Factorization in Semigroup Rings Paul Baginski Institut Camille - PowerPoint PPT Presentation

UFDs General domains Computations Factorization in Semigroup Rings Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Second Annual Iberian Meeting on Numerical Semigroups Granada, Spain February 4, 2010 Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Joint work with K. Grace Kennedy, University of California, Santa Barbara . Work in preparation. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Semigroup Rings Let S = � n 1 , . . . , n b � be a numerical monoid, with n i ≥ 1 such that they form a minimal generating set. D a UFD. The semigroup ring of S over D is the following subring of D [ x ]: � � n � � � a i x i , where a i � = 0 ⇒ i ∈ S D [ S ] := f ∈ D [ x ] � f = � � i =0 Set D 0 = { f ∈ D [ x ] | f (0) � = 0 } . Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations D a UFD. Consider a nonzero f ∈ D [ S ]. In D [ x ], f factors uniquely as f = dx n g 1 · · · g m p 1 · · · p t where: 1 n , m , t ≥ 0 2 d ∈ D × is a nonzero scalar 3 g i ∈ D 0 , g i irreducible in D [ x ], g i / ∈ D [ S ] 4 p i ∈ D 0 , p i irreducible in D [ x ], p i ∈ D [ S ] Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations D a UFD. Consider a nonzero f ∈ D [ S ]. In D [ x ], f factors uniquely as f = dx n g 1 · · · g m p 1 · · · p t where: 1 n , m , t ≥ 0 2 d ∈ D × is a nonzero scalar 3 g i ∈ D 0 , g i irreducible in D [ x ], g i / ∈ D [ S ] 4 p i ∈ D 0 , p i irreducible in D [ x ], p i ∈ D [ S ] Observations: 1 n ∈ S 2 Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations D a UFD. Consider a nonzero f ∈ D [ S ]. In D [ x ], f factors uniquely as f = dx n g 1 · · · g m p 1 · · · p t where: 1 n , m , t ≥ 0 2 d ∈ D × is a nonzero scalar 3 g i ∈ D 0 , g i irreducible in D [ x ], g i / ∈ D [ S ] 4 p i ∈ D 0 , p i irreducible in D [ x ], p i ∈ D [ S ] Observations: 1 n ∈ S 2 Lemma: Each p i is prime in D [ S ] Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example f = dx n g 1 · · · g m p 1 · · · p t Suppose g 1 · · · g m ∈ D [ S ]. Then f consists of the following subfactors: f = dx n g 1 · · · g m p 1 · · · p t We can factor each subfactor separately to get a factorization of f . Does this naive approach produce a useful factorization of f ? Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example f = dx n g 1 · · · g m p 1 · · · p t Suppose g 1 · · · g m ∈ D [ S ]. Then f consists of the following subfactors: f = dx n g 1 · · · g m p 1 · · · p t We can factor each subfactor separately to get a factorization of f . Does this naive approach produce a useful factorization of f ? Example: S = � 2 , 3 � , and consider Z [ S ]. f = x 11 − 3 x 10 + x 9 + x 8 + 4 x 6 = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example f = dx n g 1 · · · g m p 1 · · · p t Suppose g 1 · · · g m ∈ D [ S ]. Then f consists of the following subfactors: f = dx n g 1 · · · g m p 1 · · · p t We can factor each subfactor separately to get a factorization of f . Does this naive approach produce a useful factorization of f ? Example: S = � 2 , 3 � , and consider Z [ S ]. f = x 11 − 3 x 10 + x 9 + x 8 + 4 x 6 = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) • By Lemma, x 2 + 1 is a prime element of Z [ S ] • ( x + 1)( x − 2) 2 = x 3 − 3 x 2 + 4 ∈ Z [ S ]. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example So we can factor f = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) by taking a factorization of x 6 , a factorization of ( x + 1)( x − 2) 2 and throwing on the prime element x 2 + 1. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example So we can factor f = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) by taking a factorization of x 6 , a factorization of ( x + 1)( x − 2) 2 and throwing on the prime element x 2 + 1. x 6 has two factorizations: [ x 2 ][ x 2 ][ x 2 ] and [ x 3 ][ x 3 ]. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example So we can factor f = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) by taking a factorization of x 6 , a factorization of ( x + 1)( x − 2) 2 and throwing on the prime element x 2 + 1. x 6 has two factorizations: [ x 2 ][ x 2 ][ x 2 ] and [ x 3 ][ x 3 ]. ( x + 1)( x − 2) 2 is irreducible. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example So we can factor f = x 6 ( x + 1)( x − 2) 2 ( x 2 + 1) by taking a factorization of x 6 , a factorization of ( x + 1)( x − 2) 2 and throwing on the prime element x 2 + 1. x 6 has two factorizations: [ x 2 ][ x 2 ][ x 2 ] and [ x 3 ][ x 3 ]. ( x + 1)( x − 2) 2 is irreducible. So get two factorizations of f in D [ S ]: [ x 2 ][ x 2 ][ x 2 ][( x + 1)( x − 2) 2 ][ x 2 + 1] and [ x 3 ][ x 3 ][( x + 1)( x − 2) 2 ][ x 2 + 1]. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Example Factorizations of f Length Factorization [ x 2 ][ x 2 ][ x 2 ][( x + 1)( x − 2) 2 ][ x 2 + 1] 5 [ x 3 ][ x 3 ][( x + 1)( x − 2) 2 ][ x 2 + 1] 4 [ x 2 ( x + 1)][ x 2 ( x − 2)][ x 2 ( x − 2)][ x 2 + 1] 4 [ x 2 ][ x 2 ( x + 1)( x − 2)][ x 2 ( x − 2)][ x 2 + 1] 4 [ x 2 ][ x 2 ( x + 1)][ x 2 ( x − 2) 2 ][ x 2 + 1] 4 [ x 3 ( x + 1)( x − 2)][ x 3 ( x − 2)][ x 2 + 1] 3 [ x 3 ( x + 1)][ x 3 ( x − 2) 2 ][ x 2 + 1] 3 So longest factorization has length L ( f ) = 5 and the shortest factorizations have length ℓ ( f ) = 3. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Theorem (B. and Kennedy) Let S be a numerical monoid, D a UFD, and f ∈ D [ S ] , which factors as f = dx n g 1 · · · g m p 1 · · · p t in D [ x ] . If g 1 · · · g m ∈ D [ S ] , then L ( f ) = L ( d ) + L ( x n ) + L ( g 1 · · · g m ) + L ( p 1 · · · p t ) Furthermore, the longest naive factorizations are the unique longest factorizations. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Theorem (B. and Kennedy) Let S be a numerical monoid, D a UFD, and f ∈ D [ S ] , which factors as f = dx n g 1 · · · g m p 1 · · · p t in D [ x ] . If g 1 · · · g m ∈ D [ S ] , then L ( f ) = L D ( d ) + L S ( n ) + L ( g 1 · · · g m ) + t Furthermore, the longest naive factorizations are the unique longest factorizations. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Theorem (B. and Kennedy) Let S be a numerical monoid, D a UFD, and f ∈ D [ S ] , which factors as f = dx n g 1 · · · g m p 1 · · · p t in D [ x ] . If g 1 · · · g m ∈ D [ S ] , then L ( f ) = L D ( d ) + L S ( n ) + L ( g 1 · · · g m ) + t Furthermore, the longest naive factorizations are the unique longest factorizations. What about other f in D [ S ]? Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Theorem (B. and Kennedy) Let S be a numerical monoid, D a UFD, and f ∈ D [ S ] , which factors as f = dx n g 1 · · · g m p 1 · · · p t in D [ x ] . If g 1 · · · g m ∈ D [ S ] , then L ( f ) = L D ( d ) + L S ( n ) + L ( g 1 · · · g m ) + t Furthermore, the longest naive factorizations are the unique longest factorizations. Theorem (B. and Kennedy) Let S be a numerical monoid, D a UFD, and f ∈ D [ S ] , which factors as f = dx n g 1 · · · g m p 1 · · · p t in D [ x ] . If g ∈ D 0 such that gg 1 · · · g m ∈ D [ S ] , then L ( f ) ≤ L ( fg ) − 1 Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations Suppose D is just generally a domain. Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings