All Questions
13 questions
7
votes
1
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271
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Algebraic proof that the monoid ring of a torsion-free monoid is reduced
In what follows, I say that a monoid $M$ is torsion-free if the $n$-th power map is injective for all $n \geq 1$. I have a proof of the following result:
Claim: if $M$ is a torsion-free commutative ...
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7
votes
1
answer
167
views
For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup?
Let $K$ be a multiplicatively written semigroup (either commutative or not) and $H$ a subsemigroup of $K$. We say that $H$ is divisor-closed (in $K$) if $x \in H$ for all $x, y \in K$ such that $x \...
- 10.5k
4
votes
1
answer
385
views
Which monoids can be realized as the monoid of ideals of a commutative monoid?
Let $H$ be a commutative monoid (written multiplicatively). We say that a set $I \subseteq H$ is an ideal of $H$ if $IH = I$. The set $\mathcal I(H)$ of all ideals of $H$ is made into a (commutative) ...
- 10.5k
4
votes
1
answer
173
views
On the factorization of powers of atoms in the ring of integers of a number field
Let $H$ be a multiplicatively written monoid with identity $1_H$. An atom of $H$ is a non-unit element $a \in H$ that doesn't split into the product of two non-unit elements.
Given $x \in H$, we ...
- 10.5k
4
votes
0
answers
234
views
Do you know rings without involutions, auto-anti-isomorphics? In that case, what is the minimal example?
Do you know rings without involutions, but auto-anti-isomorphic (isomorphic to their opposite)? In that case, what is the minimal example?
If a ring has an involution f, then f is an anti-automorphism;...
4
votes
1
answer
364
views
Values attained by the coheight of $(H \setminus H^\times)^k$ as a function of $H$ and $k$
Edit (Apr 24, 2017). I'm updating this post in the light of the latest developments of a related thread.
Let $H$ be a multiplicatively written, commutative monoid, and set $M := H \setminus H^\times$,...
- 10.5k
3
votes
2
answers
101
views
A non-reduced, commutative BF-monoid s.t. $au = u$ for all $a \in \mathcal A(H)$ and $u \in H^\times$
Let $H$ be a monoid, and denote by $H^\times$ and $\mathcal A(H)$, respectively, the set of units (or invertible elements) and the set of atoms (or irreducible elements) of $H$ (an element $a \in H$ ...
- 10.5k
3
votes
2
answers
164
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Weak ideal systems $r$ for which the $r$-coheight satisfies a kind of triangle inequality
Let $H$ be a multiplicatively written, commutative monoid with identity $1_H$, and let $\mathcal P(H)$ be the power set of $H$. If $X, Y \subseteq H$, we will set $$XY := \{xy: x \in X,\, y \in Y\}.$$
...
- 10.5k
3
votes
0
answers
47
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Counting the monic atoms $f$ in the semiring $\mathbf N[x]$ with $f(0)=1$, bounded coefficients, and degree $k$ (in the limit as $k \to \infty$)
Let $H$ be the multiplicative monoid of the (usual) semiring of polynomials in one variable $x$ with coefficients in $\mathbf N$. Given $\alpha, k \in \mathbf N$, denote by $\mathcal A_k(\alpha)$ the ...
- 10.5k
2
votes
1
answer
122
views
If $H$ is essentially equimorphic to $K$, then is $H$ atomic only if so is $K$?
I will first state my question, and then give all the relevant definitions.
Q. Let $H$ and $K$ be monoids, and assume $H$ is essentially equimorphic to $K$. Is it true that $H$ is atomic only if so ...
- 10.5k
2
votes
0
answers
62
views
Extensions of an ideal-theoretic criterion for a monoid to be BF
Let $H$ be a multiplicatively written, commutative monoid. We denote by $H^\times$ the set of units (or invertible elements) of $H$, and by $\mathcal A(H)$ the set of atoms (or irreducible elements) ...
- 10.5k
1
vote
1
answer
96
views
If $H$ is commutative and unit-cancellative, then so is the monoid of non-empty ideals of $H$
Let $H$ be a (multiplicatively written) commutative monoid with identity $1_H$. Given $X, Y \subseteq H$, we take
$$XY := \{xy: x \in X,\, y \in Y\}.$$
We call a set $I \subseteq H$ an ideal of $H$ ...
- 10.5k
0
votes
0
answers
101
views
Spherical Rings
My question is concerned with filtered rings. It is a classical result that if $R$ is a finitely generated commutative ring graded by a semigroup $S$ then $S$ is also finitely generated.
The reverse ...
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