The Degree of the Splitting Field of a Random Polynomial over a Finite Field

The Degree of the Splitting Field of a Random Polynomial over a Finite Field John D. Dixon and Daniel Panario School of Mathematics and Statistics Carleton University, Ottawa, Canada fjdixon,danielg@math.carleton.ca Mathematics Subject Classi cations: T06, 20B99 Abstract The asymptotics of the order of a random permutation have been widely studied. P. ErdÄos and P. Tur an proved that asymptotically the distribution of the logarithm of the order of an element in the symmetric group S n is normal with mean 2 (log n)2 and variance 3 (log n)3. More recently R. Stong has shown that the mean of the order is asymptotically exp(c p n= log n + O( p n log log n= log n)) where C = 2:99047 : : :. We prove similar results for the asymptotics of the degree of the splitting eld of a random polynomial of degree n over a nite eld. Introduction We consider the following problem. Let F q denote a nite eld of size q and consider the set P n (q) of monic polynomials of degree n over F q. What can we say about the degree over F q of the splitting eld of a random polynomial from P n (q)? Because we are dealing with nite elds and there is only one eld of each size, it is well known that the degree of the splitting eld of f() 2 P n (q) is the least common multiple of the degrees of the irreducible factors of f() over F q. Thus the problem can be rephrased as follows. Let be a partition of n (denoted ` n) and write in the form k 2 k 2 :::n kn where has k s parts of size s. We shall say that a polynomial is of shape if it has k s irreducible factors of degree s for each s. Let w( ; q) be the proportion of polynomials in P n (q) which have shape. If we de ne m( ) to be the least common multiple of the sizes of the parts of, then the degree of the splitting eld over F q of a polynomial of shape is m( ). The average degree of a splitting eld is given by E n (q) := `n w( ; q)m( ): An analogous problem arises in the symmetric group S n. A permutation in S n is of type = k 2 k 2 :::n kn if it has exactly ks cycles of length s for each s, and its order is then the electronic journal of combinatorics (2004), #R00

equal to m( ). If w( ) denotes the proportion of permutations in S n which are of type, then the average order of a permutation in S n is equal to E n := `n w( )m( ): We can think of m( ) as a random variable where ranges over the partitions of n and the probability of is w( ; q) and w( ) in the respective cases. Properties of the random variable m( ) (and related random variables) under the distribution w( ) have been studied by a number of authors, notably by ErdÄos and Tur an in a series of papers [, 2, 3] and [4]. In particular, the main theorem of [3] shows that in this case the distribution of log m( ) is approximated by a normal distribution with mean (log 2 n)2 and variance (log 3 n)3 in a precise sense. In our notation the theorem reads as follows. For each real x de ne ½ ª n (x) := ` n j log m( ) 2 (log n)2 + p x ¾ (log n) 3=2 : 3 Then for each x 0 > 0: w( )! p Z x e t2 =2 dt as n! uniformly for x 2 [ x 0 ; x 0 ] : 2¼ 2ª n (x) In particular, the mean of the random variable log m( ) is asymptotic to 2 (log n)2, but this does not imply that log E n (the log of the mean of m( )) is asymptotic to (log n)2 2 and indeed it is much larger. The problem of estimating E n was raised in [4], and the rst asymptotic expression for log E n was obtained by Goh and Schmutz [6]. The result of Goh and Schmutz was re ned by Stong [9] who showed that r µp n n log log n log E n = C log n + O ; log n where C = 2:99047::: is an explicitly de ned constant. The object of the present paper is to prove analogous theorems for the random variable m( ) under the distribution w( ; q). Actually, it turns out that these theorems hold for several important classes of polynomials which we shall now describe. Consider the classes: ² M (q): the class of all monic polynomials over F q. In this class the number of polynomials of degree n is q n for each n. ² M 2 (q): the class of all monic square-free polynomials over F q. In this class the number of polynomials of degree n is ( q )q n for each n: ² M 3 (q): the class of all monic square-free polynomials over F q whose irreducible factors have distinct degrees. In this class the number of polynomials of degree n is a(n; q)q n where, for each q, a(n; q)! a(q) := Q k ( + i k(q)q k ) exp( =k) as n! where i k (q) is the number of monic irreducible polynomials of degree k over F q (see [7] Equation () with j = 0). the electronic journal of combinatorics (2004), #R00 2

