FERMAT QUOTIENTS OVER FUNCTION FIELDS
Jim Sauerberg and
Linghsueh Shu
Union College and
University of Vermont
April 25th, 1996
Abstract.
If P is an irreducible element of a polynomial ring over a finite field,
then one can define a Fermat quotient function associated to P . This is the direct
analog of the traditional Fermat quotient function defined over the rational numbers
using Fermat’s “Little” Theorem. This paper provides answers for several of the
central questions about the Fermat quotients over function fields.
Let p be a prime number and a be a positive integer. In 1640 Pierre de Fermat
showed that ap − a is divisible by p. So the quotient qa = (ap − a)/p is actually
an integer and is called the Fermat quotient of a with respect
to p. Niels Henrik
Abel was apparently the first to ask whether qa ≡ 0 mod p has any solutions with
1 < a < p, a question that Carl Jacobi answered in the affirmative by providing
the solutions a = 3 or 9 and p = 11, a = 14 and p = 29, and a = 18 and p = 37
[3]. In 1909 A. Wieferich showed that if p is an odd prime and q2 is non-zero
modulo p, then xp + y p + z p = 0 can have no solutions in the positive integers
with p - xyz, happily connecting Fermat’s “Little” Theorem and the first case of
his “Last” Theorem [11]. This and similar results motivated much study of the
Fermat quotients.
These studies can be divided into two areas. First, the vanishing of the Fermat
quotients: can one say exactly when qa is congruent to zero modulo p? Can one
give the exact power of p dividing qa [6]? Are there infinitely many pairs a, p so
that qa ≡ 0 mod p [1]? It seems not even known whether there are infinitely
many p so that q2 is not congruent to zero [7].
In the second area of study one fixes p and tries to give the order of the set
{qa : 0 ≤ a < p} considered inside Z/pZ. p
H. Vandiver proved that this set contains
√
at least p but no more than p − (1 + 2p − 5)/2 elements [10]. What if one
considers {qa : 0 ≤ a < pr } inside of Z/pr Z, for some fixed r? For r = 2 one
easily shows that there are exactly p − 1 distinct a such that qa ≡ 0 mod p2 , but
apparently little else is known.
Key words and phrases. Function Fields, Fermat Quotients.
The second author was partially supported by NSF Grant DMS-930198
2
JIM SAUERBERG AND LINGHSUEH SHU
In this paper we study the Fermat quotients of function fields defined over finite
fields. Let Fq be the field of q elements, for q a power of the prime number p.
Suppose P is an irreducible polynomial in Fq [x], the polynomial ring with constant
field Fq . Then Fq [x]/(P ) is a field of order q deg(P ) . Since the multiplicative group
deg(P )
of a finite field is cyclic, the polynomial Aq
− A is divisible by P in Fq [x] for
any A ∈ Fq [x]. Thus it makes sense to define the Fermat quotient of A with respect
to P to be the polynomial
QP (A) =
Aq
deg(P )
−A
P
.
Summary of the main results.
1) Suppose 1 ≤ deg(A) < deg(P ) for P an irreducible polynomial. Let be the
largest integer so that A = (A0 )p for some A0 in Fq [x]. Then ordP QP (A) =
p − 1.
2) There are infinitely many pairs A, P in Fq [x] with P irreducible and 1 ≤
deg(A) < deg(P ) such
that QP (A) ≡ 0 mod P . Further, for aδ fixed r ≥ 1,
r
QP (A) ≡ 0 mod P for infinitely many P if and only if A is a p r -th power in
Fq [x], where pδr −1 ≤ r < pδr .
r
3) For each r ≥ 1, the map from {A ∈ Fq [x] : deg(A) < deg(P r )} to
Fq [x]/(P )
taking A to QP (A) is Fq -linear. Its image has dimension d − 1 − (d − 1)/p if
r = 1, and at least (r − 1)d if r ≥ 2, where d = deg(P ) and [ · ] is the greatest
integer function.
