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Exponential polynomials Paola DAquino Seconda Universita di Napoli - PowerPoint PPT Presentation

Exponential polynomials Paola DAquino Seconda Universita di Napoli Cesme, May 2012 Topics Exponential rings, exponential fields and exponential polynomial ring Ritts Factorization Theorem Schanuels Conjecture and Shapiros


  1. Exponential polynomials Paola D’Aquino Seconda Universita’ di Napoli Cesme, May 2012

  2. Topics Exponential rings, exponential fields and exponential polynomial ring Ritt’s Factorization Theorem Schanuel’s Conjecture and Shapiro’s Conjecture

  3. Exponential rings Definition: An exponential ring, or E -ring, is a pair ( R , E ) where R is a ring (commutative with 1) and E : ( R , +) → ( U ( R ) , · ) a morphism of the additive group of R into the multiplicative group of units of R satisfying 1 E ( x + y ) = E ( x ) · E ( y ) for all x , y ∈ R 2 E (0) = 1 . ( K , E ) is an E -field if K is a field.

  4. Examples Examples: 1 ( R , e x ); ( C , e x ); 2 ( R , E ) where R is any ring and E ( x ) = 1 for all x ∈ R . 3 ( S [ t ] , E ) where S is E -field of characteristic 0 and S [ t ] the ring of formal power series in t over S . Let f ∈ S [ t ], where f = r + f 1 with r ∈ S ∞ � ( f 1 ) n / n ! E ( f ) = E ( r ) · n =0 4 K [ X ] E E-ring of exponential polynomials over ( K , E )

  5. Exponential polynomials Sketch of the construction: Let R be a partial E -ring, R = D ⊕ ∆, where D = dom ( E ). Let t ∆ be a multiplicative copy of ∆, and consider R [ t ∆ ]. Extend E to R by defining E ( δ ) = t δ , δ ∈ ∆. Decompose R [ t ∆ ] = R ⊕ t ∆ −{ 0 } . Iterate ω times, and get E total. 1 Let R = Z [ X ], D = (0) e ∆ = R . The limit of previous construction is [ X ] E , the free E-ring on X . 2 Let ( K , E ) be an E-field and R = K [ X ]. Decompose K [ X ] = K ⊕ ∆, where D = K and ∆ = { f : f (0) = 0 } . The limit of previous construction is the E-ring K [ X ] E of exponential polynomials in X over K .

  6. Exponential polynomials An exponential polynomial in [ x , y ] E is represented as P ( x , y ) = − 3 x 2 y − x 5 y 7 + (2 xy + 5 y 2 ) e ( − 7 x 3 +11 x 5 y 4 ) +(6 − 2 xy 5 ) e (5 x +2 x 7 y 2 ) e 5 x − 10 y 2 T HEOREM Let ( R , E ) be an E-domain. Then R [ X ] E is an integral domain whose units are uE ( f ), where u is invertible in R and f ∈ R [ X ] E . D EFINITION An element f ∈ R [ X ] E is irreducible if there are no non-units g and h in R [ X ] E such that f = gh .

  7. Exponential polynomials D EFINITION Let f = � N i =1 a i t α i be an exponential polynomial. Then the support of f = supp ( f ) = Q -space generated by α 1 , . . . , α N . D EFINITION An exponential polynomial f ( x ) is simple if dim supp ( f ) = 1 . sin(2 π x ) = e 2 π ix − e − 2 π ix 2 i

  8. Factorization theory Ritt in 1927 studied factorizations of exponential polynomial 1 + β 1 e α 1 z + . . . + β k e α k z over C , using factorizations in fractional powers of classical polynomials in many variables. Gourin (1930) and Macoll (1935) gave a refinement of Ritt’s factorization theorem for exponential polynomials of the form p 1 ( z ) e α 1 z + . . . + p k ( z ) e α k z with α i ∈ C , and p i ( z ) ∈ C [ z ] . D’A. and Terzo (2011) gave a factorization theorem for general exponential polynomials f ( X ) ∈ K [ X ] E , where K is an algebraically closed field of characteristic 0 with an exponentiation.

