\theta)=\F_{q^d}$
and $\theta^{q^d}=\theta$. Put
\[
\beta_i=\theta^{q^i}-\theta\qquad(1\le i\le d-1);
\]
these are nonzero, and $\beta_1,\dots,\beta_{d-1}$ together with $0$ are the roots of
$\wp(X+\theta)$. For a polynomial $G\in\Fq[t]$, the Hasse derivatives $G^{[k]}$ are defined by
$G(t+u)=\sum_kG^{[k]}(t)u^k$. Since $\wp$ is separable, $v_\wp(G)$ is the order of vanishing of $G$
at $\theta$, that is
\begin{equation}\label{eq:ord}
v_\wp(G)=\min\{k\ge0:\ G^{[k]}(\theta)\ne0\}.
\end{equation}

\begin{lemma}\label{lem:exp}
With $e_i=q^{d-i}-1$,
\[
F_d(\theta+u)=(-1)^d\prod_{i=1}^{d-1}\bigl(\beta_i-u+u^{q^i}\bigr)^{e_i}\qquad\text{in }\F_{q^d}[u].
\]
\end{lemma}

\begin{proof}
In characteristic $p$, $[i](\theta+u)=\theta^{q^i}+u^{q^i}-\theta-u=\beta_i-u+u^{q^i}$; apply
\eqref{eq:Fd}.
\end{proof}

\begin{lemma}\label{lem:hasse}
Let $x_i=\beta_i^{-1}$ and let $h_k=h_k(x_1,\dots,x_{d-1})$ be the complete homogeneous symmetric
polynomials, so $\sum_{k\ge0}h_ku^k=\prod_i(1-x_iu)^{-1}$. Then
\[
\sum_{k\ge0}h_k\,u^k=\frac{\wp'(\theta)}{\sum_{j\ge1}\wp^{[j]}(\theta)\,u^{j-1}}.
\]
Moreover $\wp^{[j]}(\theta)=0$ for $2\le j\le p^r-1$, and $\wp^{[p^r]}(\theta)\ne0$ if $r\ge1$.
Consequently $h_1=\dots=h_{p^r-2}=0$ and $h_{p^r-1}\ne0$ when $r\ge1$, and $h_1\ne0$ when $r=0$.
\end{lemma}

\begin{proof}
We have $\wp(u+\theta)=\sum_j\wp^{[j]}(\theta)u^j=u\prod_{i=1}^{d-1}(u-\beta_i)$, so
$Q(u):=\sum_{j\ge1}\wp^{[j]}(\theta)u^{j-1}=\prod_i(u-\beta_i)$ and $Q(0)=\wp'(\theta)$. Hence
$\prod_i(1-x_iu)=Q(u)/Q(0)$, which is the first identity.

Write $\wp=\sum_nc_nt^n$, so that $\wp^{[j]}=\sum_n\binom nj c_nt^{n-j}$. For $j\ge1$ this
polynomial has degree less than $d$, so it vanishes at $\theta$ if and only if it is zero, that is,
if and only if $p\mid\binom nj$ for every $n$ with $c_n\ne0$. Let $2\le j\le p^r-1$ and $c_n\ne0$.
The lowest $r$ base-$p$ digits of $n$ are all $0$, or are $1,0,\dots,0$; since $j<p^r$ has
$j\ge2$, some digit of $j$ exceeds the corresponding digit of $n$, and Lucas' theorem gives
$p\mid\binom nj$. If $r\ge1$, by maximality of $r$ some exponent $n$ has $n\bmod p^{r+1}\notin\{0,1\}$;
as $n\equiv0$ or $1\pmod{p^r}$, its digit $n_r$ is nonzero, and Lucas' theorem gives
$\binom n{p^r}\equiv n_r\not\equiv0$. Distinct $n$ give distinct monomials $t^{n-p^r}$, so
$\wp^{[p^r]}\ne0$. Finally $\wp'(\theta)\ne0$ by separability, and if $r=0$ some exponent has
$n\not\equiv0,1\pmod p$, so $\binom n2=n(n-1)/2\not\equiv0$ and $\wp^{[2]}(\theta)\ne0$.

If $r\ge1$, the denominator in the first identity is $\wp'(\theta)\bigl(1+cu^{p^r-1}+O(u^{p^r})\bigr)$
with $c=\wp^{[p^r]}(\theta)/\wp'(\theta)\ne0$, so the quotient is $1-cu^{p^r-1}+O(u^{p^r})$. If $r=0$,
the same computation gives $h_1=-\wp^{[2]}(\theta)/\wp'(\theta)\ne0$.
\end{proof}

\section{Proof of Theorem~\ref{thm:main}}

Let $f_k=F_d^{[k]}(\theta)$, the coefficient of $u^k$ in $F_d(\theta+u)$. By \eqref{eq:ord},
$m=\min\{k\ge1: f_k\ne0\}$ and $f_0=F_d(\theta)=-1$.

\emph{Step 1: $f_k=-h_k$ for $0\le k\le q-1$.} Modulo $u^q$ all terms $u^{q^i}$ in
Lemma~\ref{lem:exp} vanish. Moreover
$(\beta_i-u)^{e_i}=(\beta_i-u)^{q^{d-i}}/(\beta_i-u)=(\beta_i^{q^{d-i}}-u^{q^{d-i}})/(\beta_i-u)$, and
$q^{d-i}\ge q$. Hence, modulo $u^q$,
\[
F_d(\theta+u)\equiv(-1)^d\prod_{i=1}^{d-1}\frac{\beta_i^{q^{d-i}}}{\beta_i-u}
=(-1)^d\prod_{i=1}^{d-1}\beta_i^{e_i}\cdot\prod_{i=1}^{d-1}(1-x_iu)^{-1}
=F_d(\theta)\sum_kh_ku^k=-\sum_kh_ku^k .
\]

\emph{Step 2: the case $r\le s$.} If $r=0$, Step~1 and Lemma~\ref{lem:hasse} give $f_1=-h_1\ne0$,
so $m=1$. If $1\le r\le s$, then $p^r-1\le q-1$, and Step~1 with Lemma~\ref{lem:hasse} gives
$f_1=\dots=f_{p^r-2}=0$ and $f_{p^r-1}\ne0$, so $m=p^r-1$.

\emph{Step 3: the case $r>s$.} Then $p^r-1\ge pq-1\ge q+2$, so Lemma~\ref{lem:hasse} gives
$h_1=\dots=h_{q+1}=0$, and by Step~1 $f_1=\dots=f_{q-1}=0$. We compute modulo $u^{q+2}$. For
$i\ge2$ the term $u^{q^i}$ vanishes, since $q^2\ge q+2$. As in Step~1,
$\bigl(\beta_i-u+\delta_{i,1}u^q\bigr)^{e_i}=N_i/D_i$ with $D_i=\beta_i-u+\delta_{i,1}u^q$ and
$N_i=D_i^{q^{d-i}}$. Modulo $u^{q+2}$, $N_i=\beta_i^{q^{d-i}}$ for $i\le d-2$, since $q^{d-i}\ge q^2$,
while $N_{d-1}=\beta_{d-1}^q-u^q$; here $d\ge p^r\ge pq\ge9$, so the
factors $i=1$ and $i=d-1$ are distinct. Hence,
with $y=\beta_{d-1}^{-q}$,
\[
F_d(\theta+u)\equiv-(1-yu^q)\,\frac{1}{1-x_1u+x_1u^q}\prod_{i=2}^{d-1}\frac1{1-x_iu}.
\]
Now $(1-x_1u+x_1u^q)^{-1}=(1-x_1u)^{-1}-x_1u^q(1-x_1u)^{-2}+O(u^{2q})$, so
\begin{align*}
F_d(\theta+u)&\equiv-(1-yu^q)\Bigl(\sum_kh_ku^k-x_1u^q(1-x_1u)^{-1}\sum_kh_ku^k\Bigr)
\\&\equiv-(1-yu^q)\Bigl(\sum_kh_ku^k-x_1u^q\bigl(1+(x_1+h_1)u\bigr)\Bigr).
\end{align*}
Reading off coefficients,
\[
f_q=-(h_q-x_1-y),\qquad f_{q+1}=-\bigl(h_{q+1}-(x_1+y)h_1-x_1^2\bigr).
\]
Since $\theta^{q^d}=\theta$, we have $\beta_{d-1}^q=\theta^{q^d}-\theta^q=-\beta_1$, so $y=-x_1$. With
$h_q=h_{q+1}=h_1=0$ we get $f_q=0$ and $f_{q+1}=x_1^2=\beta_1^{-2}\ne0$. Hence $m=q+1$.\qed

\begin{remark}
In Step~3 the vanishing of $f_q$ depends on the identity $\beta_{d-1}^q=-\beta_1$, which links the
factors $i=1$ and $i=d-1$ of Lemma~\ref{lem:exp}: the term $-x_1$ coming from $i=1$ cancels the
term $-y$ coming from $i=d-1$.
\end{remark}

