Questions tagged [ac.commutative-algebra]
Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
541
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41
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An algebra of "integrals"
When discussing divergent integrals with people, I got curious about the following:
Is there an $\mathbb{R}$-algebra $A$ together with a map (could be defined on just a subspace)
$$\int_0^{\infty}: ...
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115
votes
5
answers
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What do epimorphisms of (commutative) rings look like?
(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "...
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3
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Guises of the Stasheff polytopes, associahedra for the Coxeter $A_n$ root system?
Richard Stanley keeps a famous running compilation of different guises of the celebrated Catalan numbers. The number of vertices of the associahedron is one instantiation among the multitude, and the ...
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1
answer
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Infinite tensor products
Let $A$ be a commutative ring and $M_i, i \in I$ be a infinite family of $A$-modules. Define their tensor product $\bigotimes_{i \in I} M_i$ to be a representing object of the functor of multilinear ...
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1
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Bijection implies isomorphism for algebraic varieties
Let $f:X\to Y$ be a morphism of algebraic varieties over $\mathbb C$. Assume that
a) $f$ is bijective on $\mathbb C$-points
b) $X$ is connected
c) $Y$ is normal.
Does it imply that $f$ is an ...
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4
votes
1
answer
460
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Infinite dimensional involutions: infinitely large sets of multivariate polynomials self-inverse under self-substitution
Examples of infinite dimensional involutions
Edit 2/25/23, as suggested by YCOR below: (Start)
The first return on a Google search on involution--from late Latin 'a rolling up'--gives the Oxford ...
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5
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Is there a "geometric" intuition underlying the notion of normal varieties?
I first got concious of the notion of normal varieties around 3 years ago and despite the fact that by now I can manipulate with it a bit, this notion still puzzles me a lot.
One thing that strikes me ...
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35
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6
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On the universal property of the completion of an ordered field
I have been trying to write up some notes on completion of ordered fields, ideally in the general case (i.e., not just completing $\mathbb{Q}$ to get $\mathbb{R}$ but considering the completion via ...
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9
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4
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Formal power series is Taylor expansion of rational function iff Hankel determinants vanish?
Let $$ u(T)=\sum_{n = 0}^\infty a_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $...
user97565
8
votes
1
answer
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Any two bivariate algebraically dependent polynomials are always in the same ring generated by some bivariate polynomial?
If $f(x,y)$ and $g(x,y)$ are two algebraically dependent polynomials over some field $k$, is it true that there exists a bivariate polynomial $p(x,y)$ such that both $f(x,y)$ and $g(x,y)$ are in the ...
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11
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Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
Is there an infinite field $k$ together with a polynomial $f \in k[x]$ such that the associated map $f \colon k \to k$ is not surjective but misses only finitely many elements in $k$ (i.e. only ...
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81
votes
30
answers
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Applications of the Chinese remainder theorem
As the title suggests I am interested in CRT applications. Wikipedia article on CRT lists some of the well known applications (e.g. used in the RSA algorithm, used to construct an elegant Gödel ...
75
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9
answers
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Irreducibility of polynomials in two variables
Let $k$ be a field. I am interested in sufficient criteria for $f \in k[x,y]$ to be irreducible. An example is Theorem A of this paper (Brindza and Pintér, On the irreducibility of some polynomials in ...
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votes
5
answers
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Does Smith normal form imply PID?
Let $R$ be a nonzero commutative ring with $1$, such that all finite matrices over $R$ have a Smith normal form. Does it follow that $R$ is a principal ideal domain?
If this fails, suppose we ...
user5810
26
votes
13
answers
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Homological algebra for commutative monoids?
Homological algebra for abelian groups is a standard tool in many fields of mathematics. How much carries over to the setting of commutative monoids (with unit)? It seems like there is a notion of ...
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25
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5
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Why does the (S2) property of a ring correspond to the Hartogs phenomenon?
Hartogs Theorem says every function whose undefined locus is of codim 2 can be extend to the whole domain. I saw people saying this corresponds to the (S2) property of a ring. But I can't see why this ...
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25
votes
2
answers
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Maximal ideals in the ring of continuous real-valued functions on ℝ
For a compact space $K$, the maximal ideals in the ring $C(K)$ of continuous real-valued functions on $K$ are easily identified with the points of $K$ (a point defines the maximal ideal of functions ...
