Splitting Fields for Characteristic Polynomials
of Matrices with Entries in a Finite Field.
Eric Schmutz
Mathematics Department, Drexel University,Philadelphia, Pennsylvania, 19104.
Abstract
Let Mn be the set of all n × n matrices with entries in the finite field Fq . Let X(A)
be the degree of the splitting field of the characteristic polynomial of A, and let µn
be the average degree:
X
1
µn =
X(A).
|Mn |
A∈Mn
A theorem of Reiner is used to prove that,as n → ∞,
√
µn = eB n/ log n(1+o(1)) ,
where B is an explicit constant.
Key words: Finite field, splitting field, random matrix, characteristic polynomial
1
Introduction
If f ∈ Fq [x],let X(f ) be the degree of the splitting field of f , i.e. the smallest
Q
d such that f factors as a product of linear factors f = (x − ri ), with all
i
the roots ri in Fqd . Mignotte and Nicolas [12],[16], and Dixon and Panario [2]
asked how large X(f ) is for a typical polynomial f . More precisely, let Pn be
the set of all monic degree n polynomials with coefficients in the finite field Fq ,
and let Pn be the uniform probability measure: Pn ({f }) = q −n for all f ∈ Pn .
They studied the asymptotic distribution of the random varaible log X, and
noted the strong analogies between this problem and the “Statistical Group
Theory”of Erdős and Turán[3],[4]. Dixon and Panario[2] also estimated the
P
the average degree q −n
X(f ), i.e. the expected value of X. Hansen and
f ∈Pn
Email address: [email protected] (Eric Schmutz).
Preprint submitted to Elsevier
1 November 2007
Schmutz compared random polynomials with the characteristic polynomials
of random invertible matrices. Based on the results in [8], it was reasonable to
conjecture that a matrix-analogue of the Dixon-Panario theorem should hold.
The number of matrices having a given characteristic polynomial depends, in
a complicated way, on the degrees of the irreducible factors that the polyno2
mial has (Reiner[17]). Select a matrix A uniform randomly from among all q n
matrices having entries in the finite field Fq , and let f be the characteristic
polynomial of A. Hence the characteristic polynomial f is being selected randomly, but not uniform randomly, from among all monic degree n polynomials
2
in Fq [x]. Let µn =the average, over all q n matrices A, of the degree of the
splitting field of the characteristic polynomial of A. We prove here that, as
n → ∞,
√
µn = eB
s
where B = 2 2
R∞ log(1+t)
0
e−t −1
n/ log n(1+o(1))
,
dt = 2.990 . . . The constant B has appeared previ-
ously in the study of random permutations[5] and random polynomials[2].
The remainder of this section specifies the paper’s symbols and notations.
Definitions are listed here in quasi-alphabetical order, and may be used later
without comment.
s
• B=2 2
R∞ log(1+t)
et −1
0
∞
Q
• c∞ =
(1 −
j=1
1
)
2j
dt = 2.990 . . .
= .288 . . .
• | · | = degree: if f is a polynomial in Fq [x], then |f | is its degree.
• F (u, r) :=
r
Q
i=1
(1 −
1
)
ui
for positive integers u, r, and F (u, 0) := 1
• gf = divisor (in Fq [x]) of f that is minimal among those monic divisors g
of f for which X(f ) = X(g).
• Gn = {gf : f ∈ Pn }.
• hf = gff
• Ik =the set of monic irreducible polynomials of degree k in Fq [x]
• Hn =
n
P
k=1
1
,
k
the n’th Harmonic number.
• Ik = |Ik,q |, the cardinality of Ik (The font distinguishes the set from its
cardinality. To save space, q is implicit.)
• I=
•
•
•
•
•
∞
S
k=1
Ik =monic polynomials in Fq [x] that are irreducible over Fq
Λm = set of partitions of m having distinct parts.
Λ̃m = partitions of m (not necessarily distinct parts.)
λ ` m The conventional notation for λ ∈ Λ̃m .
Mn =set of all all n × n matrices with entries in the finite field Fq .
Mn = probability measure on Pn defined by Mn ({f }) = the proportion of
matrices in Mn whose characteristic polynomial is f . (To save space, q is
2
implicit.)
• mφ (f ) = the multiplicity of φ in f : for φ ∈ I and f ∈ F[x], φmφ (f ) divides
f but φmφ (f )+1 does not divide f .
P
2
• µn = q −n
X(A).
A∈Mn
• Pn = set of all q n monic polynomials of degree n in Fq [x].
• Pn = uniform probability measure on Pn : Pn ({f }) = q −n .
• S = set of polynomials in Pn that factor completely, i.e. have all their roots
in Fq .
• X(f ) = degree of the splitting field of f , if f ∈ Fq [x].
