Theorems of Sylvester and Schur

Transcription

1 Theorems of Sylvester and Schur T.N. Shorey An old theorem of Sylvester states that a product of consecutive positive integers each exceeding is divisible by a prime greater than. We shall give a proof of this theorem and apply it to prove a result of Schur on the irreducibility of certain polynomials. Irreducibility of certain polynomials We begin with Theorem (Schur [6]) Let a 0, a,, a n Z with a 0, = a n =. Then x n a n x n n! + a n (n )! + + a x + a 0 is irreducible over rationals. The proof that we shall give depends on Newton polygons used by Coleman [2] and Filaseta [4]. I than Professor Michael Filaseta for sending me copies of his preprints and reprints which helped me in writing this article. The text of this paper is based on my tal in Sesquicentennial Celebrations of University of Mumbai.

2 Newton polygons Let p > 0 be prime. For non-zero integer n, let ν(n) = ν p (n) = ord p (n). Further we put Let with and ν(0) =. m f(x) = a j x j Z[x] j=0 a 0 a m 0 S = {(0, ν(a m )), (, ν(a m )), (m, ν(a 0 ))}. We consider the lower edges along the convex hull of these points. The left most edge has one end point (0, ν(a m )) and the right most edge has one end point (m, ν(a 0 )). The end points of every edge belong to S. If (i, ν(a m i )) and (j, ν(a m j )) with i < j are end points of such an edge, then every point (u, v(a m u )) of S with i < u < j lie on or above the line passing through (i, ν(a m i )) and (j, ν(a m j )). Definition The polygonal path formed by these edges is called the Newton polygon associated to f(x) with respect to p. Example: Let p = 2 and f(x) = 2x 6 + x 4 + 2x 2 + 4x + 4. Now S = {(0, ν(2)), (, ν(0)), (2, ν()), (3, ν(0)), (4, ν(2)), (5, ν(4)), (6, ν(4))} 2

3 i.e. (0, ), (, ), (2, 0), (3, ), S = (4, ), (5, 2), (6, 2) Newton polygon for f(x) (0, ) (5, 2) (4, ) (6, 2) (2, 0) We do not allow two different edges to have the same slope. Thus the slopes of the Newton polygon are increasing when calculated from the leftmost edge to the right-most edge. We shall need the following result. Theorem (Dumas [] ) Let p > 0 prime and g(x), h(x) Z[x] with and g(0)h(0) 0 u 0 3

4 be the leading coefficient of g(x)h(x) with ord p (u) = t. Then the edges of the Newton polygon of g(x)h(x) with respect to p can be formed by constructing a polygonal path beginning with (0, t) and using translates of edges in the Newton polygons of g(x) and h(x) with respect to p (using exactly one translate from each edge). Necessarily, the translated edges are translated in such a way as to form a polygonal path with the slopes of the edges increasing. Example: Let p = 3 and g(x) = x 3 + 3x 2 + 2x + 9, h(x) = 2x 2 + 9x + 3. Then f(x) = g(x)h(x) = 2x 5 + 5x x x 2 + 7x For g(x): and Newton polygon. S = {(0, 0), (, ), (2, ), (3, 2)} 2 (0, 0) (, ) (2, ) (3, 2) 4

5 For h(x): and Newton polygon. S = {(0, 0), (, 2), (2, )} (, 2) (0, 0) 2 (2, ) For f(x): and Newton polygon S = {(0, 0), (, ), (2, 3), (3, 3), (4, 2), (5, 3)} (, ) 2 (2, 3) (4, 2) (5, 3) (0, 0) u = 2 and t = 0. 5

6 u = 2 and t = 0. The edge with slope /2 of Newton polygon of f is obtained by combining the edge with slope /2 of Newton polygon of g followed by a translate of the edge with slope /2 of Newton polygon of h. (Here translation is obtained by shifting (0, 0) to (2,).) Then we continue with a translate of the edge with slope of Newton polygon of g. (Here translation is secured by shifting (2, ) to (4, 2)). As an application of Dumas theorem, we derive Eisenstein s Criterion for irreducibility: Let F(x) = n A j x j Z[x] j=0 and p prime with p A n, p A j for 0 j < n, p 2 A 0. Then F(x) is irreducible over Q. Proof. The Newton polygon of F with respect to p: Here S = {(j, ord p (A n j )) 0 j n}. (n, ) (0, 0) 6

7 The point (j, ord p (A n j )) lies above the edge since n < ord p(a n j ). j Assume that F is reducible. Then F(x) = G(x)H(x) with and G(x), H(x) Z[x] degg > 0, degh > 0. Thus, by Dumas result, the edge of the Newton polygon of F is obtained by translating the edges of Newton polygons of G and H. In particular, the edge of Newton polygon of F has a lattice point other than (0, 0) and (n, ). This is a contradiction. The proof also depends on another result on the greatest prime factor of a product of consecutive positive integers stated in the beginning. For an integer ν >, we write and we put Then we have P(ν) = max p ν P() =. p Theorem 2 (Sylvester [7]). For n > > 0, we have P(n(n + ) (n + )) >. Thus a product of consecutive positive integers each exceeding is divisible by a prime greater than. The assumption n > is necessary since P(.2 ). We use Newton polygons for the following result. 7

8 Lemma Let > 0. Suppose that g(x) = m b j x j Z[x] j=0 and p > 0 prime with p b m, p b j for j = 0,,, m. Further assume that the right most edge of the Newton polygon of g(x) with respect to p has slope <. Let a 0, a,, a m Z with a 0 = a m =. Then f(x) = cannot have a factor of degree. m a j b j x j j=0 Lemma implies Schur Theorem: We tae in Lemma and g(x) = b j = m! j! m b j x j, f(x) = j=0 m a j b j x j. It suffices to show that f(x) is irreducible. Assume that f is reducible. Let be the smallest degree of an irreducible factor of f. Then Thus m 2. j=0 m m 2. 8

9 Now we apply Theorem 2 to find a prime p with p + such that Thus and p (m + ) m = b m. p b j for j = 0,,, m p b m since b m =. Now we consider Newton polygon of g(x) with respect to p. Here S = {(j, ν(b m j )) 0 j m} and (0, ν(b m )) = (0, 0) is the left most end point and is the right most end point. (m, ν(b 0 )) = (m, ν(m!)) For contradicting Lemma, we show that the slope of the right most edge is < /. For this, we show max j m ν(b 0 ) ν(b j ) m (m j) < i.e. i.e. max j m ν(j!) j max j m ν(m!) ν( m! j! ) j < <. ( ) 9

10 Now ν(j!) = [ j p ] + [ j p 2] + < j( p + p 2 + ) p = j p since p +. This proves (*). = j p j Proof of Lemma : It is easy to see that there is no loss of generality in restricting to a j = for 0 j m. Thus f(x) = g(x). Assume that f(x) has a factor or degree. Then there exist u(x), v(x) Z[x] such that f(x) = u(x)v(x). We consider Newton polygon of f(x) with respect to p. Here S = {(0, ν(b m )) = (0, 0), (, ν(b m )), (, ν(b m + )), (, ν(b m )), (m, ν(b 0 ))}. We observe that y coordinates of last m + points in S are positive. Since the slopes of the edges are increasing when calculated from the left most edge to the right most edge, we see that the slope of each edge is < /. Further the left most edge may have slope zero. Now we consider an edge with positive slope. Let (a, b) and (c, d) be lattice points on this edge. 0

11 Then the slope of the edge is the slope of the line passing through (a, b) and (c, d). Thus implying c a d b c a < c a >. Since deg u(x) =, we observe that the translates of the edges of u(x) do not lie on the edges of f(x) with positive slope. Therefore, by Theorem, the left most edge of the Newton polygon of f(x) must have slope zero and length. On the other hand, since the y-coordinates of all other than the first points in S are positive, its length <. This is a contradiction. This completes the proof of Theorem. The above method has been applied for showing the irreducibility of several polynomials. For example, it has been used in [ 5 ] to prove that Bessel polynomials are irreducible. Proof of Theorem 2 We give a proof due to Erdös [3]. We put (n, ) = n(n + ) (n + ). Let n + = x. Then x 2 if n >. and (n, )! = x(x ) (x + )! ( ) x =.

