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3 votes
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Levelled trees and the homotopy transfer theorem

In section 10.3.12 of Loday-Vallette's book "Algebraic operads", given a $P_\infty$-algebra $(A,d,\alpha)$ the Homotopy Transfer Theorem applied to $H_*(A,d)$ is studied. There, because the ...
groupoid's user avatar
  • 215
8 votes
1 answer
322 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively ...
Salvo Tringali's user avatar
6 votes
3 answers
551 views

Conjecture about commutative semigroups

Conjecture: given any commutative semigroup $S$ of order $n \ge 4$, there exist $a, b \in S$ with $a \ne b$, an integer $m \ge \lfloor (n-1)/2 \rfloor$, and two $m$-element subsets $X = \{x_1, \ldots, ...
Fabius Wiesner's user avatar
1 vote
0 answers
47 views

Examples of graded subrings of $\mathbb Q(T)$

The following question came up in some discussion on some very unrelated matters. A graded algebra $A$ is an algebra $A$ with a decomposition $A = \oplus_{i \in \mathbb Z} A_i$ such that $A_i A_j \...
user5831's user avatar
  • 2,029
-2 votes
1 answer
77 views

integral ring extension implies algebraicity of their fraction fields extension?

$\DeclareMathOperator\Fr{Fr}$There is something I don't get about the following : Start with the well known fact that if $A\subset B\subset C$ are rings with $B$ the integral closure of $A$ in $C$, ...
huurd's user avatar
  • 1,031
8 votes
2 answers
596 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups,...
Salvo Tringali's user avatar
5 votes
0 answers
216 views

Lifting a morphism between quasi-projective varieties

Let $\mathcal{V}$ be an affine algebraic variety over $\mathbb{R}$, $G$ be a finite group acting freely on $\mathcal{V}$. Consider the quotient space $Y:=\mathcal{V}/G$, which itself is a quasi-...
Math_Newbie's user avatar
7 votes
2 answers
488 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ...
Salvo Tringali's user avatar
4 votes
0 answers
267 views

If $\mathbb{C}[a,b,c] \subsetneq \mathbb{C}[x]$, then there exist $f,g$ s.t. $\mathbb{C}[a,b,c] \subseteq \mathbb{C}[f,g] \subsetneq \mathbb{C}[x]$

I ran into this MSE question and would like to ask about its answer and plausible generalizations. The quoted MSE question asks if the following claim is true or false and why: Claim: Let $a,b,c \in \...
user237522's user avatar
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0 votes
1 answer
223 views

On the Irreducibility of Cyclotomic polynomials

Let $F$ be any field with $\operatorname{Char} F=q$. Let $p$ be a prime such that $p\neq q$. Suppose $F$ has no $p$-th root of unity except $1$. Is it true that the cyclotomic polynomial $X^{p-1}+\...
S.D.'s user avatar
  • 494
1 vote
0 answers
110 views

How large can the Krull dimension of the Rees algebra be?

Let $d$ be a natural number. How large can the Krull dimension of the Rees algebra $A[It]$ be, where $A$ is a commutative ring of Krull dimension $d$, and $I$ is an ideal of $A$. Currently, I know the ...
Ryota Kuroki's user avatar
3 votes
0 answers
250 views

Action (of a graded monoid) required

Reference request: Did the construction below appear anywhere before? Any mentions of it or especially any links to something commonly known would be really helpful. I feel that it might be related to ...
Nikita Safonkin's user avatar
3 votes
0 answers
91 views

Quillen-Suslin theorem for non-strict polydiscs in the sense of Berkovich

Let $K$ be a complete non-archimedean field of mixed characteristic $(0,p)$. Choose $\rho_1,\dots,\rho_n\in \mathbb{R}_{>0}$ and let $P$ be a finite projective module over $K\langle\rho_1^{-1}t_1,\...
Peter Wong's user avatar
2 votes
1 answer
327 views

Completion of a local ring is noetherian (under some hypothesis)

I was reading the proof of Lemma 10.12 in this paper. In the second sentence, the following fact is used implicitly: Let $(R,\mathfrak{m})$ be a commutative local ring. Let $\widehat{R}$ be its $\...
Don's user avatar
  • 293
11 votes
1 answer
513 views

When does derived tensor product commute with arbitrary products?

Let $R$ be a commutative Noetherian ring. Let $M$ be an $R$-module. It is well-known that $M$ is finitely generated if and only if the functor $M\otimes_R (-)$ preserves arbitrary products (for ...
uno's user avatar
  • 412
7 votes
0 answers
202 views

Finite generation and finite presentation over a truncated valuation ring: is there an easier proof?

Let $K^+$ be a valuation ring which is $\pi$-adically complete for some pseudouniformizer $\pi$. Nagata 053E proved that every finitely generated and flat (equivalently, torsion-free) $K^+$-algebra is ...
Piotr Achinger's user avatar
5 votes
0 answers
128 views

What are the conditions for the dual of the exterior algebra to be isomorphic to the exterior algebra of the dual?

The exterior algebra $\Lambda^*_kM$ can be defined for a $k$-module $M$, where $k$ is a commutative ring. A number of sources mention, without condition or proof, a (canonical) isomorphism $$(\Lambda^*...
Cameron's user avatar
  • 171
4 votes
0 answers
176 views

What is the equivalent of Artin gluing for quasicoherent sheaves?

Given a topological space or locale $X$ and an open $j : U \hookrightarrow X$ with closed complement $i : K \hookrightarrow X$, the inverse image functor $\langle i^*, j^* \rangle : \textbf{Sh} (X) \...
Zhen Lin's user avatar
  • 15.9k
3 votes
0 answers
120 views

Derived tensor by perfect complex preserves exact triangle in singularity category?

Let $R$ be a commutative Noetherian ring. Let $\operatorname{D}_{sg}(R)$ be the singularity category of $R$, i.e., the Verdier localization of $D_b(\text{mod } R)$ by the thick subcategory of perfect ...
Snake Eyes's user avatar
3 votes
0 answers
95 views

History of the notion of integral ring extension?

[I asked that question in "history of science and mathematics" but didn't get any answer so I take my chance here. I hope this is not out of context] Can anyone give me references about the ...
huurd's user avatar
  • 1,031
6 votes
1 answer
307 views

Hochschild cohomology and differential operators

The Hochschild-Kostant-Rosenberg theorem says, that for a commutative algebra $R$ over a field $k$ with certain smoothness and finiteness, we have an identification $\mathrm{HH}^\bullet(R)\cong \...
Qwert Otto's user avatar
3 votes
2 answers
255 views

Is being graded commutative a necessary condition on $A$ such that $H^*(A)$ is commutative?

If we consider any differential graded algebra $A^\bullet$, then its homology is a graded algebra, since the tensor product interacts well with homology. A sufficient condidtion for the homology to be ...
curious math guy's user avatar
4 votes
1 answer
132 views

Zero dimensional complete intersection ring of length a power of $p$

Let $k$ be an algebraically closed field of characteristic $p>0$ and let $C$ be a $k$-algebra of finite dimension over $k$ such that $k[C^p]=k$. Under these hypothesis it is known by results of L. ...
Diego Sulca's user avatar
1 vote
0 answers
207 views

When is the derived category of a ring generated by injective modules

Are there any equivalent conditions on a ring to the condition that the localizing subcategory of $D(R)$ generated by injective modules is the entire category? Are there any non-examples in Boolean ...
Andy Jiang's user avatar
  • 2,356
5 votes
0 answers
176 views

Example of a Boolean Ring with infinite injective dimension over itself

It is known that Boolean rings can have infinite global dimension (free Boolean algebra on a large enough number of generators) [ see The Global Dimension of Boolean Rings by Pierce]. Are there any ...
Andy Jiang's user avatar
  • 2,356
1 vote
0 answers
136 views

Local cohomology and image of $1$ under the canonical map from Ext to local cohomology

Let $R$ be a commutative Noetherian local ring, and $S$ be an $R$-algebra. Let $x_1,\dots,x_t$ be elements, in the maximal ideal of $R$, which is a regular sequence on both $R$ and $S$, and let $I$ be ...
uno's user avatar
  • 412
3 votes
2 answers
271 views

