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I'm given $(n_1,n_2,n_3)$, with $\operatorname{gcd}(n_1,n_2,n_3)=1$. Then, I need to find $c_1$, the least positive integer such that $c_1n_1=n_2\mathbb{N}+n_3\mathbb{N}$. I additionally need the specific coefficients $r_{12}$ and $r_{13}$ such that $c_1n_1=r_{12}n_2+r_{13}n_3$.

Note that every number mentioned so far is non-negative.

So far, I'm calculating the Apery set of $n_2\mathbb{N}+n_3\mathbb{N}$ (with respect to it's multiplicity), and checking for membership through this. While this works to find my $c_1$, it leaves me nowhere with finding $r_{12}$ or $r_{13}$. While I could immediatally solve the diophantine equation $c_1n_1=r_{12}n_2+r_{13}n_3$, I can't help but think this isn't a very efficient way to do it.

Should I just solve the linear diophantine equation, or is there some better way to find the particular way that $c_1n_1$ is written in terms of $n_2$ and $n_3$?

Mark Schultz-Wu
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  • How can a number $c_1n_1$ be equal to a set? Do you mean a generator for $n_2\mathbb{N}+n_3\mathbb{N}$? – AmirHosein Sadeghimanesh May 31 '17 at 15:30
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    @AmirHoseinSadeghiManesh that was notation from some paper I was copying. The notation is supposed to mean what you specified. – Mark Schultz-Wu May 31 '17 at 15:33
  • are you sure taking $g.c.d.(n_1,n_2,n_3)=1$ is enough to have existence of a natural number $c_1$ such that $c_1n_1$ generates $n_2\mathbb{N}+n_3\mathbb{N}$. For example take $(n_1,n_2,n_3)=(4,2,3)$. Then obviously for no choice of $c_1\in\mathbb{N}$, $c_1n_1\mathbb{N}$ is equal to $n_2\mathbb{N}+n_3\mathbb{N}$. – AmirHosein Sadeghimanesh Jun 12 '17 at 13:30

1 Answers1

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A numerical semigroup is a subset of $\mathbb{I}=\mathbb{N}\cup\{0\}$ which is a subsemigroup of $(\mathbb{I},+)$ containing $0$ and its complement is finite. The only cyclic numerical semigroup is the trivial one generated by $1$ which is the whole $\mathbb{I}$. For any other element $x\neq 1$, the semigroup generated by $\{x\}$ plus zero has an infinite complement since it is precisely $x\mathbb{I}$ or you can write $\{0\}\cup x\mathbb{N}$. If $x=0$ then the complement is $\mathbb{N}$, infinite. If $x$ is nonzero, since it is not $1$, it is greater than or equal to $2$, and there are infinite natural numbers not multiple of $x$.

Now knowing this, the only cases that $\langle n_2,n_3\rangle\cup\{0\}$ can be cyclic is when one of $n_2$ and $n_3$ is $1$ or $(n_2,n_3)=(2,3)$. Otherwise it won't be the whole $\mathbb{I}$ and hence it is not possible to generate it by one element. No matter what is $n_1$ and what you choose for $c_1$. Very simple example $(n_1,n_2,n_3)=(2,3,4)$. The set $(2c_1)\mathbb{I}$ never contains an odd number, no matter of what $c_1$ is! And $3\mathbb{I}+4\mathbb{I}=\{0,3,4,6,7,8,9,\cdots\}$. Of course you have $g.c.d(2,3,4)=1$ but it won't make the statement in your question to be true.

  • $\langle A\rangle$ is used for the subalgebraic structure generated by a set $A$, like subsemigroup, linear subspaces, ideals, etc.
  • A semigroup is cyclic if it is generated by one element.
AmirHosein Sadeghimanesh
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