Recursive constructions of irreducible polynomials over finite fields
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1 Recursive constructions of irreducible polynomials over finite fields Carleton FF Day Ottawa Lucas Reis (UFMG - Carleton U) September 2017
2 F q : finite field with q elements, q a power of p.
3 F q : finite field with q elements, q a power of p. GL 2 (F q ): 2x2 non-singular matrices with entries in F q.
4 F q : finite field with q elements, q a power of p. GL 2 (F q ): 2x2 non-singular matrices with ( entries ) in F q. a b Given f (x) F q [x] of degree n and A = GL c d 2 (F q ),
5 F q : finite field with q elements, q a power of p. GL 2 (F q ): 2x2 non-singular matrices with ( entries ) in F q. a b Given f (x) F q [x] of degree n and A = GL c d 2 (F q ), A f := (bx + d) n f ( ) ax + c. bx + d
6 F q : finite field with q elements, q a power of p. GL 2 (F q ): 2x2 non-singular matrices with ( entries ) in F q. a b Given f (x) F q [x] of degree n and A = GL c d 2 (F q ), For B = A f := (bx + d) n f ( ) ax + c. bx + d ( ) 0 1, B f = x 1 0 n f ( ) 1 x is the reciprocal of f (x).
7 M := {f F q [x] f has no root in F q }.
8 M := {f F q [x] f has no root in F q }. Basic Properties. For A, B be elements of GL 2 (F q ) and f, g M, the following hold: (i) A f M and deg(a f ) = deg f, (ii) If E denotes the identity element of GL 2 (F q ), then E f = f, (iii) (AB) f = A (B f ), (iv) A (f g) = (A f ) (A g), (v) f is irreducible if and only if A f is irreducible.
9 I n := irreducible monic polynomials of degree n.
10 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant).
11 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q.
12 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q. * From the basic properties, PGL 2 (F q ) acts on I n, n 2 via the compositions [A] f.
13 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q. * From the basic properties, PGL 2 (F q ) acts on I n, n 2 via the compositions [A] f. How about the invariants?
14 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q. * From the basic properties, PGL 2 (F q ) acts on I n, n 2 via the compositions [A] f. How about the invariants? C A (n) := {f I n [A] f = f },
15 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q. * From the basic properties, PGL 2 (F q ) acts on I n, n 2 via the compositions [A] f. How about the invariants? C A (n) := {f I n [A] f = f }, N A (n) = C A (n)
16 I n := irreducible monic polynomials of degree n. PGL 2 (F q ): GL 2 (F q )/ (matrices up to a constant). Definition For [A] PGL 2 (F q ) and f I n, n 2, [A] f is the only monic polynomial = λ (A f ) with λ F q. * From the basic properties, PGL 2 (F q ) acts on I n, n 2 via the compositions [A] f. How about the invariants? C A (n) := {f I n [A] f = f }, N A (n) = C A (n) C A := n 2 C A (n).
17 A characterization of C A :
18 A characterization of C A : Theorem (Stichtenoth, Topuzoglu - FFA 2012) ( ) a b Let A = be an element of GL c d 2 (F q ). For each nonnegative integer r, set F r (x) = bx qr +1 ax qr + dx c. For any f I n with n 2, the following are equivalent: (i) f (x) divides F r (x) for some r 0, (ii) [A] f = f.
19 Set D = ord([a]): any element of C A has degree 2 or degree Dm for some m 1.
20 Set D = ord([a]): any element of C A has degree 2 or degree Dm for some m 1. In particular, N A (n) = 0 if n > 2 and n is not divisible by D.
21 Set D = ord([a]): any element of C A has degree 2 or degree Dm for some m 1. In particular, N A (n) = 0 if n > 2 and n is not divisible by D.Also, N A (Dm) Φ(D) Dm qm.