• X(A) = X(f ), if A is a matrix with characteristic polynomial f.
• X(λ) = least common multiple of the parts of λ, if λ is an integer partition.
The last three definitions overload the symbol X. However this is natural and
consistent: the degrees of the irreducible factors of a polynomial f ∈ Fq [x]
form a partition of |f |, and it is well known that the degree of the splitting
field of f is the least common multiple of the degrees of its irreducible factors.
2
Comparison of the probability measures
There is an explicit formula for the number of matrices with a given charactersitic polynomial:
Theorem 1 (Reiner[17])If f =
Q
φmφ (f ) is a polynomial in Pn , then
φ
Mn ({f }) = Q
q −n F (q, n)
F (q |φ| , mφ (f ))
φ∈I
(See also Crabb[1], Fine-Herstein[6]), and Gerstenhaber[10]). In order to apply
Theorem 1, we need a simple lemma:
Lemma 1 For all non-negative integers a, b, and all prime powers q,
F (q, a + b) ≥ F (q, a)F (q, b)
Proof. Since q a+j ≥ q j for all j, we have
F (q, b) =
b
Y
j=1
(1 −
b
Y
1
1
)
≤
(1 − a+j ).
j
q
q
j=1
3
But then
F (q, a + b) = F (q, a)
b
Y
(1 −
j=1
1
q a+j
) ≥ F (q, a)F (q, b).
•
In one direction, there is a simple relationship between the probability measures Pn and Mn :
Proposition 1 For all A ⊆ Pn ,
Mn (A) ≥ c∞ Pn (A).
Proof. It is obvious from the definition of F that, for all u > 1 and all
non-negative integers r,
0 < F (u, r) ≤ 1.
(1)
If f ∈ A, then by Theorem 1 and (1),
Mn ({f }) ≥ F (q, n)q −n ≥ c∞ q −n .
Summing over f ∈ A we get Proposition 1.
•
It is interesting to note that the inequality in Proposition 1 has no analogue
in the other direction:
n ({f })
= ∞.
Proposition 2 lim sup max M
Pn ({f })
f ∈Pn
n→∞
Proof. Consider f =the product of all irreducible polynomials of degree less
m
P
than or equal to m. In this case n = nm =
kIk , where Ik is the number of
k=1
monic irreducible polynomials of degree k , and
Mn ({f }) = q −n Q
m
F (q, n)
k=1
(1 −
(2)
1 Ik
)
qk
4
Since F (q, n) ≥ c∞ , it suffices to prove that
m
Q
(1 −
k=1
1 Ik
)
qk
= o(1) as m → ∞.
The following bounds appear on page 238 of Mignotte[11]:
1X
1
2
µ(d)q k/d ≥ q k ( − k/2 ).
k d|k
k kq
q k ≥ Ik =
(3)
Using first the inequality log(1 − x) ≤ −x, and then the inequality on the
right side of (3), we get
m
X
1
1
1
+ O(1) = O( ).
(1 − k )Ik ≤ exp −
q
m
k=1 k
k=1
m
Y
!
•
3
Non-existence of Jordan forms.
Neumann and Praeger[13] estimated the probability that the characteristic
polynomial of a random matrix has none of its roots in Fq . In this section
we estimate the probability that the characteristic polynomial of a random
matrix has all of its roots in Fq .
Theorem 2 For all prime powers q and all positive integers n,
!
c∞ q
−n
!
n+q−1
n+q−1
≤ Mn (S) ≤ q −n
q−1
q−1
Proof. Suppose f ∈ Fq [x]. Then f ∈ S iff two conditions are satisfied:
(1) The multiplicities of the linear factors form composition of n into nonP
negative integer parts:
mx−α (f ) = n, and
α∈Fq
(2) mφ (f ) = 0 for all φ ∈
Id ; no irreducible factor has degree larger than
d≥2
one.
S
It is well known that there are exactly n+q−1
compositions of n into q nonq−1
negative parts. It therefore suffices to prove that, for any f ∈ S, c∞ q −n ≤
Mn ({f }) ≤ q −n .
Q
Suppose f =
Mn ({f }) = Q
(x − α)mx−α (f ) and
α∈Fq
q −n F (q,n)
.
F (q,mx−α (f ))
P
mx−α (f ) = n. By Theorem 1,
α∈Fq
Lemma 1 implies that
Q
α∈Fq
α∈Fq
5
F (q, mx−α (f )) ≥ F (q, n).
Therefore, Mn ({f }) ≤ q −n .
For the other direction, apply Proposition 1 with A = {f }.
•
Note that, for fixed q,
∞. Hence we have
n+q−1
q−1
q −n approaches zero exponentially fast as n →
Corollary 1 For almost every matrix A ∈ Mn , there is no matrix B ∈ Mn
such that B is in Jordan canonical form and is similar to A.
