Questions tagged [ac.commutative-algebra]

Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.

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6 answers
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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}: ...
36min's user avatar
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118 votes
5 answers
13k views

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, "...
Anton Geraschenko's user avatar
18 votes
3 answers
2k views

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 ...
Tom Copeland's user avatar
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36 votes
1 answer
9k views

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 ...
Martin Brandenburg's user avatar
16 votes
1 answer
5k views

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 ...
Alexander Braverman's user avatar
74 votes
5 answers
13k views

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 ...
aglearner's user avatar
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77 votes
9 answers
25k views

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 ...
Hailong Dao's user avatar
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35 votes
6 answers
3k views

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 ...
Pete L. Clark's user avatar
9 votes
4 answers
2k views

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 $...
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8 votes
1 answer
346 views

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 ...
Adam's user avatar
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5 votes
1 answer
560 views

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 ...
Tom Copeland's user avatar
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95 votes
11 answers
6k views

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 ...
Philipp Lampe's user avatar
81 votes
30 answers
68k views

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 ...
28 votes
5 answers
3k views

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 ...
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27 votes
5 answers
5k views

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 ...
Yuhao Huang's user avatar
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27 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 ...
Chris Schommer-Pries's user avatar
25 votes
2 answers
8k views

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 ...
Alon Amit's user avatar
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22 votes
6 answers
7k views

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 ...
aglearner's user avatar
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21 votes
5 answers
6k views

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 ...
warsaga's user avatar
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15 votes
1 answer
1k views

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 ...
Berry's user avatar
  • 269
12 votes
3 answers
4k views

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 ...
Ali Reza's user avatar
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52 votes
3 answers
5k views

What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra?

On page 98 of Weibel's An Introduction to Homological Algebra he mentions that the ring $R = \prod_{i=1}^\infty \mathbb{C}$ has global dimension $\geq 2$ with equality iff the continuum hypothesis ...
David White's user avatar
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132 votes
3 answers
20k views

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 ...
106 votes
0 answers
16k views

A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?

Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization below. If you post something that doesn't ...
Martin Brandenburg's user avatar
63 votes
4 answers
20k views

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 ...
evgeniamerkulova's user avatar
53 votes
3 answers
6k views

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}...
Maharana's user avatar
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52 votes
2 answers
3k views

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 $M$ ...
user avatar
51 votes
2 answers
7k views

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 $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. ...
Manny Reyes's user avatar
  • 5,102
49 votes
4 answers
4k views

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 ...
Yonatan Harpaz's user avatar
42 votes
2 answers
3k views

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 ...
Siddharth Venkatesh's user avatar
39 votes
5 answers
4k views

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 ...
John Baez's user avatar
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37 votes
2 answers
3k views

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 ...
Qiaochu Yuan's user avatar
31 votes
7 answers
5k views

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 $\mathfrak{S}_n$ ...
babubba's user avatar
  • 1,953
30 votes
2 answers
3k views

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-...
Russell Easterly's user avatar
30 votes
6 answers
7k views

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
6 answers
11k views

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 ...
jkramerm's user avatar
  • 510
30 votes
1 answer
13k views

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 ...
ashpool's user avatar
  • 2,837
25 votes
5 answers
2k views

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 ...
23 votes
6 answers
6k views

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, ...
user2330's user avatar
  • 1,310
20 votes
2 answers
4k views

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 ...
user18107's user avatar
  • 101
17 votes
1 answer
1k views

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 ...
Robert Bruner's user avatar
17 votes
4 answers
4k views

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 $\mathcal{O}_p(X)$ ...
Jérémy Blanc's user avatar
16 votes
5 answers
5k views

When are dual modules free?

Let $A$ be a commutative integral domain, with fraction field $K$. Let $T$ be a torsion-free finitely generated $A$ module, so $T \otimes_A K$ is a finite dimensional vector space $V$. Let $T^*$ be ...
David E Speyer's user avatar
14 votes
0 answers
909 views

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 ...
Vít Tuček's user avatar
  • 8,159
13 votes
3 answers
2k views

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 ...
Mark C. Wilson's user avatar
12 votes
5 answers
5k views

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 $...
darij grinberg's user avatar
11 votes
0 answers
603 views

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) ...
11 votes
1 answer
1k views

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 ...
Michael  Barr's user avatar
8 votes
2 answers
2k views

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:...
Issam Ibnouhsein's user avatar
8 votes
1 answer
233 views

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 ...
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