For x > 0 de ne n (x) := ½ ` n j log m( ) 2 (log n)2 > x p 3 (log n) 3=2 ¾ : Then for each of the classes of polynomials described above we have a weak analogue of the theorem of ErdÄos and Tur an quoted above, and an exact analogue of Stong's theorem. Theorem Fix one of the classes M i (q) described above. For each ` n, let w( ; q) denote the proportion of polynomials in this class whose factorizations have shape. Then there exists a constant c 0 > 0 (independent of the class) such that for each x there exists n 0 (x) such that w i ( ; q) c 0 e x=4 for all q and all n n 0 (x): () 2 n(x) In particular, almost all f() of degree n in M i (q) have splitting elds of degree exp(( 2 + o())(log n) 2 ) over F q as n!. Theorem 2 Let C be the same constant as in the Goh-Schmutz-Stong theorem. Then in each of the classes described above the average degree E n (q) of a splitting eld of a polynomial of degree n in that class satis es r log E n (q) = C n log n + O 2 Properties of w( ; q) µp n log log n uniformly in q. log n First consider the value of w( ; q) for each of the three classes. Let i s = i s (q) denote the number of monic irreducible polynomials of degree s over F q. Then (see, for example, [8]) we have q s = P djs di d so a simple argument shows that q s s i s qs s ( + 2q s=2 ) : Let ` n have the form k :::n kn. Since Pn (q) contains q n polynomials, and there are ways to select k irreducible factors of degree s, we have is +k k w( ; q) = q n µ is + k s = k s q sk s µ is + k s in M (q): Similarly, since there are ( q )q n polynomials of degree n in M 2 (q), and there are i s k ways to select k distinct irreducible factors of degree s, in this case we have µ is µ is w( ; q) = = q sks in M ( q )q n k s ( q 2 (q): ) k s the electronic journal of combinatorics (2004), #R00 3 k s

Finally, since there are a(n; q)q n polynomials of degree n in M 3 (q) and each of these polynomials has at most one irreducible factor of each degree, we get w( ; q) = a(n; q)q n µ i ks s = µ q i sks ks s a(n; q) k s when each part in has multiplicty, and w( ; q) = 0 otherwise. we also have k s w( ) = k 2 k 2:::n k nk!k 2!:::k n! : in M 3 (q) As is well known We shall use the notation n to denote the set of all partitions of n, n;k to denote the set of partitions k 2 k 2 :::n n k in which each ki < k and 0 n;k to denote the complementary set of partitions. It is useful to note that in M (q) and M 2 (q) we have w( ; q)! w( ) as q!. However, this behaviour is not uniform in. Indeed for each of these two classes the ratio w( ; q)=w( ) is unbounded above and below for xed q if we let range over all partitions of n and n!. This means we have to be careful in deducing our theorems from the corresponding results for w( ). In M 3 (q), we have w( ; q) = 0 whenever 2 0 n;2, and a simple computation shows that a(n; q)w( ; q)! w( ) as q! whenever 2 n;2 : Lemma 3 There exists a constant a 0 > 0 such that for all n and all prime powers q >. q a 0 and a(n; q) a 0 Proof. The rst inequality is satis ed whenever a 0 2, so it is enough to prove that the set of all a(n; q) has a strictly positive lower bound. We shall use results from [7, Theorems and 2]. In our notation [7] shows that a(q) increases monotonically with q starting with a(2) = 0:3967 : : :, and that for some absolute constant c we have ja(n; q) a(q)j c=n for all n. In particular, a(n; q) a(q) c=n a(2) c=n. Thus a(n; q) a(2) > 0 for all q whenever n > n 2 0 := b2c=a(2)c. On the other hand, as we noted above, in M 3 (q), a(n; q)w( ; q)! w( ) as q! whenever 2 n;2 and is 0 otherwise. Thus a(n; q) = 2 a(n; q)w( ; q)! w( ) = b(n), say, as q! : n 2 k;2 Evidently, b(n) > 0 (it is the probability that a permutation in S n has all of its cycles of di erent lengths). De ne b 0 := min fb(n) j n = ; 2; :::; n 0 g. Then the limit above shows that there exists q 0 such that a(n; q) b 2 0 whenever n = ; 2; :::; n 0 and q > q 0. Finally, choose a 0 2 such that =a 0 is bounded above by 2 a(2), 2 b 0 and all a(n; q) with n = ; 2; :::; n 0 and q q 0. This value of a 0 satis es the stated inequalities. the electronic journal of combinatorics (2004), #R00 4