  9. Ritt’s basic idea Ritt: reduce the factorization of an exponential polynomial to that of a classical polynomial in many variables in fractional powers. If Q ( Y 1 , . . . , Y n ) ∈ K [ Y 1 , . . . , Y n ] is an irreducible polynomial over K , it can happen that for some q 1 , . . . , q n ∈ N + , Q ( Y q 1 1 , . . . , Y q n n ) becomes reducible: X − Y irreducible, but X 3 − Y 3 = ( X − Y )( X 2 + XY + Y 2 ) Ex: D EFINITION A polynomial Q ( Y ) is power irreducible (over K ) if for each q ) is irreducible. q ∈ N n + , Q ( Y q ) A factorization of Q ( Y ) gives a factorization of Q ( Y q ) = Q ( Y q 1 1 , . . . , Y q n A factorization of Q ( Y n ) gives a factorization of Q ( Y 1 , . . . , Y n ) in fractional powers of Y 1 , . . . , Y n .

  10. Associate polynomial Let f ( X ) = � m h =1 a h t b h , where a h ∈ K [ X ] and b h ∈ ∆ and let { β 1 , . . . , β l } be a Z -basis of supp ( f ). Modulo a monomial we consider f as polynomial in e β 1 , . . . , e β l , with coefficients in K [ X ] . Let Y i = e β i , for i = 1 , . . . , l . f ( X ) ∈ K [ X ] E � Q ( Y 1 , . . . , Y l ) ∈ K [ X ][ Y 1 , . . . , Y l ] monomial: Y m 1 · . . . · Y m n n , where m 1 , . . . , m n ∈ Z , i.e. an 1 invertible element in K [ X ] E Simple exponential polynomials correspond to a single variable classical polynomials

  11. Factorization theorem If Q ( Y ) = Q 1 ( Y ) · . . . · Q r ( Y ) then f ( X ) = f 1 ( X ) · . . . · f r ( X ) and for any q positive integers q ) = R 1 ( Y ) · . . . · R p ( Y ) then f ( X ) = g 1 ( X ) · . . . · g p ( X ) . if Q ( Y All the factorizations of f ( X ) are obtained in this way. L EMMA Let f ( X ) and g ( X ) be in ∈ K [ X ] E . If g ( X ) divides f ( X ) then supp ( ag ) is contained in supp ( bf ), for some units a , b . Remark: If f is a simple polynomial and g divides f then g is also simple.

  12. Factorization theorem q ) is Problem: How many tuples q are there such that Q ( Y reducible? 1 1 k − Z k is a factor for all k > 0 Have to avoid Y − Z since Y T HEOREM There is a uniform bound for the number of irreducible factors of Q ( Y q 1 1 , . . . , Y q l l ) for Q ( Y 1 , . . . , Y l ) irreducible with more than two terms and arbitrary q 1 , . . . , q l ∈ N + . The bound depends only on M = max { d Y 1 , . . . , d Y l } .

  13. Factorization Theorem T HEOREM (Ritt) Let f ( z ) = λ 1 e µ 1 z + ... + λ N e µ N z , where λ i , µ i ∈ C . Then f can be written uniquely up to order and multiplication by units as f ( z ) = S 1 · . . . · S k · I 1 · . . . · I m where S j are simple polynomials with supp ( S j 1 ) � = supp ( S j 2 ) for j 1 � = j 2 , and I h are irreducible exponential polynomials.