Notice that part 2) follows from part 1), and that our results depend only on the
degree of the polynomial P , not on the polynomial itself. The reason our results
are more complete than those known over number fields is in large part because
the characteristic of our field is finite, so the P -adic expansion of QP (A) is easier
to handle.
n
In the first section we fix n ≥ 1 and consider the (xq − x)-adic expansion of
n
Aq − A, which is relatively simple and can be given explicitly.
This leads us to
n
n
consider what we call the Q-quotients, Qn (A) = Aq − A / xq − x , and prove
results about them similar to those summarized above. In the second, shorter
n
section we use these results, an d the fact that xq − x is the product of the monic
irreducible polynomials in Fq [x] of degree dividing n, to study the Fermat quotient
over rational function fields.
The authors both wish to express their deep appreciation to Professor Y. Hellegouarch for his generous attitude and many useful suggestions while they were
preparing this article.
§1. Q-Quotients
Let Fq be the finite field with q elements, with q a power of the prime p. Fix a
positive integer n, and for A an element of the polynomial ring Fq [x] consider the
quotient
n
Aq − A
.
xq n − x
FERMAT QUOTIENTS OVER FUNCTION FIELDS
A=
3
P
Aq
ai xi , for ai ∈ Fq , then we have
X
n
n
−A=
ai q xiq − ai xi
i=0
n
i≥0
=
X
i≥1
=
X
n
ai xiq − xi
n
n
n
n
ai xq − x xi−1 x(i−1)(q −1) + x(i−2)(q −1) + · · · + xq −1 + 1 ,
i≥1
n
which is divisible by xq − x. So it is reasonable to let Qn be the function on Fq [x]
defined by
n
Aq − A
Qn (A) = qn
.
x −x
To give a title to this section, we call Qn the Q-quotient function (leaving the fixed
n to be understood). We first give the basic properties of Qn , all of which may be
proven with trivial calculations.
Proposition 1.1. For all polynomials A and B in Fq [x] and constants c in Fq ,
1) Qn (A + B) = Qn (A) + Qn (B).
2) Qn (cA) = cQn (A).
n
3) Qn (AB) = (xq − x)Qn (A)Qn (B) + BQn (A) + AQn (B).
n
n
Remark: if σ is the Frobenius action on Fq [x]/(xq − x), i.e. σ(A) = Aq , then
Qn is a σ-derivation, that is, Qn is an Fq -vector space map on Fq [x] and Qn (AB) =
Qn (A)B + σ(A)Qn (B). The quotient discussed here is called a standard Galois
derivative of the first order [5].
Pm
n
Corollary 1.2. If A = i=0 Ai (xq − x)i ∈ Fq [x] for any Ai in Fq [x], then
A0 − A X qn
iq n −1
qn
iq n
Qn (A) = Qn (A0 ) + qn
+
Ai (x − x)
+ Qn (Ai )(x − x)
.
x − x i=1
m
n
n
n
In particular, Qn A + B(xq − x) ≡ Qn (A) − B mod (xq − x)q −1 .
For r ≥ 1, define the map Qn,r to be the composition
Qn
n
Qn,r : Fq [x]−−→Fq [x] −→ Fq [x]/(xq − x)r ,
and let En,r be the Fq -vector space {A ∈ Fq [x] : deg(A) < rq n }. So Qn,r (A) = 0
n
means that Qn (A) is divisible by (xq − x)r .
Corollary 1.3.
1) Qn,r is a σ-derivation on Fq [x], and so is a Fq -vector space map.
2) The kernel of Qn,1 is (En,1 )p and Qn,1 En,1 ∼
= En,1 /(En,1 )p is a vector subspace
n
of Fq [x]/(xq − x) of dimension q n − q n /p.
