Mixing
Characterizing commutative semigroups with a factorization property.
Let $(N, times)$ be a commutative semigroup and assume that a countably infinite subset $P$ of $N$ algebraically generates $N$, and let ${mathcal F}(P)$ denote the set of all non-empty finite subsets of $P$. Assume that there exists a mapping
$$quad quad mathtt P: N to {mathcal F}(P)$$
satisfying the following properties:
$$tag 1 forall , p in P, mathtt P (p) = {p}$$
$$tag 2 forall , a,b,c in N, ; text{If } c = ab text{ then } mathtt P(c) = mathtt P(a) cup mathtt P(b)$$
Example: The function that maps every integer in $(mathbb N^{ge 2}, *)$ to its prime factors.
Question 1: Is every such structure the quotient of a universal one
generated by the 'alphabet of letters' in $P$ creating 'words'?
Question 2: If the answer is yes how to we create the quotients? Are
they all defined by factoring out relations, satisfying some rules, so
that the conditions of $mathtt P$ are still guaranteed to hold?
My Work
It looks like $(mathbb N^{ge 2}, *)$ is the universal structure with free generators the set of prime numbers.
abstract-algebra prime-factorization semigroups
edited Nov 22 '18 at 13:21
Klangen
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
add a comment |
Let $(N, times)$ be a commutative semigroup and assume that a countably infinite subset $P$ of $N$ algebraically generates $N$, and let ${mathcal F}(P)$ denote the set of all non-empty finite subsets of $P$. Assume that there exists a mapping
$$quad quad mathtt P: N to {mathcal F}(P)$$
satisfying the following properties:
$$tag 1 forall , p in P, mathtt P (p) = {p}$$
$$tag 2 forall , a,b,c in N, ; text{If } c = ab text{ then } mathtt P(c) = mathtt P(a) cup mathtt P(b)$$
Example: The function that maps every integer in $(mathbb N^{ge 2}, *)$ to its prime factors.
Question 1: Is every such structure the quotient of a universal one
generated by the 'alphabet of letters' in $P$ creating 'words'?
Question 2: If the answer is yes how to we create the quotients? Are
they all defined by factoring out relations, satisfying some rules, so
that the conditions of $mathtt P$ are still guaranteed to hold?
My Work
It looks like $(mathbb N^{ge 2}, *)$ is the universal structure with free generators the set of prime numbers.
abstract-algebra prime-factorization semigroups
edited Nov 22 '18 at 13:21
Klangen
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
add a comment |
Let $(N, times)$ be a commutative semigroup and assume that a countably infinite subset $P$ of $N$ algebraically generates $N$, and let ${mathcal F}(P)$ denote the set of all non-empty finite subsets of $P$. Assume that there exists a mapping
$$quad quad mathtt P: N to {mathcal F}(P)$$
satisfying the following properties:
$$tag 1 forall , p in P, mathtt P (p) = {p}$$
$$tag 2 forall , a,b,c in N, ; text{If } c = ab text{ then } mathtt P(c) = mathtt P(a) cup mathtt P(b)$$
Example: The function that maps every integer in $(mathbb N^{ge 2}, *)$ to its prime factors.
Question 1: Is every such structure the quotient of a universal one
generated by the 'alphabet of letters' in $P$ creating 'words'?
Question 2: If the answer is yes how to we create the quotients? Are
they all defined by factoring out relations, satisfying some rules, so
that the conditions of $mathtt P$ are still guaranteed to hold?
My Work
It looks like $(mathbb N^{ge 2}, *)$ is the universal structure with free generators the set of prime numbers.
abstract-algebra prime-factorization semigroups
edited Nov 22 '18 at 13:21
Klangen
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
Let $(N, times)$ be a commutative semigroup and assume that a countably infinite subset $P$ of $N$ algebraically generates $N$, and let ${mathcal F}(P)$ denote the set of all non-empty finite subsets of $P$. Assume that there exists a mapping
$$quad quad mathtt P: N to {mathcal F}(P)$$
satisfying the following properties:
$$tag 1 forall , p in P, mathtt P (p) = {p}$$
$$tag 2 forall , a,b,c in N, ; text{If } c = ab text{ then } mathtt P(c) = mathtt P(a) cup mathtt P(b)$$
Example: The function that maps every integer in $(mathbb N^{ge 2}, *)$ to its prime factors.
Question 1: Is every such structure the quotient of a universal one
generated by the 'alphabet of letters' in $P$ creating 'words'?
Question 2: If the answer is yes how to we create the quotients? Are
they all defined by factoring out relations, satisfying some rules, so
that the conditions of $mathtt P$ are still guaranteed to hold?
My Work
It looks like $(mathbb N^{ge 2}, *)$ is the universal structure with free generators the set of prime numbers.
abstract-algebra prime-factorization semigroups
abstract-algebra prime-factorization semigroups
edited Nov 22 '18 at 13:21
Klangen
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
edited Nov 22 '18 at 13:21
Klangen
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
edited Nov 22 '18 at 13:21
Klangen
1,65411334
edited Nov 22 '18 at 13:21
Klangen
1,65411334
edited Nov 22 '18 at 13:21
Klangen
1,65411334
1,65411334
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
asked Nov 20 '18 at 21:10
CopyPasteIt
4,0451627
4,0451627
add a comment |
add a comment |
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