  14. Factorization Theorem T HEOREM (D’A-Terzo) Let f ( X ) ∈ K [ X ] E , where ( K , E ) is an algebraically closed E -field of char 0 and f � = 0 . Then f factors, uniquely up to units and associates, as finite product of irreducibles of K [ X ] , a finite product of irreducible polynomials Q i in K [ X ] E with support of dimension bigger than 1, and a finite product of polynomials P j where supp ( P j 1 ) � = supp ( P j 2 ) , for j 1 � = j 2 and whose supports are of dimension 1. C OROLLARY If f is irreducible and the dimension of supp ( f ) > 1 then f is prime

  15. Schanuel’s conjecture Let α 1 , . . . , α n ∈ C , l . d . ( α 1 , . . . , α n ) =linear dimension of < α 1 , . . . , α n > Q tr . d . Q ( α 1 , . . . , α n ) =transcendence degree of Q ( α 1 , . . . , α n ) over Q . tr . d . Q ( α 1 , . . . , α n , e α 1 , . . . , e α n ) ≥ l . d . ( α 1 , . . . , α n ) ( SC ) Generalized Schanuel Conjecture Assume ( R , E ) is an E -ring and char ( R ) = 0. Let λ 1 , . . . , λ n ∈ R then tr . d . Q ( λ 1 , . . . , λ n , e λ 1 , . . . , e λ n ) − l . d . ( λ 1 , . . . , λ n ) ≥ 0

  16. Known cases Generalization of Lindemann-Weierstrass Theorem: Let α 1 , . . . , α n be algebraic numbers which are linearly independent over Q .Then e α 1 , . . . , e α n are algebraic independent over Q . 1 λ = 1 transcendence of e (Hermite 1873) 2 λ = 2 π i transcendence of π (Lindemann 1882) 3 λ = ( π, i π ) then tr.d.( π, i π, e , e i π ) = 2 , i.e. π, e π are algebraically independent over Q (Nesterenko 1996) 4 (SC) is true for power series C [[ t ]] (Ax 1971)

  17. The Schanuel machine λ = (1 , π i ) , SC ⇒ t . d (1 , i π, e , e i π ) ≥ l . d . (1 , i π ) . Then e , π, are algebraically independent over Q ; λ = (1 , i π, e ) SC ⇒ t . d (1 , i π, e , e , e i π , e e ) ≥ l . d . (1 , i π, e ) . Then π, e , e e are algebraically independent over Q ; λ = (1 , i π, i π 2 , e , e e , e i π 2 ) , SC ⇒ t . d (1 , i π, i π 2 , e , e e , e i π 2 , e , e i π , e i π 2 , e e , e e e , e e i π 2 ) ≥ l . d . (1 , i π, i π 2 , e , e e , e i π 2 ) . Then π, e , e e , e e e , e i π 2 , e e i π 2 are algebraically independent / Q .

  18. Algebraic consequences of Schanuel’s Conjecture T HEOREM (Macintyre) Suppose S is an E -ring satisfying (SC), and S 0 is the E -subring of S generated by 1. Then the natural E-morphism ϕ : [ ∅ ] E → S 0 is an E -isomorphism, i.e. S 0 is isomorphic to E -free ring on the empty set. C OROLLARY (SC) There is an algorithm which decides if two exponential constants coincide.

  19. Algebraic consequences of Schanuel’s Conjecture T HEOREM (Terzo) (SC) Let [ x , y ] E be the free E -ring generated by { x , y } and let ψ be the E -morphism: ψ : [ x , y ] E → ( C , exp ) defined by ψ ( x ) = π and ψ ( y ) = i . Then there exists a unique isomorphism f : [ x , y ] E / Ker ψ → � i , π � E and Ker ψ = � e xy + 1 , y 2 + 1 � E . C OROLLARY (SC) The only algebraic relations among π , e and i over C are e i π = − 1 and i 2 = − 1

  20. Algebraic consequences of Schanuel’s Conjecture T HEOREM (Terzo) (SC) Let [ x ] E be the free E-ring generated by { x } and let R be the E-subring of ( R , exp ) generated by π . Then the E-morphism ϕ : [ x ] E → ( R , exp ) x �→ π is an E-isomorphism. C OROLLARY (SC) 1 There is an algorithm for deciding if two exponential polynomials in π and i are equal in C . 2 There is an algorithm for deciding if two exponential polynomials in π are equal in R .

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