Proof. Part 1 is clear. For part 2, by Proposition 1.1, Qn,1 (x) = 1 and Qn,1 (xj ) =
n
n
qX
qX
−1
−1
j
j−1
jx , and so Qn,1
aj x =
aj jxj−1 . This is zero if and only if p divides
j=0
j=1
j for all j with aj 6= 0. The result then follows. 4
JIM SAUERBERG AND LINGHSUEH SHU
Lemma 1.4. If A =
m
X
ai xi is a polynomial in Fq [x] of degree m < q n , then the
i=0
(x
q
n
qn
− x)-adic expansion of A
−A is
m
X
j
Pm
n
bj xq − x , where bj = i=j ai ji xi−j
j=1
n
is a polynomial of degree less than q . In particular, Qn (A) =
j−1
qn
x
−x
b
.
j
j=1
Pm
Proof. By the binomial theorem,
iq n
x
i X
k
i i−k qn
−x =
x
x −x .
k
i
k=1
n
So, since Aq =
X
n
ai xiq , we have
qn
A
−A=
m
X
n
ai xiq − xi
i=1
X
i i i−k qn
k
=
ai
x (x − x)
k
i=1
k=1
m
m X
X
j
n
i i−j
=
ai
x
xq − x .
j
j=1 i=j
m
X
The rest of the proof is then clear. n
We can now find the exact power of xq − x dividing A, for deg(A) < q n . For
an integer m let ordp (m) be the largest power of p dividing m.
m
X
Theorem 1.5. Let A =
ai xi be a polynomial in Fq [x] with degree m < q n , and
i=0
n
n
let = (A) = ordp gcd i |ai 6= 0 . Then (Aq − A)/(xq − x)p is a polynomial
in Fq [x] and
n
X i
n
Aq − A
ai xi−p mod (xq − x)p .
n
≡
q
p
(x − x)
p
p |i
Further, the polynomial appearing on the right hand side of the equivalence is nonn
zero and has degree less than q n . In particular, p is the largest power of xq − x
n
dividing Aq − A.
Proof. By Lemma 1.4 we have
qn
A
−A=
m
X
n
bj xq − x
j
j=1
Pm
for bj = i=j ai ji xi−j . For each j, the polynomial bj has degree less than q n . So
n
each bj is either zero or is not divisible by xq − x. Further, since the terms in each
FERMAT QUOTIENTS OVER FUNCTION FIELDS
5
n
n
ai ji is non-zero. Thus the exact power of xq − x dividing Aq − A equals j for
the smallest j such that ai ji 6= 0 for some i ≥ j.
Because we are in a ring with characteristic p, we may
make the last statement
more explicit. Let = (A) = ordp gcd{ i |ai 6= 0} and consider ai ji . By
definition of , if p does not divide i, then ai = 0. On the other hand, if p does
divide i, and 1 ≤ j < 2p with j 6= p , then ordp (j) < , so ji = 0 in Fq . Therefore
the first two possibly non-zero bj ’s are bp and b2p and
n
n
n
Aq − A = bp (xq − x)p + b2p (xq − x)2p + · · ·
X i
n
qn
p
≡ (x − x)
ai e xi−p mod (xq − x)2p .
p
p |i
To complete the proof note that, again by definition of, there is some i with ai 6= 0
and ordp (i) = . Since pi is not divisible by p, ai pi is non-zero, and so the sum
is non-zero. P
By definition of = (A), A has the form p |i ai xi . Since raising to the p -th
power is an automorphism of Fq , we see A is actually a p -th power. Thus we have
Corollary 1.6. Suppose 1 ≤ deg(A) < q n . Let be the largest integer so that
n
A = (A0 )p for some A0 in Fq [x]. Then p − 1 is the exact power of xq − x dividing
Qn (A).
When r ≥ 2 to find the dimension of the image of Qn,r we must combine Corollary 1.2 with Lemma 1.4. So we will have two cases: r < q n and r ≥ q n . As the
reader may already have noticed, just as in the classical case, the p-divisibility of
the binomial coefficients will now play an important role.
Proposition 1.7. For 2 ≤ r < q n , Qn,r En,r is a Fq -vector space of dimension
(r − 1)q n + #{i ∈ N : r ≤ i ≤ q n − 1, ri 6= 0}.
Pr−1
n
Proof. For A in En,r uniquely express A as j=0 Aj (xq − x)j , with deg(Aj ) < q n .
Then by Corollary 1.2, we have
Qn,r (A) = Qn,r (A0 ) −
r−1
X
n
Aj (xq − x)j−1 .
j=1
Writing A0 =
Pqn −1
i=0
ai xi then, by Lemma 1.4, there are bj ’s so that


r−1
X
qn
n
Qn,r (A) = 
bj − Aj (x − x)j−1  + br (xq − x)r−1 ,
j=1
n
in Fq [x]/(xq − x)r . Since the Aj ’s are arbitrary polynomials of degree less than
q n , the sum in the previous equality varies over all polynomials of degree less than
(r − 1)q n . Finally,
n
qX
−1 i i−r
br =
ai
x
6
JIM SAUERBERG AND LINGHSUEH SHU
is non-zero if and only if ri 6= 0 or ai 6= 0. Since r is fixed and the ai are arbitrary
for all i the result is obtained. One may, in principle, obtain the precise dimension of the image for any given
r by using ordp (k!) = (k − sk )/(p − 1), where sk is sum of the digits of the base p
expansion of k. For example
Corollary 1.8.
1) For r = pt < q n , the dimension of Qn,r (En,r ) is q n r − q n /p.
2) For r = pt − 1 < q n , the dimension of Qn,r (En,r ) is q n r − q n r/(r + 1).
For r ≥ q n we give only a lower bound for the dimension of the image of Qn,r .
Proposition 1.9. For q n ≤ r, Qn,r En,r is a Fq -vector space of dimension at
least (r − 1)q n .
Pr−1
n
Proof. Write A = i=0 Ai X i for deg(Ai ) < q n and X = xq − x. Define bj (Ai )
to be the polynomial of degree less than q n constructed from Ai as in Lemma 1.4,
n
qX
−1
so b0 (Ai ) = Ai and Qn (Ai ) =
bj (Ai )X j−1 . Finally, write r = mq n + s for
j=1
0 ≤ s < q n and take A0 = 0. Then one combines this formula for Qn (Ai ) with
Corollary 1.2 to see
Qn,r (A) =
n
qX
−1
−Aj X j−1 +
n
bj (Ai ) − Aiqn +j X iq +j−1
i=1 j=0
j=1
+
n
−1
m−1
X qX
s−1
X
n
bj (Am ) − Amqn +j X mq +j−1 + bs (Am )X r−1 .
j=0
Now suppose that the Ai ’s are chosen starting with i = 1, then i = 2, 3, 4, . . . . For
1 ≤ j ≤ q n −1, the coefficient of X j−1 in the expansion of Qn,r (A) is Aj and so may
be any polynomial of degree less than q n . Next, for any j the value bj (A1 ) is now
n
determined, and so Aqn +j controls the coefficient of X q +j−1 . Thus this coefficient
may be any polynomial of degree less than q n . One continues in this way to see that
the coefficient of each X t , for 1 ≤ t ≤ r − 2, is At+1 plus a previously determined
constant, with At+1 arbitrary. Therefore, for 0 ≤ t ≤ r − 2 the coefficient of X t
is arbitrary and so the dimension of the set is at least (r − 1)q n . This finishes the
proof. n
The proof really only shows that the affine sub-space {Qn,r (xq A0 ) : deg(A0 ) <
(r − 1)q n } has dimension (r − 1)q n over Fq . A little experimentation will quickly
convince the reader that giving a more accurate bound for the dimension in the
theorem is quite complicated. In any case, we will see that it is only the lower
bound that is useful in the next section.
§2. Fermat Quotients
Now we look at the generalization of the traditional Fermat quotients to rational
FERMAT QUOTIENTS OVER FUNCTION FIELDS
7
d
d in Fq [x]. For any polynomial A in Fq [x] the quotient (Aq − A)/P is actually a
polynomial in Fq [x], and so we may consider the function QP on Fq [x] defined by
d
Aq − A
Aq
QP (A) =
=
P
deg(P )
−A
P
.
We call this function the Fermat quotient for rational function fields. G. Eisenstein
noted that
qa enjoys the properties qa+pb ≡ qa −
the traditional Fermat quotient
b mod p and qab ≡ aqb + bqa mod p [3, page 105]. Proposition 2.1 gives the
similar, easily proven, properties of QP .
Proposition 2.1. For all polynomials A and B in Fq [x] and constants c in Fq
1) QP (A + B) = QP (A) + QP (B).
2) QP (cA) = cQP (A).
3) QP (AB) = P QP (A)QP (B) + BQP (A) + AQP (B).
deg(P )
4) QP (AB) = QP (A)B +σ(A)QP (B) for σ(A) = Aq
, so QP is a σ-derivation.
Pm
q d −1
5) QP (A + BP ) ≡ QP (A) − B mod P
. More generally, for A = i=0 Ai P i ,
m
QP (A) = QP (A0 ) +
d
d
A0 − A X
+
Ai P iq −1 + QP (Ai )P iq .
P
i=1
Similar to Section 1, define QP,r to be the composition
QP
QP,r : Fq [x]−−→Fq [x] −→ Fq [x]/(P r ),
and let Er be the Fq -vector space {A ∈ Fq [x] : deg(A) < deg(P r )}. Below we will
study the size of QP,r (Er ), giving an exact answer for r = 1 and a lower bound,
otherwise.
One reason why it is difficult to give the size of the set {qa : 0 ≤ a < pr } in Z/pr Z
is because the traditional Fermat quotient is not even additive: qa+b is generally
not congruent to qa + qb modulo p. H. Vandiver √
was able to show that when r = 1
√
the set in question has between p and p − (1 + 2p − 5)/2 distinct elements. The
next corollary gives the function field version of Vandiver’s result.
Corollary 2.2.
1) For each r ≥ 1, QP,r is a Fq -vector space map.
p
p
∼
2) The kernel
h ofiQP,1 is E1 , and QP,1 (E1 ) = E1 /E1 is a vector space of dimension
d−1−
d−1
p
, for d = deg(P ) and [·] the greatest integer function.
d−1
This shows that there are exactly q [ p ]+1 − q distinct polynomials A ∈ Fq [x]
with 1 ≤ deg(A) < d so that QP (A) is divisible by P . This is not known for the
traditional Fermat quotients.
Thanks to Johnson [7], one does have a practical method for determining the
traditional Fermat quotient for any given a. Let m be the smallest integer so that
at
am ≡ ±1 mod p , and let am = ±1 + tp for some t. Then qa ≡ ∓
mod p . In
m
our case, if m is the smallest integer so that Am ≡ c mod P for some c ∈ Fq × ,
and Am = c + T P for some polynomial T , then one easily shows that
AT
8
JIM SAUERBERG AND LINGHSUEH SHU
It seems much easier to compute t than to compute T and c. Theorem 2.3 below
will provide us with a somewhat more explicit representation for QP (A).
Q
d
Recall that xq − x = f (x), where the product is over all monic irreducible
d
polynomials in Fq [x] of degree dividing d. So we have ordP xq − x = 1. Applying
this to Theorem 1.5, we prove
m
X
Theorem 2.3. Let P be an irreducible polynomial of degree d, A =
ai xi be
i=0
a polynomial in Fq [x] of degree m < d, and let = (A) = ordp gcd i |ai 6= 0 .
d
Then P p divides Aq − A in Fq [x] and
X d
Aq − A
i
p
i−p
≡V
ai x
mod P p ,
p
P
p
p |i
d
where V P = xq − x. Further, the polynomial on the right hand side of the equivalence is relatively prime to P . In particular, p is the exact power of P dividing
d
Aq − A.
Proof. By Theorem 1.5 we have
X
m
p
d
i
i−p
qd
ai x
xq − x
mod P 2p .
A −A≡
p
p |i
As p is positive, all terms in the sum have degree less that d, so the polynomial
the sum constitutes is relatively prime to P . Since V is also relatively prime to P
the rest of the theorem then follows. Using the theorem we can now answer several of the questions in the introduction.
Corollary 2.4. Let P be an irreducible polynomial.
1) Suppose 1 ≤ deg(A) < deg(P ) and is
the largest integer so that A = (A0 )p for
some A0 in Fq [x]. Then ordP QP (A) = p − 1.