Comment: The corollary is stated rather glibly. A clumsier, but more precise
statement is that, for every > 0 and every prime power q, there is an integer
< . The subscripts in the number N,q
N,q such that, for all n > N,q , q|S|
n2
are included to emphasize the fact that N,q depends on both and q. It is
interesting to compare
thefixed-q-large-n limit, namely zero, with the fixedn-large-q limit: lim n+q−1
q −n = n!1 > 0.
q−1
q→∞
4
Average degree
An easy consequence of Proposition 1 is a lower bound for the average degree:
√
√ log n ))
B n/ log n(1+O( log
log n
Lemma 2 µn ≥ e
Proof. By Theorem 1,
µn =
X
f ∈Pn
Q
q −n F (q, n)
X(f )
F (q |φ| , mφ (f ))
φ
Again using the inequality (1), we get
Q
F (q |φ| , mφ (f )) ≤ 1 and F (q, n) ≥ c∞ .
φ
Therefore µn ≥ c∞
P
q −n X(f ). The lower bound then follows directly from
f ∈Pn
the results of Dixon and Panario [2].
•
The upper bound for µn is harder because, as Proposition 2 suggests, we don’t
have convenient upper bounds the Mn -probabilities of events. Two lemmas are
needed for the proof.
6
Let D(f ) = {g : g divides f in Fq [x] and X(g) = X(f )}. Then D(f ) is a
non-empty finite set that is partially ordered by divisibility. For each f , we
can choose a minimal element gf ∈ D(f ).
Lemma 3 The irreducible factors of gf appear with multiplicity one and have
different degrees.
Proof. Suppose that, on the contrary, φ1 and φ2 are irreducible polynomials
g
of degree d and that φ1 φ2 divides gf . Let g = φf1 . Then X(g) = X(f ) and g
divides gf . This contradicts the minimality of gf .
•
Lemma 4 If |gf | = d, then Mn ({f }) ≤ 4Md ({gf })Mn−d ({hf }).
Proof. Since f = gf hf , we have mφ (f ) = mφ (gf )+mφ (hf ). It therefore follows
from Lemma 1 that
F (q |φ| , mφ (f )) ≥ F (q |φ| , mφ (gf ))F (q |φ| , mφ (hf )).
(4)
Combining (4) with Theorem 1, we get
Mn ({f }) =
qn
Q
F (q, n)
F (q, n)
≤ nQ
|φ|
|φ|
q
F (q , mφ (f ))
F (q , mφ (gf ))F (q |φ| , mφ (hf ))
φ
=
Finally,
φ
F (q, n)
Md ({gf })Mn−d ({hf }).
F (q, d)F (q, n − d)
F (q,n)
F (q,d)F (q,n−d)
≤
1
F (q,d)
≤
1
c∞
≤ 4.
•
q
Theorem 3 µn = exp B
n
(1
log n
+
√ log n ))
O( log
log n
.
Proof.
µn = E(X) =
X
Mn (f )X(f )
f ∈Pn
=
X
g∈Gn
X(g)
X
Mn ({gh}),
h
where the inner sum is over all h for which ggh = g. By Lemma 4, this is less
than
X
X
X(g)4M|g| ({g})
Mn−|g| ({h})
g∈Gn
h
7
Since the inner sum is less that one, we have
µn ≤ 4
X
X(g)Mn ({g}).
(5)
g∈Gn
If g ∈ Gn , then the degrees of the irreducible factors of g form a partition
of |g| into distinct parts. Grouping together polynomials that have the same
partition, we see that the right side of (5) is less than or equal to
4
n
X
X
LCM (λ1 , λ2 , . . .)q −m
m=1 λ∈Λm
Y
i
If λ has distinct parts λ1 , λ2 , . . ., then
Iλi
.
(1 − q1λi )
Q
i
known that Iλi ≤
(6), we get
µn ≤
q λi
.
i
(1 − q1λi ) ≥
(6)
m
Q
i=1
(1 − q1i ) ≥ c∞ . It is well
Putting these two estimates back into the right side of
n
n
X LCM (λ1 , λ2 , . . .)
X LCM (λ1 , λ2 , . . .)
4 X
4 X
≤
(7)
c∞ m=1 λ∈Λm
λ1 λ2 · · ·
c∞ m=1 λ∈Λ̃
λ1 λ2 · · ·
m
This last quantity has appeared previously in the study of random permutations [9],[18] where it was approximated by coefficient of xn in the generating
function
∞
Y
x2p x3p
1
p
(1
+
x
+
+
+ · · ·).