Orbits under the automorphism group of projective space

Let $\mathbb{P}^d_K$ be projective space of dimension $d\geq 1$ over an infinite field $K$. Let $x\in\mathbb{P}^d_K$ with $\dim\overline{\lbrace x\rbrace}=n\leq d-1$. My question: is the set $\lbrace ...
Vector's user avatar
  • 133
5 votes
1 answer
264 views

Non-existence of power divided structure on a maximal ideal of truncated polynomial rings (example from Koblitz)

In 3.3 of Berthelot-Ogus's book Notes on Crystalline Cohomology, an example from Koblitz is exhibited without proof. Let $k$ be a field (or a commutative ring) with characteristic $p>0$, $$A:=k[x_1,...
XYC's user avatar
  • 441
2 votes
1 answer
159 views

Field extensions and completions at possibly infinite places

In Serre's Corps Locaux, Chapter 2 §3, is presented a classical proof. We are in an "ABKL" setup, where $K/L$ is finite, $A$ is Dedekind, $B$ is the integral closure of $A$ and $B$ is $A$-...
Adrien Zabat's user avatar
3 votes
0 answers
89 views

Ordering the elements of a semigroup by $a \le b$ iff $a=b$ or $b=ab=ba$

Let $S$ be a semigroup, written multiplicatively. The binary relation $\le$ on (the underlying set of) $S$, whose graph consists of all pairs $(a,b) \in S \times S$ such that $a = b$ or $b = ab = ba$, ...
Salvo Tringali's user avatar
6 votes
0 answers
152 views

Can Harrison cohomology be written using Ext?

Just like Hochschild cohomology for associative algebras and Chevalley-Eilenberg cohomology for Lie algebras, it'll be nice (or disappointing?) if Harrison cohomology can be expressed in terms of Ext'...
Qwert Otto's user avatar
2 votes
1 answer
149 views

Finitely generated modules over completion

Let $k$ be a field, $A$ a finitely generated $k$-algebra and $I \subset A$ an ideal with $I$-adic completion $\hat{A} = \varprojlim A/I^n$. Is every finitely generated $\hat{A}$-module the completion ...
jba's user avatar
  • 23
3 votes
1 answer
180 views

How to find equations of $\mathbb{C}^*$-curves

Fix positive integers $t_1,t_2,t_3$. Suppose we have a $\mathbb{C}^*$ action on $\mathbb{C}^3\setminus\{0\}$ defined by $$\mathbb{C}^* \times \mathbb{C}^3 \setminus \{0\} \to \mathbb{C}^3 \setminus \{...
user43198's user avatar
  • 1,981
4 votes
1 answer
330 views

An assertion of Mahler

Let $\rho$ be an integer greater than $1$. In the article "Uber das Verschwinden von Potenzreihen mehrerer Veränderlichen in speziellen Punktfolgen" https://link.springer.com/article/10.1007/...
joaopa's user avatar
  • 3,998
5 votes
1 answer
151 views

Dimension from Hilbert series with variable-weighted grading?

Suppose $R = F[x_1,...,x_n]/I$ is a finitely generated ring over the field $F$, and suppose there is a function $d:[n] \to \mathbb{Z}$ such that defining the degree of a monomial $x_1^{e_1} ... x_n^{...
Joshua Grochow's user avatar
1 vote
0 answers
104 views

Is a normal domain a filtered colimit of Noetherian normal domains?

As described in the title, is any normal domain a filtered colimit of Noetherian normal domains? It will be great if one can show this, even with additional conditions, or if one can provide a ...
GTA's user avatar
  • 1,024
5 votes
1 answer
156 views

The inverse limit of a sequence of ring surjections commutes with taking difference subsets of the respective units & gluing in some primes?

Define $R_n := \Bbb{Z}/p_n\#$ the ring of integers modulo primorial $p_n\# = p_n p_{n-1} \cdots p_1$. Let $U_n$ denote the group of units modulo $p_n\#$ in these rings. Then if $f_{n,n+1}: \Bbb{Z}/p_{...
Daniel Donnelly's user avatar
10 votes
1 answer
243 views

If $E_\text{sep}/F$ is normal, then must $E/F$ be normal?