2) There are infinitely many pairs A, P in Fq [x]
with P irreducible and 1 ≤
deg(A) < deg(P ) such that QP (A) ≡ 0 mod P .
3) There are no irreducible polynomials A with deg(A) < deg(P ) and QP (A) ≡
0 mod P .
4) For a given A ∈ Fq [x] and r ≥ 1, QP (A) is divisible by P r for infinitely many
irreducible polynomials P if and only if A is a pδr -th power in Fq [x] for pδr −1 ≤
r < p δr .
Proof. Part 1) follows from the theorem, and parts 2) and 3) from part 1). For part
4), if QP,r (A) = 0 for infinitely many P , then QP,r (A) must be zero for infinitely
many P with deg(A) < deg(P ). By part 1), this occurs exactly when A is a pδr -th
power in Fq [x]. The polynomial V played an innocuous role in the proof of the theorem above.
However, when r ≥ 2, it does not. To explain this, let us consider QP,r (A0 ) for
Pd−1
A0 = i=0 ai xi . By Lemma 1.4,
QP (A0 ) =
d−1
X
bj V j P j−1
FERMAT QUOTIENTS OVER FUNCTION FIELDS
9
where the bj = bj (A0 ) are polynomials of degree at most d − 1 − j. Since V is a
polynomial of degree q d − d, this is no longer the P -adic expansion of QP (A0 ). To
find this expansion, we would need to know the P -adic expansion of V . However,
part 1) of Corollary 2.4 and a translation of the proof of Proposition 1.6 prove
Proposition 2.5. For 2 ≤ r, the image of QP,r : Er → Fq [x]/(P r ) has dimension
)−1 at least (r − 1) deg(P ) and the kernel has dimension at least deg(P
.
δ
r
p
In some sense the imperfection of the proposition indicates that the function
field Fermat quotient is one step easier than the traditional Fermat quotient, as the
difficulties do not arise modulo P but modulo P 2 .
We finish by mentioning possible generalizations. Over the rational numbers
there is a theory of Fermat quotients modulo composite numbers (see [1]). If
we were to take P to be non-irreducible, theQdifficulties with V would become
much more pronounced. For instance, if P = i Pi ei is a factorization of P into
d
D
irreducibles, then we must replace xq − x by (xq − x)e , for D the least common
multiple of {deg(Pi )} and e the largest of the ei . Doing this is not necessarily
n
impossible, though, since if P = xq − x then QP is the Q-quotient studied in
Section 1.
It is also possible to use the Carlitz module action on Fq [x] to define and study
“Fermat-Carlitz” quotients. This study was undertaken by V. Mauduit and we
refer the reader to her paper [9]. Similarly, the analog of Fermat’s Last Theorem
using Carlitz modules is known as the Fermat-Goss theorem and was solved by L.
Denis [2]. Finally, in view of the subject of this paper, for studies of the classical
version of Fermat’s Last Theorem over function fields we direct the reader to [4]
and [8].
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
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L. Dickson, “History of the Theory of Numbers”, vol. I, Chelsea Publishing, New York,
1952, Chapter IV.
N. Greenleaf, On Fermat’s equation in (t), Amer. Math. Monthly 76 (1969), 808–809.
Y. Hellegouarch, Galois Calculus and Carlitz Exponentials, in “The Arithmetic of Function
Fields” (D. Goss, D. Hayes, and M. Rosen, eds.), Walter de Gruyter, 1992, pp. 33–50.
W. Johnson, On the p-divisibility of the Fermat Quotients, Math. Comp. 32 (1978), no. 141,
297–301.
W. Johnson, On the non-vanishing of the Fermat quotients (mod p), J. reine angew. Math.
292 (1977), 196–200.
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of the Third International Conference”, Lecture Note Series of the London Math. Soc. (S.D.
Cohen and H. Niederreiter, ed.), Cambridge University Press, 1995.
H. Vandiver, An aspect of the linear congruence with applications to the theory of Fermat’s
quotient, Bull Amer Math Soc 22 (1915), Nov, 61–67.
A. Wieferich, Zum letzten Fermatschen Theorem, J. reine angew. Math. 136 (1909), 293–
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C