Timeline for For which abelian groups $G$ does the monoid of zero-sum sequences over $G$ embed into a ring as a divisor-closed subsemigroup?
Current License: CC BY-SA 3.0
23 events
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| Jul 28, 2017 at 11:22 | vote | accept | Salvo Tringali | ||
| Jul 28, 2017 at 8:21 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added 7 characters in body
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| Jul 28, 2017 at 8:14 | vote | accept | Salvo Tringali | ||
| Jul 28, 2017 at 8:14 | |||||
| Jul 28, 2017 at 8:06 | comment | added | Salvo Tringali | I've just realized that the question in the first of my comments above has an obvious (negative) answer if I don't specify that $f$ and $g$ have non-empty support. But even so, the inequality is definitively wrong, as noted by @DavidHandelman in another thread (now deleted): For all $n \in \mathbf N$, we have $x^{n+1}-1 = (x-1) \sum_{i=0}^n x^i$ in the polynomial ring $K[x]$ (which is the same as $K[H]$, when $H$ is the monoid of non-negative integers under addition). | |
| Jul 27, 2017 at 23:15 | answer | added | Salvo Tringali | timeline score: 4 | |
| Jul 26, 2017 at 15:24 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
fixed a detail
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| Jul 26, 2017 at 15:12 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
updated to include some late developments
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| Jul 26, 2017 at 14:18 | comment | added | Benjamin Steinberg | I had forgotten about negatives. If you have a monoid with trivial unit group it can't embed in a ring of characteristic not 2 in a divisor closed way because of units. | |
| Jul 26, 2017 at 14:09 | comment | added | Salvo Tringali | On the other hand, doesn't your remark entail that, if $R$ is a commutative unital ring, then for a reduced monoid $H$ to embed as a divisor-closed submonoid into the multiplicative monoid of the monoid ring $R[H]$ we need $1_R = -1_R$? Indeed, the existence of the embedding implies, from what I noted before about the atoms and units of divisor-closed submonoids, that $R^\times \subseteq R[H]^\times \cong H^\times$. Yet, $H$ being reduced means that $H^\times$ is trivial (by definition). So $1_R$ and its additive inverse in $R$ must coincide. What am I missing? | |
| Jul 26, 2017 at 13:56 | comment | added | Salvo Tringali | Divisor-closedness: Thanks for the clarification. Coefficients: I agree. But my sub-question is partially independent from the OP. Let me rephrase it as follows: "Let $D$ be an integral domain and $H$ a commutative, cancellative, torsion-free monoid. Pick $f, g \in D[H]$ and etc." I had worded it in terms of a field, because it shouldn't make any difference (for the sub-question I'm asking about the support of a product) to work in the field of fractions of $D$ rather than in $D$. | |
| Jul 26, 2017 at 13:32 | comment | added | Benjamin Steinberg | I meant divisor closed. But then you can't allow the coefficients to be a field in the monoid ring | |
| Jul 26, 2017 at 12:04 | comment | added | Salvo Tringali | As for 2nd to last of yours comments: Divisor-free = divisor-closed? Sorry, I'm not familiar with the former term, while the use of the latter is widespread (at least in factorization theory). As for your last comment: No, I don't want divisor-closedness to be defined up to associates. Therefore, if $H$ and $K$ are monoids and $H$ is divisor-closed (in $K$), then $H^\times = K^\times$ and $\mathcal A(H) = \mathcal A(K) \cap H$, where $\mathcal A(M)$ denotes the set of atoms (aka irreducible elements) of a given monoid $M$. | |
| Jul 26, 2017 at 11:36 | comment | added | Benjamin Steinberg | Do you not want factor closed up to units? | |
| Jul 26, 2017 at 11:00 | comment | added | Benjamin Steinberg | I am not sure the support can be controlled that way but H is divisor free in the monoid ring ZH | |
| Jul 25, 2017 at 17:52 | comment | added | Salvo Tringali | @BenjaminSteinberg Let me answer with another question: Let $K$ be a field and $H$ a commutative, cancellative, torsion-free monoid. Pick $f, g \in K[H]$, and denote by $\ast$ the multiplication in the monoid ring $K[H]$. Is it true that $|{\rm supp}(f \ast g)| \ge |{\rm supp}(f)| + |{\rm supp}(g)|-1$? Assume, if necessary, that $K$ has characteristic zero. | |
| Jul 25, 2017 at 11:53 | history | edited | YCor | CC BY-SA 3.0 |
fixed typo/abbreviation in title
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| Jul 25, 2017 at 9:32 | comment | added | Benjamin Steinberg | Does the semigroup ring work? | |
| Jul 25, 2017 at 9:23 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
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| Jul 25, 2017 at 9:17 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added some motivation
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| Jul 25, 2017 at 9:10 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added some motivation
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| Jul 25, 2017 at 9:04 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added some motivation
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| Jul 25, 2017 at 8:39 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
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| Jul 25, 2017 at 8:31 | history | asked | Salvo Tringali | CC BY-SA 3.0 |