This question has been asked in Math.StackExchange (see here) for more than a week and I even put a bounty on it. But still it hasn't been correctly answered (the current answer there was written by ...
Z Wu's user avatar
  • 452
1 vote
0 answers
216 views

Dimension under change of ground field

I apologize if this question is "too elementary" - I am not an algebraic geometer. Are the following statements true? Let $k\subset K$ an extension of algebraically closed fields of ...
Scheman's user avatar
  • 19
0 votes
0 answers
114 views

Clarifications sought on the paper on the semigroup associated with a free polynomial by Ali Abbas and Abdallah Assi

I have three questions regarding the proof of Proposition 4 on page 4 of this paper here. For those interested in addressing these questions, please refer to some definitions in the first two or three ...
Mousa hamieh's user avatar
3 votes
0 answers
161 views

On generators of the multiplicative semigroup $\{r\in\mathbb Q:\ r>1\}$

The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the ...
Zhi-Wei Sun's user avatar
  • 15.6k
3 votes
0 answers
92 views

Reference for the monoidal category structure $X \otimes Y = X + Y + X \times Y$ on a distributive category

Given a distributive category $\mathscr C$ (more generally a rig category), we can define a (semicocartesian) monoidal category structure on $\mathscr C$ with tensor product given by $X \otimes Y := X ...
varkor's user avatar
  • 10.7k
2 votes
1 answer
262 views

Randomly fixing elements and transcendence degree

Given $f_1,\ldots,f_n \in \mathbb{F}_q[x_1,\ldots,x_m]$ such that $\deg(f_i) \leq d < q$. Suppose we have for some $1 \leq j \leq m$ $$ \operatorname{trdeg}_{\mathbb{F}(x_1,\ldots,x_j)}\{f_1,\ldots,...
Rishabh Kothary's user avatar
5 votes
0 answers
211 views

On the natural map $\mathrm{Br}(R) \rightarrow \mathrm{Br}(S)$ of Brauer groups

$\DeclareMathOperator\Br{Br}$Let $R$ be a commutative ring, and let $\Br(R)$ be the Brauer group of $R$ as defined by Auslander and Goldman. Let $S$ be a commutative $R$-algebra, and consider the ...
Moutand Mohammed's user avatar
3 votes
3 answers
387 views

Basic question about completion of local ring

Let $(A,m)$ be a Noetherian local ring and $\tilde{A}$ its completion by the maximal ideal $m$. Are the following three statements true? (i) If $\tilde{A}$ is free over $A$, then $A\cong \tilde{A}$ (...
George's user avatar
  • 328
3 votes
0 answers
181 views

Conditions for an open mapping between spectra

Let $(A,\mathfrak{m})$ be a Locally Noether Ring, and $\hat{A} = \varprojlim A/ \mathfrak{m}^{n}$ .Furthermore, let $f : A \to \hat{A}$ be a canonical morphism, and consider the mapping $f^{*} : Spec(\...
mathle's user avatar
  • 41
9 votes
1 answer
326 views

Nonzero module with vanishing derived fibers

What's an example of a nonzero $R$-module with vanishing derived fibers at all points of $\mathrm{Spec}(R)$? This was asked in When does a quasicoherent sheaf vanish? but the answer there only says ...
Andy Jiang's user avatar
  • 2,356
1 vote
1 answer
67 views

An example of a commutative ring $R$ which has a proper right ideal which is not a right SIP $R$-module

Recall that a module $M_R$ ($R$ is a ring with unity) is called SIP if the intersection of any two summands of $M$ is also a summand. I asked before if there exists a commutative ring which is not an ...
Hussein Eid's user avatar
2 votes
1 answer
232 views

Reference for surjectivity of the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$

Let $R$ be a commutative ring, $G$ a finite group with an action over $R$. Let $G_1, G_2 \subset G$ be two subgroups. Then the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$ is ...
Zheming Xu's user avatar
2 votes
1 answer
181 views

Idempotent algebras over absolutely flat ring

Is it possible to classify all idempotent algebras over an absolutely flat ring? Are there any idempotent $E_{\infty}$ algebras which are not discrete? I am particularly interested in the special case ...
Andy Jiang's user avatar
  • 2,356