(1 − x)2 primes p
2
3
The conclusion (see appendix) was that the right side of (7) is
s
!
n
log log n
exp B
)) .
(1 + O( √
log n
log n
(8)
•
References
[1] M.C.Crabb,Counting nilpotent endomorphisms, Finite Fields and Their
Applications, 12 (2006) 151–154.
[2] John Dixon and Daniel Panario, The degree of the splitting field of a random
polynomial over a finite field, Electron. J. Combin. 11 (2004), no. 1, Research
Paper 70, 10 pp. (electronic).
[3] Paul Erdős and Paul Turán, On some problems of a statistical group theory
III, Acta Math. Acad. Sci. Hungar. 18 (1967) 309–320.
8
[4] Paul Erdős and Paul Turán, On some problems of a statistical group theory IV,
Acta Math. Acad. Sci. Hungar. 19 (1968) 413–435.
[5] Steven R. Finch, Mathematical constants. Encyclopedia of Mathematics and its
Applications, 94. Cambridge University Press, Cambridge, 2003 ISBN 0-52181805-2, page 287.
[6] N.J. Fine and I.N. Herstein, The probability that a matrix be nilpotent, Illinois
J.Math2(1958) 499-504.
[7] Jason Fulman, Random matrix theory over finite fields, Bull. Amer. Math. Soc.
(N.S.) 39 (2002), no. 1, 51–85
[8] Jennie C. Hansen and Eric Schmutz, How random is the characteristic
polynomial of a random matrix? Math. Proc. Cambridge Philos. Soc. 114
(1993), no. 3, 507–515.
[9] William M.Y. Goh and Eric Schmutz, The expected order of a random
permutation.Bull. London Math. Soc.23 (1991), no. 1, 34–42.
[10] Murray Gerstenhaber, On the number of nilpotent matrices with coefficients in
a finite field. Illinois J. Math. 5 1961 330–333.
[11] Maurice Mignotte, Mathematics for Computer Algebra, Springer Verlag 1992
ISBN 3-540-97675-2.
[12] M.Mignotte and J.L.Nicolas, Statistiques sur Fq [X]. Ann. Inst. H. Poincar Sect.
B (N.S.)19 (1983), no. 2, 113–121.
[13] Peter M. Neumann and Cheryl E. Praeger, Derangements and eigenvalue-free
elements in finite classical groups.
[14] Peter M. Neumann and Cheryl E. Praeger, Cyclic Matrices over Finite Fields
J. London Math. Soc. (2) 58 (1998), no. 3, 564–586.
[15] Daniel Panario, What Do Random Polynomials over Finite Fields Look Like?
Finite fields and applications, 89–108, Lecture Notes in Comput. Sci.,
2948, Springer, Berlin, 2004.
[16] J.L. Nicolas A Gaussian law on FQ [X], Topics in classical number theory, Vol.
I, II (Budapest, 1981), 1127–1162, Colloq. Math. Soc. Jnos Bolyai, 34, NorthHolland, Amsterdam, 1984.
[17] Irving Reiner, On the number of matrices with given characteristic polynomial.
Illinois J. Math. 5 1961 324–329.
[18] Richard Stong, The average order of a permutation. Electron. J. Combin.5
(1998), Research Paper 41, 6 pp. (electronic).
9
5
Appendix
At the suggestion of a referee, further details on the derivation of (8) from (7)
P LCM (λ1 ,λ2 ,...)
, so that the
are appended. For positive integers m, let Um =
λ1 λ2 ···
λ`m
n
√
P
right side of (7) is c4∞
Um . Let z = n/ log2 n, and let Bn be the coefficient
1
1−x
n
of x in G(x) =
m=1
∞
Q
(1 + xp +
primes p
x2p
2
+
x3p
3
+ · · ·).
In the middle of page 39 of [9] begins the proof that
Un = O(n)T1 T2 T3 ,
(9)
where
√
T1 ≤ n
z
Hnz
T3 ≤ Hnlog
4
!
n
= exp O(
) ,
log n
n
(10)
= exp O(log4 n log log n) ,
(11)
and
T2 ≤ Bn .
(12)
Thus
√
!
n
Un ≤ Bn exp O(
) .
log n
(13)
It is precisely the numbers Bn that were estimated, using a Tauberian then
P
orem[3], in §4 of [18]. (In Stong’s notation, Bn =
ak , and h(t) = (1 −
k=1
e−t )G(e−t )). His estimate was
√
!
n log log n
Bn = exp B n/ log n + O(
) .
log n
q
(14)
Due to the rapid growth of the numbers Bm , summation does not change the
error term in our estimate: combining (14) with (13), we get
√
n
q
X
n log log n
Um ≤ n max Um = exp(B n/ log n + O(
)).
m≤n
log n
m=1
10