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 = n1, . . . , nb be a numerical monoid, with ni ≥ 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]: D[S] :=

• f ∈ D[x]
• f =

n

• i=0

aixi, where ai = 0 ⇒ i ∈ S

• Set D0 = {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 = dxng1 · · · gmp1 · · · pt where:

1 n, m, t ≥ 0 2 d ∈ D× is a nonzero scalar 3 gi ∈ D0, gi irreducible in D[x], gi /

∈ D[S]

4 pi ∈ D0, pi irreducible in D[x], pi ∈ 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 = dxng1 · · · gmp1 · · · pt where:

1 n, m, t ≥ 0 2 d ∈ D× is a nonzero scalar 3 gi ∈ D0, gi irreducible in D[x], gi /

∈ D[S]

4 pi ∈ D0, pi irreducible in D[x], pi ∈ 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 = dxng1 · · · gmp1 · · · pt where:

1 n, m, t ≥ 0 2 d ∈ D× is a nonzero scalar 3 gi ∈ D0, gi irreducible in D[x], gi /

∈ D[S]

4 pi ∈ D0, pi irreducible in D[x], pi ∈ D[S]

Observations:

1 n ∈ S 2 Lemma: Each pi 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 = dxng1 · · · gmp1 · · · pt Suppose g1 · · · gm ∈ D[S]. Then f consists of the following subfactors: f = dxng1 · · · gmp1 · · · pt 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 = dxng1 · · · gmp1 · · · pt Suppose g1 · · · gm ∈ D[S]. Then f consists of the following subfactors: f = dxng1 · · · gmp1 · · · pt 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 = x11 − 3x10 + x9 + x8 + 4x6 = x6(x + 1)(x − 2)2(x2 + 1)

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Example

f = dxng1 · · · gmp1 · · · pt Suppose g1 · · · gm ∈ D[S]. Then f consists of the following subfactors: f = dxng1 · · · gmp1 · · · pt 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 = x11 − 3x10 + x9 + x8 + 4x6 = x6(x + 1)(x − 2)2(x2 + 1)

• By Lemma, x2 + 1 is a prime element of Z[S]
• (x + 1)(x − 2)2 = x3 − 3x2 + 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 = x6(x + 1)(x − 2)2(x2 + 1) by taking a factorization of x6, a factorization of (x + 1)(x − 2)2 and throwing

• n the prime element x2 + 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 = x6(x + 1)(x − 2)2(x2 + 1) by taking a factorization of x6, a factorization of (x + 1)(x − 2)2 and throwing

• n the prime element x2 + 1.

x6 has two factorizations: [x2][x2][x2] and [x3][x3].

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 = x6(x + 1)(x − 2)2(x2 + 1) by taking a factorization of x6, a factorization of (x + 1)(x − 2)2 and throwing

• n the prime element x2 + 1.

x6 has two factorizations: [x2][x2][x2] and [x3][x3]. (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 = x6(x + 1)(x − 2)2(x2 + 1) by taking a factorization of x6, a factorization of (x + 1)(x − 2)2 and throwing

• n the prime element x2 + 1.

x6 has two factorizations: [x2][x2][x2] and [x3][x3]. (x + 1)(x − 2)2 is irreducible. So get two factorizations of f in D[S]: [x2][x2][x2][(x + 1)(x − 2)2][x2 + 1] and [x3][x3][(x + 1)(x − 2)2][x2 + 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 5 [x2][x2][x2][(x + 1)(x − 2)2][x2 + 1] 4 [x3][x3][(x + 1)(x − 2)2][x2 + 1] 4 [x2(x + 1)][x2(x − 2)][x2(x − 2)][x2 + 1] 4 [x2][x2(x + 1)(x − 2)][x2(x − 2)][x2 + 1] 4 [x2][x2(x + 1)][x2(x − 2)2][x2 + 1] 3 [x3(x + 1)(x − 2)][x3(x − 2)][x2 + 1] 3 [x3(x + 1)][x3(x − 2)2][x2 + 1] 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 = dxng1 · · · gmp1 · · · pt in D[x]. If g1 · · · gm ∈ D[S], then L(f ) = L(d) + L(xn) + L(g1 · · · gm) + L(p1 · · · pt) 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 = dxng1 · · · gmp1 · · · pt in D[x]. If g1 · · · gm ∈ D[S], then L(f ) = LD(d) + LS(n) + L(g1 · · · gm) + 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 = dxng1 · · · gmp1 · · · pt in D[x]. If g1 · · · gm ∈ D[S], then L(f ) = LD(d) + LS(n) + L(g1 · · · gm) + 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 = dxng1 · · · gmp1 · · · pt in D[x]. If g1 · · · gm ∈ D[S], then L(f ) = LD(d) + LS(n) + L(g1 · · · gm) + 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 = dxng1 · · · gmp1 · · · pt in D[x]. If g ∈ D0 such that gg1 · · · gm ∈ 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

UFDs General domains Computations

D a UFD. Consider a nonzero f ∈ D[S]. In D[x], f factors uniquely as f = dxng1 · · · gmp1 · · · pt where:

1 n, m, t ≥ 0 2 d ∈ D× is an element of D 3 gi ∈ D0, gi irreducible in D[x], gi /

∈ D[S]

4 pi ∈ D0, pi irreducible in D[x], pi ∈ D[S]

Observations:

1 n ∈ S 2 Lemma: Each pi is prime in D[S] Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

D a domain. Consider a nonzero f ∈ D[S]. In D[x], f factors uniquely as f = dxng1 · · · gmp1 · · · pt where:

1 n, m, t ≥ 0 2 d ∈ D× is an element of D 3 gi ∈ D0, gi irreducible in D[x], gi /

∈ D[S]

4 pi ∈ D0, pi irreducible in D[x], pi ∈ D[S]

Observations:

1 n ∈ S 2 Lemma: Each pi is prime in D[S] Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

So cannot really distinguish constant factors, prime factors, and just general irreducibles g of D0. Generally have to consider f = xng1 · · · gm, where each gi irreducible in D0.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

So cannot really distinguish constant factors, prime factors, and just general irreducibles g of D0. Generally have to consider f = xng1 · · · gm, where each gi irreducible in D0. However...

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 domain, and f ∈ D[S], which factors as f = xng1 · · · gm in D[x], each gi irreducible element of

• D0. If g1 · · · gm ∈ D[S], then

L(f ) = LS(n) + L(g1 · · · gm) 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 domain, and f ∈ D[S], which factors as f = xng1 · · · gm in D[x], each gi irreducible element of

• D0. If g1 · · · gm ∈ D[S], then

L(f ) = LS(n) + L(g1 · · · gm) Furthermore, the longest naive factorizations are the unique longest factorizations. Theorem (B. and Kennedy) Let S be a numerical monoid, D a domain, and f ∈ D[S], which factors as f = xng1 · · · gm in D[x], each gi irreducible element of

• D0. If g ∈ D0 such that gg1 · · · gm ∈ 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

Two questions: 1. 2.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Two questions:

• 1. In Theorem 1, L(f ) = LS(n) + L(g1 · · · gm), assuming

g1 · · · gm ∈ D[S]. How can you compute L(g1 · · · gm)? 2.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Two questions:

• 1. In Theorem 1, L(f ) = LS(n) + L(g1 · · · gm), assuming

g1 · · · gm ∈ D[S]. How can you compute L(g1 · · · gm)?

• 2. In Theorem 2, given some g1 · · · gm, can we hope to find

g ∈ D0 such that gg1 . . . gm ∈ D[S]? Is this g reasonable?

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

• 2. Given some g1, . . . , gm irreducibles in D0, can we hope to find

g ∈ D0 such that gg1 . . . gm ∈ D[S]? Is this g reasonable?

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

• 2. Given some g1, . . . , gm irreducibles in D0, can we hope to find

g ∈ D0 such that gg1 . . . gm ∈ D[S]? Is this g reasonable? Case 1. Suppose D has characteristic > 0. Since it is a domain, D has char p for some prime p. Assuming S is primitive, let t ≥ 0 be the least nonnegative integer such that pt ∈ S. By Freshman’s Dream, n

i=0 aixipt

= n

i=0 apt i xipt. So for any

g ∈ D[x], gpt ∈ D[S].

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

• 2. Given some g1, . . . , gm irreducibles in D0, can we hope to find

g ∈ D0 such that gg1 . . . gm ∈ D[S]? Is this g reasonable? Case 1. Suppose D has characteristic > 0. Since it is a domain, D has char p for some prime p. Assuming S is primitive, let t ≥ 0 be the least nonnegative integer such that pt ∈ S. By Freshman’s Dream, n

i=0 aixipt

= n

i=0 apt i xipt. So for any

g ∈ D[x], gpt ∈ D[S]. So given g1, . . . , gm ∈ D[x], take g = gpt−1

1

· · · gpt−1

m

. Reasonable?

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

• 2. Given some g1, . . . , gm irreducibles in D0, can we hope to find

g ∈ D0 such that gg1 . . . gm ∈ D[S]? Is this g reasonable? Case 1. Suppose D has characteristic > 0. Since it is a domain, D has char p for some prime p. Assuming S is primitive, let t ≥ 0 be the least nonnegative integer such that pt ∈ S. By Freshman’s Dream, n

i=0 aixipt

= n

i=0 apt i xipt. So for any

g ∈ D[x], gpt ∈ D[S]. So given g1, . . . , gm ∈ D[x], take g = gpt−1

1

· · · gpt−1

m

. Reasonable? If D is a UFD, , because L(dxng1 · · · gm) ≤ LD(d) + LS(n) + m (“in principle”). Even if D is not a UFD, for similar reasons our method of bounding probably is not profitable.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Case 2. D is a UFD, with quotient field K. Consider the multiplicative submonoid D[S] ∩ D0 of D[S].