22 Set D = ord([a]): any element of C A has degree 2 or degree Dm for some m 1. In particular, N A (n) = 0 if n > 2 and n is not divisible by D.Also, Enumeration formulas: N A (Dm) Φ(D) Dm qm. 1. Garefalakis (JPAA ): upper triangular elements. 2. Mattarei and Pizzato (FFA ): involutions, following a work of O. Ahmadi. 3. R. (Arxiv ): general elements of PGL 2 (F q ).
23 Alternative characterization of the invariants.
24 Alternative characterization of the invariants. An irreducible polynomial f (x) of degree 2m is self-reciprocal if and only if f (x) is an irreducible of the form x m g ( x + x 1) for some g(x) of degree m.
25 1. R. (JPAA ): ( ) 1 0 A =, i.e., [A] f = (x)f (x + 1). 1 1
26 1. R. (JPAA ): ( ) 1 0 A =, i.e., [A] f = (x)f (x + 1). 1 1 The invariants appear as f (x) = g(x p x).
27 1. R. (JPAA ): ( ) 1 0 A =, i.e., [A] f = (x)f (x + 1). 1 1 The invariants ( ) appear as f (x) = g(x p x). a 0 A =, i.e., [A] f (x) = f (ax). 0 1
28 1. R. (JPAA ): ( ) 1 0 A =, i.e., [A] f = (x)f (x + 1). 1 1 The invariants ( ) appear as f (x) = g(x p x). a 0 A =, i.e., [A] f (x) = f (ax). 0 1 The invariants appear as f (x) = g(x k ), where k = ord(a).
29 1. R. (JPAA ): ( ) 1 0 A =, i.e., [A] f = (x)f (x + 1). 1 1 The invariants ( ) appear as f (x) = g(x p x). a 0 A =, i.e., [A] f (x) = f (ax). 0 1 The invariants appear as f (x) = g(x k ), where k = ord(a). 2. Mattarei and Pizzato (FFA ): involutions. The invariants apperar as f (x) = h2 n g(h 1/h 2 ), where h 1 /h 2 F q (x) is a quadratic rational function.
30 Theorem (R., August 2017) Let [A] PGL 2 (F q ) with ord([a]) = D > 1. There exists a rational function R(A) = g A ha of degree D such that f I Dm satisfies [A] f = f if and only if f (x) is an irreducible monic polynomial of the form h m A F ( g A ha ) for some F of degree m. Moreover, the rational function R(A) can be computed from the element A.
31 Conjugacy classes in PGL 2 (F q ):
32 Conjugacy classes in PGL 2 (F q ): ( ) a 0 1. type 1: A(a) :=, R(A) = x 0 1 k,
33 Conjugacy classes in PGL 2 (F q ): ( a 0 1. type 1: A(a) := 0 1 ( type 2: B := 1 1 ), R(A) = x k, ), R(A) = x p x
34 Conjugacy classes in PGL 2 (F q ): ( a 0 1. type 1: A(a) := 0 1 ( type 2: B := 1 1 ( type 3: C(b) := b 0 ), R(A) = x k, ), R(A) = x p x ), R(A) = x2 +b 2x
35 Conjugacy classes in PGL 2 (F q ): ( a 0 1. type 1: A(a) := 0 1 ( type 2: B := 1 1 ( type 3: C(b) := b 0 ( 0 c 4. type 4: D(c) := 1 1 ), R(A) = x k, ), R(A) = x p x ), R(A) = x2 +b 2x ), R(A) = D Ψ A (x) = 1 cx + 1. i=1 Ψ(i) A (x), where
36 Conjugacy classes in PGL 2 (F q ): ( a 0 1. type 1: A(a) := 0 1 ( type 2: B := 1 1 ( type 3: C(b) := b 0 ( 0 c 4. type 4: D(c) := 1 1 ), R(A) = x k, ), R(A) = x p x ), R(A) = x2 +b 2x ), R(A) = D Ψ A (x) = 1 cx + 1. i=1 Ψ(i) A (x), where The R(A) s above are called canonical rational functions.