\section{Examples and verification}\label{sec:ex}

We write $\F_9=\F_3[a]/(a^2-a-1)$. The following polynomials are irreducible, with the stated depth
and multiplicity:

\medskip
\begin{center}
\begin{tabular}{llccc}
\toprule
$q$ & $\wp$ & $r$ & regime & $m$\\
\midrule
$3$ & $t^3-t-1$ & $1$ & $r\le s$ & $2$\\
$3$ & $t^{18}-t^{10}+1$ & $2$ & $r>s$ & $4$\\
$5$ & $t^5-t-1$ & $1$ & $r\le s$ & $4$\\
$5$ & $t^{51}-2t^{50}-t^{26}+1$ & $2$ & $r>s$ & $6$\\
$9$ & $t^3-(a+1)t-(a+1)$ & $1$ & $r\le s$ & $2$\\
$9$ & $t^{10}+(a+1)t^9+t+1$ & $2$ & $r=s$ & $8$\\
$9$ & $t^{54}-a\,t^{28}+a\,t^{27}+a\,t+1$ & $3$ & $r>s$ & $10$\\
\bottomrule
\end{tabular}
\end{center}
\medskip

The ancillary file \texttt{verify\_wilson\_multiplicity.py} (Python~3, no dependency, exact
arithmetic in $\Fq[t]$, about fifteen seconds) checks:
\begin{itemize}
\item the product formula \eqref{eq:Fd} against the definition of $F_d$ over $\F_3$ for $d=2,3$;
\item for every monic irreducible $\wp$ of degree $2\le d\le7$ over $\F_3$, $2\le d\le4$ over $\F_5$
and $2\le d\le3$ over $\F_9$ (981 primes in all): the valuation $v_\wp(F_d+1)$, computed in
$\Fq[t]$ modulo a power of $\wp$ from \eqref{eq:Fd}, equals the value of Theorem~\ref{thm:main},
and $v_\wp(H_d)$ equals $m-1$ (Corollary~\ref{cor:H});
\item 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 $18$;
\item for all the examples, the coefficients $f_k$ of $F_d(\theta+u)$ computed in the residue
field $\Fq[t]/\wp$ from Lemma~\ref{lem:exp}: the first nonzero one is $f_m$; $f_k=-h_k$ for
$k<q$, with $h_k$ computed from the first identity of Lemma~\ref{lem:hasse}; the vanishing of
$\wp^{[j]}(\theta)$ for $2\le j\le\min(p^r-1,q+3)$, and $\wp^{[p^r]}\ne0$; $\beta_{d-1}^q=-\beta_1$; and, when $r>s$, $f_q=0$ and $f_{q+1}=\beta_1^{-2}$.
\end{itemize}
In small degree only the depths $0$ and $1$ occur, and the multiplicity $p-1$ of \cite{Th13} is then
exact. These computations check the statements; the proofs do not depend on them.

\subsection*{Acknowledgement}
Technical assistance: Claude (Anthropic).

\begin{thebibliography}{SSTT}
\bibitem[SSTT]{SSTT} J.~Sauerberg, L.~Shu, D.~S.~Thakur and G.~Todd, Infinitude of Wilson primes
for $\Fq[t]$, \emph{Acta Arith.} \textbf{157} (2013), 91--100.
\bibitem[Th13]{Th13} D.~S.~Thakur, Differential characterization of Wilson primes for $\Fq[t]$,
\emph{Algebra Number Theory} \textbf{7} (2013), 1841--1848.
\bibitem[Th24]{Th24} D.~S.~Thakur, Fermat--Wilson supercongruences, arithmetic derivatives and
strange factorizations, \emph{J. Th\'eor. Nombres Bordeaux} \textbf{36} (2024), no.~2, 481--491,
doi:10.5802/jtnb.1285; arXiv:2211.01076.
\end{thebibliography}

\end{document}