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Case 2. D is a UFD, with quotient field K. Consider the multiplicative submonoid D[S] ∩ D0 of D[S]. Consider U (K) = 1 + xK[[x]] and HS(K) = {h ∈ U (K) | h =

∞

• i=0

aixi, where ai = 0 ⇒ i ∈ S} Then G(S, D) = U (K)/HS(K) is a group.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Case 2. D is a UFD, with quotient field K. Consider the multiplicative submonoid D[S] ∩ D0 of D[S]. Consider U (K) = 1 + xK[[x]] and HS(K) = {h ∈ U (K) | h =

∞

• i=0

aixi, where ai = 0 ⇒ i ∈ S} Then G(S, D) = U (K)/HS(K) is a group. We have a surjective homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. The image of g ∈ D0 under σ is easy to describe in terms of g.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. The image of g ∈ D0 under σ is easy to describe in terms of g. So if D a UFD and g1, . . . gm ∈ D0 with g1 · · · gm ∈ D[S], then L(g1 · · · gm) = LB(σ(g1) · · · σ(gm)). This handles Question 1.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. For Question 2: given g1, . . . gm ∈ D0, we can consider σ(g1) · · · σ(gm), which is a sequence of elements of G(S, D). There is a γ ∈ G(S, D) such that γ · σ(g1) · · · σ(gm) ∈ B(G(S, D)). By surjectivity, have g ∈ D0 such that σ(g) = γ, so have our g such that gg1 · · · gm ∈ D[S].

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. For Question 2: given g1, . . . gm ∈ D0, we can consider σ(g1) · · · σ(gm), which is a sequence of elements of G(S, D). There is a γ ∈ G(S, D) such that γ · σ(g1) · · · σ(gm) ∈ B(G(S, D)). By surjectivity, have g ∈ D0 such that σ(g) = γ, so have our g such that gg1 · · · gm ∈ D[S]. Is this g reasonable?

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. For Question 2: given g1, . . . gm ∈ D0, we can consider σ(g1) · · · σ(gm), which is a sequence of elements of G(S, D). There is a γ ∈ G(S, D) such that γ · σ(g1) · · · σ(gm) ∈ B(G(S, D)). By surjectivity, have g ∈ D0 such that σ(g) = γ, so have our g such that gg1 · · · gm ∈ D[S]. Is this g reasonable? For arbitrary UFDs D, can always find g ∈ D0 with

• rd(g) ≤ Frob(S).

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Homomorphism σ : D0 → F(G(S, D)), such that f ∈ D[S] iff σ(f ) ∈ B(G(S, D)). In particular, σ : D[S] → B(G(S, D)) is a transfer homomorphism. For Question 2: given g1, . . . gm ∈ D0, we can consider σ(g1) · · · σ(gm), which is a sequence of elements of G(S, D). There is a γ ∈ G(S, D) such that γ · σ(g1) · · · σ(gm) ∈ B(G(S, D)). By surjectivity, have g ∈ D0 such that σ(g) = γ, so have our g such that gg1 · · · gm ∈ D[S]. Is this g reasonable? For arbitrary UFDs D, can always find g ∈ D0 with

• rd(g) ≤ Frob(S). For D a finite field, can always find g ∈ D0

with ord(g) = 1.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Example application:

Theorem (B. and Kennedy) If K is a finite field and S = e, e + 1, . . . , 2e − 1. Suppose f ∈ K[S], where f = uxng1 . . . gm, with each gi irreducible in K0.

1 If g1 . . . gm ∈ K[S], then

L(f ) = LS(n) + LB(σ(g1) · · · σ(gm)).

2 Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Example application:

Theorem (B. and Kennedy) If K is a finite field and S = e, e + 1, . . . , 2e − 1. Suppose f ∈ K[S], where f = uxng1 . . . gm, with each gi irreducible in K0.

1 If g1 . . . gm ∈ K[S], then

L(f ) = LS(n) + LB(σ(g1) · · · σ(gm)).

2 If g1 . . . gm /

∈ K[S], there is a ∈ K such that (1 + ax)g1 . . . gm ∈ K[S] and L(f ) = L(f · (1 + ax)) − 1

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Theorem (B. and Kennedy) If K is a finite field and S = e1, e2. Suppose f ∈ K[S], where f = uxng1 . . . gm, with each gi irreducible in K0.

1 If g1 . . . gm ∈ K[S], then

L(f ) = LS(n) + LB(σ(g1) · · · σ(gm)).

2 If g1 . . . gm /

∈ K[S], there is a ∈ K such that (1 + ax)g1 . . . gm ∈ K[S] and L(f ) + 1 ≤ L(f · (1 + ax)) ≤ L(f ) + n2 − n2 n1

• Paul Baginski

Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

So naive approach to calculating longest factorization can be very useful, even for bounding the longest factorizations where naive approach doesn’t apply.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

So naive approach to calculating longest factorization can be very useful, even for bounding the longest factorizations where naive approach doesn’t apply. D Krull, or even D a domain, still can construct the same group G(S, D). Don’t get a transfer homomorphism, BUT does correspond partially to factorization.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings

UFDs General domains Computations

Thank you.

Paul Baginski Institut Camille Jordan, Universit´ e Claude Bernard Lyon 1 Factorization in Semigroup Rings