37 Rational transformations:
38 Rational transformations: For f F q [x] irreducible with deg f = n and Q(x) F q (x) of degree D, Q(x) = F (x)/g(x), set f Q = G n f ( ) F. G
39 Rational transformations: For f F q [x] irreducible with deg f = n and Q(x) F q (x) of degree D, Q(x) = F (x)/g(x), set f Q = G n f ( ) F. G Also, set f 0 = f and f i = f Q i 1,
40 Rational transformations: For f F q [x] irreducible with deg f = n and Q(x) F q (x) of degree D, Q(x) = F (x)/g(x), set f Q = G n f ( ) F. G Also, set f 0 = f and f i = f Q i 1, deg f i = D deg f i 1.
41 Rational transformations: For f F q [x] irreducible with deg f = n and Q(x) F q (x) of degree D, Q(x) = F (x)/g(x), set f Q = G n f ( ) F. G Also, set f 0 = f and f i = f Q i 1, deg f i = D deg f i 1. Given f irreducible of degree n, we want to obtain an infinite sequence of irreducibles {f i } i 0 of degree D i n, via Q(x)-transformations, where Q is a canonical rational function.
42 Theorem (Cohen) Let f (x) be irreducible of degree n over F q and α F q n roots. Then f Q = G n f ( F G ) is irreducible if and only if F (x) αg(x) is irreducible over F q n. one of its
43 Theorem (Cohen) Let f (x) be irreducible of degree n over F q and α F q n roots. Then f Q = G n f ( F G ) is irreducible if and only if F (x) αg(x) is irreducible over F q n. one of its Fact: If D = ord([a]) is prime, Q = R(A) = f A /g A, then f A αg A is either irreducible or splits completely over F q n.
44 Theorem (Cohen) Let f (x) be irreducible of degree n over F q and α F q n roots. Then f Q = G n f ( F G ) is irreducible if and only if F (x) αg(x) is irreducible over F q n. one of its Fact: If D = ord([a]) is prime, Q = R(A) = f A /g A, then f A αg A is either irreducible or splits completely over F q n. In particular, if D is prime, f Q is either irreducible or split into D irreducible factors, each of degree n.
45 Theorem (Cohen) Let f (x) be irreducible of degree n over F q and α F q n roots. Then f Q = G n f ( F G ) is irreducible if and only if F (x) αg(x) is irreducible over F q n. one of its Fact: If D = ord([a]) is prime, Q = R(A) = f A /g A, then f A αg A is either irreducible or splits completely over F q n. In particular, if D is prime, f Q is either irreducible or split into D irreducible factors, each of degree n. The roots of f A αg A can be explored through the dynamics of the map x f A(x) g A (x) in F q: in general, the functional graph is full of symmetries.
46 Methods:
47 Methods: 1. Deterministic: initial conditions on f for f Q to be irreducible. For instance, Q = x p x, f (x) must be of non-zero trace and Q = x k, some conditions on the order ord(f ) of f (x).
48 Methods: 1. Deterministic: initial conditions on f for f Q to be irreducible. For instance, Q = x p x, f (x) must be of non-zero trace and Q = x k, some conditions on the order ord(f ) of f (x). 2. Iterated trials: works for D prime; if f Q is not irreducible, it splits into D irreducible factors of degree n. Pick one of those irreducibles, apply Q again. Eventually we find an irreducible.
49 Methods: 1. Deterministic: initial conditions on f for f Q to be irreducible. For instance, Q = x p x, f (x) must be of non-zero trace and Q = x k, some conditions on the order ord(f ) of f (x). 2. Iterated trials: works for D prime; if f Q is not irreducible, it splits into D irreducible factors of degree n. Pick one of those irreducibles, apply Q again. Eventually we find an irreducible. For self-reciprocals, D = 2, (Ugolini - DCC 2015).