12 Therefore Theorem 2 implies (( )) x () P > if x 2. In fact proving Theorem 2 is equivalent to showing (). Now we prove (). It is enough to prove () when is prime. Let < 2 where and 2 are consecutive primes and () is valid with replaced by. Let x 2. Then x 2 and (( )) x P >. Therefore Also and (( )) x p := P 2 >. p (x +, ) (x +, ) (x +, ). Hence ( ) x p The proof is by contradiction. We assume that (( )) x P Therefore ( ) x P( (x +, )) = P(! ) 2

13 Let = 2. Then Let = 3. Then P((x )x) 2, contradiction P((x 2)(x )x) 3 We delete the terms in which 2 and 3 appear to a maximal power. We are left with a term which is at most 2. Thus x 2 2 i.e. Let = 5. Then x 4, contradiction. P(x(x ) (x 4)) 5. We remove the terms divisible by 5 and 3. Also we remove the term in which 2 appears to maximal power. Thus x 4 4, contradiction. Let = 7. Then P(x(x )(x 2) (x 6)) 7 We remove the terms in which 7 and 5 appear. Further we remove the terms in which 2 and 3 appear to maximal powers. We are left with two terms. Then x = 2 Therefore 4 x 7 and we chec that P( (x 6, 7)) > 7, contradiction. 3

14 Hence Let and s be given by Then Hence a =. ( ) x p a = x!!(x )! p s x < p s+ ([ ] [ ] [ ]) x x s ν= p ν p ν p a p s x p ν Next we show that For this, we observe On the other hand, = x Therefore 2 if < 37 x < 3/2 if 37 = p a x π() p p a ( x ) x x + ( x ) >. ( x ) < x π(). Thus and further π() x < π(). 2 if < 37 3/2 if 37 4

15 This proves the assertion. Now we may assume that x < 3/2 otherwise the assertion follows by computations. Then x 2l < l for l 2. The remaining proof depends on the following result which we assume p p x p x /2 p p x /3 p < 4 x. By applying the above inequality with x =, Further Also Hence Next we prove (a) x < 4. p p p /2 p p p p x /2 p (b) 03 if 5 2 < x < 4. (c) 3 if 2 x 5 2. p x /3 p p x /4 p p /3 p < 4. p x /5 p < 4. p x /6 p < 4 < 4 + x. x. Then 3 and the assertion follows by computations. First we prove (a). Let x 4. 5

16 We observe that implying We have Thus On the other hand Thus ( 4 ( 2 ) ) = 4 x < 3/2 7. 4(4 ) (3 + ) 2(2 ) ( + ) > 2. ( ) 4 > = 8 2. < 4 + x < 4 +3/ /4 < 2 ( ) 2 2 (2) implying 29 and the cases 7 29 are excluded by (2). Now we show ( ) 2 > 4 2 which has been used in the proof of (a). We have > ( 3 2 )( 5 2 ) ( (2 ) 2 ) = (2 2) (2 ) 2 = 4 ( 2! 3.5 (2 ) )2 = 4 (4 (!) 2 (2!) )2. Therefore ( ) 2 = 2! (!) >

17 The proof of (b) is similar. Now we turn to the proof of (c). We have 2 x 5 2. We write We show Let Then Also since Further since and Thus Hence = x!!(x )!. P( ) x 3. x 3 < p. ord p (x!) 2. ord p (!) = 2 3 < p. ord p ((x )!) = x 2 = p x x 2 5 x = 3 5 x < 9 5 p < 2p. p. P( ) x 3. 7

18 Then 4 x 3 + x. Now the proof for (c) is similar to that of (a). References [] G. Dumas, Sur quelques cas d irréductibilité des polynômes á coefficients rationnels, Journal de Math. Pure et Appl. 2 (906), [2] F. Coleman, On the Galois groups of the exponential Taylor polynomials, Enseignement Math. 33 (987), [3] P. Erdös, A theorem of Sylvester and Schur, J. London Math. Soc. 9 (934), [4] M. Filaseta, The irreducibility of all but finitely many Bessel polynomials, Acta Math. 74 (995), [5] M. Filaseta and O. Trifonov, The irreducibility of the Bessel polynomials, Journal Reine Angew. Math. 550 (2002), [6] I. Schur, Einige Sätze über Primzahlen mit Anwendungen auf Irreduzibilitätsfragen, I, Sitzungsbser. Preuss. Aad. Wiss. Berlin Phys.- Math. Kl., 4 (929), [7] J.J. Sylvester, On arithmetical series, Messenger of Mathematics, XXI (892), -9, 87-20, and Mathematical Papers 4 (92), School of Mathematics Tata Institute of Fundamental Research Homi Bhabha Road 8

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