We next examine some properties of the w( ; q) which we shall need later. In what follows, if := k :::n kn 2 n and ¹ := l :::m lm 2 m, then the join _ ¹ denotes the partition of m + n with k s + l s parts of size s. We shall say that and ¹ are disjoint if k s l s = 0 for each s: Lemma 4 Let a 0 be a constant satisfying the conditions in Lemma 3. Then for each class M i (q) we have w( _ ¹; q) a 0 w( ; q)w(¹; q) for all and ¹. On the other hand, if and ¹ are disjoint, then w( _ ¹; q) a 2 0 w( ; q)w(¹; q). We also have w( _ ¹) w( )w(¹), with equality holding when and ¹ are disjoint. Proof. First note that in each of the classes, w( _¹; q) is 0 if either w( ; q) or w(¹; q) is 0: Suppose neither of the latter is 0 and put r := w( _ ¹; q)=w( ; q)w(¹; q). First consider the class M (q). Then r can be written as a product of terms of the form µ is + k s + l s k s + l s = µ µ is + k s is + l s The numerator of this ratio counts the number of ways of placing k s + l s indistinguishable items in i s distinguishable boxes. The denominator counts the number of ways of doing this when k s of the items are of one type and l s are another, and so is at least as great as the numerator. Hence we conclude that r < a 0 in this case. Moreover, when and ¹ are disjoint then each term is equal to and so r = a 2 0. This proves the claim for the class M (q). Taking limits as q! also gives a proof of the nal statement. Now consider the class M 2 (q). In this case r=( q ) can be written as a product of terms of the form µ is = k s + l s k s µ µ is is The numerator counts the number of ways to choose k s + l s out of i s items, whilst the denominator is at least as large as i s is k s k s l s which counts the number of ways to choose k s +l s items when k s are of one type and l s are another type. This shows that each term is at most and so r ( q ) a 0 as required. Again, in this case, when the partitions are disjoint, each term is equal to and so r = q a 2 0. This proves the claim for the class M 2 (q), and the proof for the class M 3 (q) is similar (in this case w( _ ¹; q) is 0 unless and ¹ are disjoint). Lemma 5 For all partitions of the form [s k ] and all q we have in each of the classes M i (q). k s l s : w( s k ; q) a 0 k + (2s) k l s : the electronic journal of combinatorics (2004), #R00 5

Proof. Fix q and k and de ne v s := q sk k! ky µ q s s + j : j=0 Since i s q s =s we have w( s k ; q) a 0 v s for each of the three classes. We also note that Finally since v = k Y ( + j=q) k Y ( + j=2) = k + : k! k! 2 k k Y v s+ =v s = q k j=0 j=0 q s+ =(s + ) + j q s =s + j we obtain w( s k ; q) a 0 v s a 0 s k v so the result follows. j=0 k Y µ k q k qs s s + = ; s + Lemma 6 Let = k :::n kn be a partition of n. The following are true for each of the classes M i (q). (a) If each k s k for some xed integer k > 0, then w( ; q) a 0 e 2k(k ) w( ): (b) There exists a constant c such that, if each k s, then w( ; q) c w( ). Proof. (a) For each of the classes we have w( ; q) a 0 n Y j=0 q sks k s! (i s + k s ) ks : Using the bound i s q s =s we obtain Y n µ w( ; q) a 0 + s(k ks s ) s ks k s! q s à n! a 0 w( ) exp sk s (k s )q s : Since P sq s P s2 s = 2, this proves (a): (b) Similarly, for partitions with no two parts of the same size we have (for any of the classes) w( ; q) q sk s iks s à w( ) exp 2 µ s k s ks!! n k s q s=2 + 2q s=2 so the lower bound follows with c := exp 2 P 2 s=2 = 0:007999. Recall that the set n;k consists of all partitions of n in which each part has multiplicity < k, and 0 n;k consists of the remaining partitions. ks the electronic journal of combinatorics (2004), #R00 6

Lemma 7 For all classes M i (q), and all n and q 2 0 n;k w( ; q) a 2 0 k + whenever k 2: 2k Similarly 2 0 n;k w( ) k + whenever k 2: 2k Proof. Each 2 0 n;k can be written in the form [sk ] _ ¹ for some ¹ ` n ks in at least one way. Hence using Lemmas 4 and 5 we obtain 2 0 n;k w( ; q) ¹`n ks = a 0 w([s k ]; q) a 2 0 w([s k ] _ ¹; q) a 0 w([s k ]; q) k + (2s) k + k a2 0 2 : k ¹`n ks w(¹; q) This proves the stated inequality. The corresponding inequality for w( ) is similar. 3 Proof of Theorem Since n (x) and ª n (x)nª n ( x) are complementary sets for x > 0, and the error function is even, the theorem of ErdÄos and Tur an quoted in the Introduction shows that for xed x > 0: W n (x) := w( )! (x) as n! ; where (x) := p 2¼ ½Z x 2 n(x) e t2 =2 dt + Z x ¾ e t2 =2 dt = p 2 Z e t2 =2 dt: 2¼ x A simple integration by parts shows (see, for example, [5, Chap. 7]) that (x) < 2e x2 =2 p 2¼x for x > 0: Thus, for x, there exists n 0 (x) > 0 such that W n (x) < e x2 =2 whenever n > n 0 (x). De ne n;k (x) := n (x) \ n;k and 0 n;k (x) := n(x) \ 0 n;k. Now using Lemma 6 we have, for each of the classes M i (q), that W n;k (x; q) := w( ; q) a 0 e 2k(k ) w( ) a 0 e 2k(k ) W n (x): 2 n;k (x) 2 n;k (x) the electronic journal of combinatorics (2004), #R00 7

On the other hand Lemma 7 shows that for k : W 0 n;k(x; q) := 2 0 n;k (x) w( ; q) 2 0 n;k (x) w( ; q) a 2 0 k + 2 k < 8a2 0e (k+)=2 : Thus for x, k and n n 0 (x) we have w( ; q) = W n;k (x; q) + Wn;k 0 (x; q) < a 0e 2k(k ) e x2 =2 + 8a 2 0 e (k+)=2 : 2 n (x) If x 2, then we can choose k := bx=2c and obtain e 2k(k ) e x2 =2 + 8a 0 e (k+)=2 < e x + 8a 0 e x=4 < ( + 8a 0 )e x=4 ; uniformly in x. Thus taking c 0 := a 0 ( + 8a 0 ) we obtain () for x 2. However, by adjusting the value of c 0 if necessary we can ensure that the inequality () is also valid for x with x < 2. Then the inequality is valid for all x. Finally, we prove the last assertion of the theorem. Given any " > 0 and ± > 0, choose x so that c 0 e x=4 < ±, and then choose n n 0 (x) so that x < " p 3 log n. Now () shows that for all n n the proportion of f() of degree n in M i (q) which have splitting elds whose degree lies outside of the interval exp(( 2 ")(log n)2 ); exp(( 2 + ")(log n)2 ) is bounded by c 0 e x=4 < ±. This is equivalent to what is stated. 4 Proof of Theorem 2 We start by proving an upper bound for E n (q). De ne ~E n := max fe m j m = ; 2; :::; ng : (It seems likely that ~ E n = E n but we have not been able to prove this.) Lemma 8 There exists a constant c 2 > 0 such that, in each of the classes M i (q), E n (q) c 2 ~ En for all q and all n. Proof. Let k 2 be the least integer such that We shall de ne c 2 := 2a 0 e 2k(k ). a 2 0 (k + )s =2: (2s) k the electronic journal of combinatorics (2004), #R00 8