50 Methods: 1. Deterministic: initial conditions on f for f Q to be irreducible. For instance, Q = x p x, f (x) must be of non-zero trace and Q = x k, some conditions on the order ord(f ) of f (x). 2. Iterated trials: works for D prime; if f Q is not irreducible, it splits into D irreducible factors of degree n. Pick one of those irreducibles, apply Q again. Eventually we find an irreducible. For self-reciprocals, D = 2, (Ugolini - DCC 2015). 3. Probabilistic: pick a random irreducible f of degree n and check if f Q is irreducible or not.
51 Efficiency of iterations: if A is of type 1, 3 or 4 and Q = R(A), once f i is irreducible, f j is irreducible for any j i.
52 Efficiency of iterations: if A is of type 1, 3 or 4 and Q = R(A), once f i is irreducible, f j is irreducible for any j i. The case A of type 1 is a classical result.
53 Efficiency of iterations: if A is of type 1, 3 or 4 and Q = R(A), once f i is irreducible, f j is irreducible for any j i. The case A of type 1 is a classical result. The case A of type 4 or 3, we can verify that the map x f A(x) g A (x) is conjugated to map x x D in F q, via Mobius permutations.
54 Efficiency of iterations: if A is of type 1, 3 or 4 and Q = R(A), once f i is irreducible, f j is irreducible for any j i. The case A of type 1 is a classical result. The case A of type 4 or 3, we can verify that the map x f A(x) g A (x) is conjugated to map x x D in F q, via Mobius permutations. * The case A of type 2 is more complicated: if f i = f i 1 (x p x) is irreducible, f i+1 is reducible.
55 Iterated trial: related to the functional graph of the map x f A(x) g A (x).
56 Iterated trial: related to the functional graph of the map x f A(x) g A (x). Insert a functional graph.
57 Suppose that Q = R(A) = f A /g A is a canonical rational function associated to A, with ord([a]) = D.
58 Suppose that Q = R(A) = f A /g A is a canonical rational function associated to A, with ord([a]) = D. N A (Dn) Φ(D) Dn qn, n >> 1
59 Suppose that Q = R(A) = f A /g A is a canonical rational function associated to A, with ord([a]) = D. N A (Dn) Φ(D) Dn qn, n >> 1 We know that the invariants of degree Dn arise from f Q, with f of degree n.
60 Suppose that Q = R(A) = f A /g A is a canonical rational function associated to A, with ord([a]) = D. N A (Dn) Φ(D) Dn qn, n >> 1 We know that the invariants of degree Dn arise from f Q, with f of degree n. Necessary condition: f must be irreducible. How many they are?
61 Suppose that Q = R(A) = f A /g A is a canonical rational function associated to A, with ord([a]) = D. N A (Dn) Φ(D) Dn qn, n >> 1 We know that the invariants of degree Dn arise from f Q, with f of degree n. Necessary condition: f must be irreducible. How many they are? Close to qn n, n >> 1.
62 Random Method:
63 Random Method: Pick f irreducible of degree n. If f Q is irreducible, proceed with the iterations f i = f Q i 1. If not, pick another irreducible of degree n.
64 Random Method: Pick f irreducible of degree n. If f Q is irreducible, proceed with the iterations f i = f Q i 1. If not, pick another irreducible of degree n. In particular, for a random irreducible of degree n, f Q is also irreducible with probability p A Φ(D) D.
65 Random Method: Pick f irreducible of degree n. If f Q is irreducible, proceed with the iterations f i = f Q i 1. If not, pick another irreducible of degree n. In particular, for a random irreducible of degree n, f Q is also irreducible with probability p A Φ(D) D. Geometric Distribution with p = p A.
66 Random Method: Pick f irreducible of degree n. If f Q is irreducible, proceed with the iterations f i = f Q i 1. If not, pick another irreducible of degree n. In particular, for a random irreducible of degree n, f Q is also irreducible with probability p A Φ(D) D. Geometric Distribution with p = p A. In particular, the expected number of trials is 1 p A D Φ(D).