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21
votes
6
answers
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A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite
I was trying to understand completely the post of Terrence Tao on Ax-Grothendieck theorem. This is very cute. Using finite fields you prove that every injective polynomial map $\mathbb C^n\to \mathbb ...
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15
votes
1
answer
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Are there non-reflexive modules isomorphic to their bi-dual?
Let $M$ be an $R$-module. We say that $M$ is reflexive if the natural map $M\rightarrow M^{**}$ is an isomorphism.
I'd like to know if there exists a module isomorphic to its bi-dual but not ...
- 269
11
votes
0
answers
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Inversion, Koszul duality, combinatorics and geometry
According to this MO answer Koszul duality is related to operations on generating series;
1) multiplicative inversion for quadratic algebras,
2) compositional inversion for quadratic operads,
3) ...
130
votes
3
answers
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When is the tensor product of two fields a field?
Consider two extension fields $K/k, L/k$ of a field $k$.
A frequent question is whether the tensor product ring $K\otimes_k L$ is a field. The answer is "no" and this answer is often ...
59
votes
4
answers
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When is the product of two ideals equal to their intersection?
Consider a ring $A$ and an affine scheme $X=\operatorname{Spec}A$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds ...
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3
answers
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Is it true that, as $\Bbb Z$-modules, the polynomial ring and the power series ring over integers are dual to each other?
Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z}...
- 1,692
51
votes
2
answers
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a categorical Nakayama lemma?
There are the following Nakayama style lemmata:
(the classical Nakayama lemma) Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$-module. If $m_1, \ldots, m_n$ generate $M$ ...
user19475
50
votes
2
answers
6k
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Ring-theoretic characterization of open affines?
Background
Recall that, given two commutative rings $A$ and $B$, the set of morphisms of rings $A\to B$ is in bijection with the set of morphisms of schemes $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. ...
- 4,937
49
votes
4
answers
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Why is there a duality between spaces and commutative algebras?
1) The category of affine varieties over $\mathbb{C}$ is equivalent to the opposite category of finitely generated reduced algebras over $\mathbb{C}$. The equivalence associates to an affine variety ...
- 9,026
42
votes
2
answers
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Is every Noetherian Commutative Ring a quotient of a Noetherian Domain?
This was an interesting question posed to me by a friend who is very interested in commutative algebra. It also has some nice geometric motivation.
The question is in two parts. The first, as stated ...
39
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5
answers
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What are the main structure theorems on finitely generated commutative monoids?
I should read J. C. Rosales and P. A. García-Sánchez's book Finitely Generated Commutative Monoids and L. Redei's book The Theory of Finitely Generated Commutative Semigroups. I haven't. But here's ...
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votes
2
answers
3k
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How can I define the product of two ideals categorically?
Given a commutative ring $R$, there is a category whose objects are epimorphisms surjective ring homomorphisms $R \to S$ and whose morphisms are commutative triangles making two such epimorphisms ...
- 112k
30
votes
1
answer
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Rank of a module
What's wrong with defining the rank of a finitely generated module over any (commutative) ring to be just the smallest number of generators? All books I know define rank only locally this way. But why ...
- 2,809
30
votes
6
answers
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Algebraic stacks from scratch [closed]
I have a pretty good understanding of stacks, sheaves, descent, Grothendieck topologies, and I have a decent understanding of commutative algebra (I know enough about smooth, unramified, étale, and ...
30
votes
7
answers
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Invariant polynomials under a group action (hidden GIT)
Let's say I start with the polynomial ring in $n$ variables $R = \mathbb{Z}[x_1,...,x_n]$ (in the case at hand I had $\mathbb{C}$ in place of $\mathbb{Z}$).
Now the symmetric group $\mathfrak{S}_n$ ...
- 1,943
29
votes
6
answers
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Duals and Tensor products
Let $A$ be a commutative ring with a unit element. Let $M$ and $N$ be $A$-modules. Let $M^v$ and $N^v$ be the dual modules. In general, do we have $M^v \otimes N^v \cong (M\otimes N)^v$? It is ...
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votes
2
answers
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Even XOR Odd Infinities?
Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$.
(http://en.wikipedia.org/wiki/Peano_axioms#First-...