We shall prove the lemma by induction on n. Note that E (q) = c 2 = c 2 ~ E. Assume n 2 and that E m (q) c 2 ~ Em for all m < n. Now Lemma 4 shows that En;k 0 (q) := 2 0 n;k w( ; q)m( ) a 0 sw( s k ; q) ¹`n ks ¹`n ks = a 0 sw( s k ; q)e n ks (q): w(¹; q)m(¹) w( s k _ ¹; q)m( s k _ ¹) Thus using Lemma 5, the choice of k and the induction hypothesis, we obtain E 0 n;k (q) a2 0 (k + )s (2s) k c 2 ~ E n ks 2 c 2 ~ E n n o because the sequence ~En is monotonic. On the other hand, Lemma 6 shows by the choice of c 2. E n;k (q) := 2 n;k w( ; q)m( ) a 0 e 2k(k ) w( )m( ) a 0 e 2k(k ) E n 2 c 2 E ~ n 2 n;k Hence and the induction step is proved. E n (q) = E n;k (q) + E 0 n;k (q) c 2 ~ E n To complete the proof of the theorem we must prove a lower bound for E n (q). Let n denote the set of partitions ¼ of the form: (i) ¼ is a partition of some integer m with n r < m n where r is the smallest prime > p n; (ii) the parts of ¼ are distinct and each is a multiple of a di erent prime > p n. Note that if the parts of ¼ are k r ; :::; k t r t where r ; :::; r t are distinct primes > p n then w(¼)m(¼) Qi r i= (k i r i )! = Q i =k i. Consider the partitions of n which can be written in the form ¼ _! where ¼ 2 n and! 2 n j¼j. In Sect. 3 of [9] (see especially the bottom of page 3) Stong notes (in our notation) that since ¼ and! are disjoint: E n w(¼ _!)m(¼ _!) ¼2 n!`n j¼j ¼2 n w(¼)m(¼)!`n j¼j w(!) = ¼2 n w(¼)m(¼): the electronic journal of combinatorics (2004), #R00 9

³ He then proves that the last sum is greater than E n exp O Similarly, using Lemma 4 we obtain E n (q) w(¼ _!; q)m(¼ _!) ¼2 n!`n j¼j a 2 0 ¼2 n w(¼; q)m(¼)!`n j¼j w(!; q) = a 2 0 ³ pn log log n. log n ¼2 n w(¼; q)m(¼): Since each ¼ 2 n has all its parts of di erent sizes, Lemma 6 shows that w(¼; q) c w(¼), and so from the result due to Stong quoted above µ µp n log log n E n (q) a 2 0 c w(¼)m(¼) E n exp O : log n ¼2 n The lower bound in our theorem now follows from Stong's theorem. References [] P. ErdÄos and P. Tur an, On some problems of a statistical group theory I, Z. Wahrschein. Verw. Gebeite 4 (965) 75{86. [2] P. ErdÄos and P. Tur an, On some problems of a statistical group theory II, Acta Math. Acad. Sci. Hungar. 8 (967) 5{63. [3] P. ErdÄos and P. Tur an, On some problems of a statistical group theory III, Acta Math. Acad. Sci. Hungar. 8 (967) 309{320. [4] P. ErdÄos and P. Tur an, On some problems of a statistical group theory IV, Acta Math. Acad. Sci. Hungar. 9 (968) 43{435. [5] W. Feller, \An Introduction to Probability Theory and its Applications", Vol. (3rd. ed.), Wiley, New York, 968. [6] W. Goh and E. Schmutz, The expected order of a random permutation, Bull. London Math. Soc. 23 (99) 34{42. [7] A. Knopfmacher and R. Warlimont, Distinct degree factorizations for polynomials over a nite eld, Trans. Amer. Math. Soc. 347 (995) 2235{2243. [8] R. Lidl and H. Niederreiter, \Finite Fields", Cambridge Univ. Press, 997. [9] R. Stong, The average order of a permutation, Electronic J. Combinatorics 5 (998) #R4. the electronic journal of combinatorics (2004), #R00 0