67 Random Method: Pick f irreducible of degree n. If f Q is irreducible, proceed with the iterations f i = f Q i 1. If not, pick another irreducible of degree n. In particular, for a random irreducible of degree n, f Q is also irreducible with probability p A Φ(D) D. Geometric Distribution with p = p A. In particular, the expected number of trials is 1 p A D Φ(D). For D prime, D Φ(D) = D D 1 2.
68 Example 1: A = ( ) GL 2 (F 2 )
69 Example 1: A = ( ) GL 2 (F 2 ) Q = R(A) = 1 x x + 1 x + x = x 3 + x + 1 x 2. + x
70 Example 1: A = ( ) GL 2 (F 2 ) Q = R(A) = 1 x x + 1 x Set f 0 = x + 1 and f i = f Q i 1, i 1. + x = x 3 + x + 1 x 2. + x
71 Example 1: A = ( ) GL 2 (F 2 ) Q = R(A) = 1 x x + 1 x Set f 0 = x + 1 and f i = f Q i 1, i 1. f 1 = x 3 + x x = x 3 + x + 1 x 2. + x
72 Example 1: A = ( ) GL 2 (F 2 ) Q = R(A) = 1 x x + 1 x Set f 0 = x + 1 and f i = f Q i 1, i 1. f 1 = x 3 + x f 2 = x 9 + x x = x 3 + x + 1 x 2. + x
73 Example 1: A = ( ) GL 2 (F 2 ) Q = R(A) = 1 x x + 1 x Set f 0 = x + 1 and f i = f Q i 1, i 1. f 1 = x 3 + x f 2 = x 9 + x x = x 3 + x + 1 x 2. + x f 3 = x 27 + x 26 + x 24 + x 18 + x 17 + x 11 + x 9 + x 8 + x 3 + x 2 + 1
74 f 4 = x 81 + x 64 + x 16 + x + 1
75 f 4 = x 81 + x 64 + x 16 + x + 1 f 5 = x x x x x x x x x x x x x x x x x x x x 131 +x 129 +x 128 +x 114 +x 113 +x 99 +x 98 +x 96 +x 83 +x 81 +x 80 + x 66 +x 65 +x 51 +x 50 +x 48 +x 35 +x 33 +x 32 +x 18 +x 17 +x 3 +x 2 +1.
76 Example 2: A = ( ) GL 2 (F 5 )
77 Example 2: A = ( ) GL 2 (F 5 ) Q = R(A) = 3x 6 + x + 4 x 5. x
78 Example 2: A = ( ) GL 2 (F 5 ) Set f 0 = x and f i = f Q i 1, i 1. Q = R(A) = 3x 6 + x + 4 x 5. x
79 Example 2: A = ( ) GL 2 (F 5 ) Set f 0 = x and f i = f Q i 1, i 1. f 1 = x 6 + 2x + 3 Q = R(A) = 3x 6 + x + 4 x 5. x
80 Example 2: A = ( ) GL 2 (F 5 ) Set f 0 = x and f i = f Q i 1, i 1. f 1 = x 6 + 2x + 3 Q = R(A) = 3x 6 + x + 4 x 5. x f 2 = x 36 + x 31 + x x x x 10 + x 6 + 4x 5 + x + 4
81 Example 2: A = ( ) GL 2 (F 5 ) Set f 0 = x and f i = f Q i 1, i 1. f 1 = x 6 + 2x + 3 Q = R(A) = 3x 6 + x + 4 x 5. x f 2 = x 36 + x 31 + x x x x 10 + x 6 + 4x 5 + x + 4 f 3 = x x x x x x x x x x x x x x x x x x x x x x x x x x x x 86 + x x x x x x x x x x x 51 + x x x 36 + x x 31 + x x 26 + x x x x 11 + x x 6 + 4x 5 + 4x + 1.
82 Thank you!
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