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24
votes
5
answers
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Exotic principal ideal domains
Recently I realized that the only PIDs I know how to write down that aren't fields are $\mathbb{Z}, F[x]$ for $F$ a field, integral closures of these in finite extensions of their fraction fields that ...
21
votes
6
answers
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Noether's normalization lemma over a ring A
Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, ...
- 1,250
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votes
5
answers
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Fast computation of a Groebner basis. What is possible?
I need to compute a Groebner basis of 18 polynomials in 19 variables the terms of which have degree at most 3. My aim is to exploit a symmetry in a PDE problem and I am not an expert in algebra or ...
- 1,116
18
votes
2
answers
3k
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Ideals of the ring of smooth functions
The ring $C^\infty(M)$ of smooth functions on a smooth manifold $M$ is a topological ring with respect to the Whitney topology and the usual ring operations. Is it possible to describe, maybe under ...
- 81
17
votes
1
answer
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Smith Normal Form of powers of a matrix
What invariants of a matrix determine the Smith Normal Form (SNF) of all the powers of a matrix?
The question makes sense over any PID $R$. If we let $M = M_n(R)$ and $G=Gl_n(R)$, then SNF is a ...
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votes
4
answers
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Completion of a local ring of a curve
Let $X$ be a smooth projective irreducible curve defined over an algebraically closed field $\mathbb{K}$ (of arbitrary characteristic), and let $p\in X$ be a closed point. Denote by $\mathcal{O}_p(X)$ ...
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14
votes
0
answers
861
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What is the state of art in Groebner bases
How big polynomial systems can we deal with? How do you know when you don't even have to try?
Motivation:
Recently I tried to solve a problem posed in another MO question and ultimately I got stuck ...
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13
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3
answers
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which homogeneous polynomials split into linear factors?
Let $R$ be the set of homogeneous polynomials of degree $n$ in $d$ variables over $\mathbb{C}$. When $n>2$, the set of elements of $R$ that split into a product of linear factors forms a proper ...
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5
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reduced ⊗ reduced = reduced; what about connected?
Several questions actually.
All rings and algebras are supposed to be commutative and with $1$ here.
(1) Let $k$ be a field, and let $A$ and $B$ be two $k$-algebras. I need a proof that if $A$ and $...
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3
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Existence of prime ideals and Axiom of Choice.
One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple ...
- 1,768
11
votes
1
answer
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Homomorphisms from powers of Z to Z
I believe it is known that if I is a set of non-measurable cardinality, then any homomorphism $Z^I\to Z$ factors through a finite power. Here $Z$ is the group of integers. Can anyone give a ...
- 877
8
votes
2
answers
2k
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von neumann algebras and measurable spaces
I've read some pages on links between von neumann (VN) algebras and measurable spaces (Spectra of $C^*$ algebras and Non-commutative geometry from von Neumann algebras?), but I can't get the following:...
- 575
8
votes
1
answer
222
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Is the restriction of a graded automorphism linearizable in characteristic zero?
This question follows up a previous one which was answered by Todd Leason. I want to impose two new requirements on the setup.
Let $k$ be a characteristic zero field. Let $A=k[x_1,\dots,x_n]$ be the ...
- 2,811
7
votes
0
answers
477
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"Consecutive" irreducible polynomials
If $P\in {\mathbb Z}[X]$ is a polynomial of degree $2$, then
it is easy to see that for any integer $m$, at least one of the polynomials
$P-(m+1),P-(m+2),P-(m+3),P-(m+4)$ is irreducible in ${\mathbb Z}...
- 3,535
6
votes
1
answer
438
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$f,g \in \mathbb{Z}[x,y]$ satisfying: $\operatorname{Jac}(f,g)=0$ and $f,g \notin \mathbb{Z}[h]$ for every $h \in \mathbb{Z}[x,y]$?
Is it possible to find $f,g \in \mathbb{Z}[x,y]$ (with $\deg(f),\deg(g) \geq 1$) such that the following two conditions are satisfied:
(1) $\operatorname{Jac}(f,g)=f_xg_y-f_yg_x = 0$.
(2) ...
- 2,613
6
votes
2
answers
415
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Divisibility of the degree of an extension by the degree of its residual field
Let $A$ be an integrally closed domain whose quotient field is $K$, $L$ be a finite Galois extension of $K$, and $B$ be the integral closure of $A$ in $L$. Let $M_A$ be a maximal ideal of $A$, and $...
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