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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101
votes
5answers
9k 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, "...
30
votes
6answers
5k views

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}: ...
30
votes
1answer
7k 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 ...
61
votes
5answers
9k 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 ...
15
votes
3answers
1k 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 ...
6
votes
1answer
2k 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 ...
7
votes
1answer
277 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 ...
69
votes
9answers
18k 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 ...
91
votes
11answers
5k 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 ...
27
votes
5answers
2k 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 ...
22
votes
12answers
3k views

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 ...
15
votes
1answer
845 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 ...
6
votes
1answer
219 views

If $f,g \in D[x,y]$ are algebraically dependent over $D$, then $f,g \in D[h]$ for some $h\in D[x,y]$?

This question asks: If $f,g \in k[x,y]$ are two algebraically dependent polynomials over an arbitrary field $k$, is it true that there exists a polynomial $h \in k[x,y]$ such that $f,g \in k[h]$, ...
43
votes
2answers
4k 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 $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. ...
51
votes
2answers
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 $M$ ...
29
votes
6answers
6k 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 ...
21
votes
5answers
4k 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 ...
25
votes
6answers
7k 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 ...
21
votes
2answers
6k 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 ...
21
votes
6answers
5k 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, ...
28
votes
7answers
4k 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 $\mathfrak{S}_n$ ...
35
votes
6answers
2k 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 ...
27
votes
1answer
10k 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 ...
13
votes
5answers
4k 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 $...
40
votes
2answers
2k 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 ...
19
votes
5answers
5k 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 ...
16
votes
4answers
3k 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 $\mathcal{O}_p(X)$ ...
10
votes
3answers
3k 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 ...
7
votes
2answers
1k 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:...
10
votes
1answer
734 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 ...
9
votes
0answers
434 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) ...
15
votes
1answer
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 ...
13
votes
3answers
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 ...
6
votes
2answers
350 views

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 $...
8
votes
1answer
198 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 ...
5
votes
1answer
222 views

$f,g \in \mathbb{Z}[x,y]$ 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) ...
0
votes
2answers
326 views

Rank of a $ \mathbb{Z}_{p}[[T]] $ module

Let $p$ be a prime and $M$ is a finitely generated $ \mathbb{Z}_{p}[[T]] $ module. Suppose $M[p]$ denotes the $p$-torsion of $M$. Then $M[p]$ and $M/(p)$ are both $ F_{p}$ vector spaces. So we can ...
99
votes
2answers
8k views

How would you solve this tantalizing Halmos problem?

$1-ab$ invertible $\implies$ $1-ba$ invertible has a slick power series "proof" as below, where Halmos asks for an explanation of why this tantalizing derivation succeeds. Do you know one? Geometric ...
118
votes
3answers
14k 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 justified by ...
95
votes
1answer
11k views

A short proof for $\dim(R[T])=\dim(R)+1$

For a commutative ring $R$ we clearly have $\dim(R[T]) \geq \dim(R)+1$. If $R$ is noetherian, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial ...
46
votes
4answers
14k views

When is the product of two ideals equal to their intersection?

Consider a ring $A$ and an affine scheme $X=SpecA$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds to the union $V(...
40
votes
4answers
3k 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 ...
74
votes
1answer
3k views

$R$ is isomorphic to $R[X,Y]$, but not to $R[X]$

Is there a commutative ring $R$ with $R \cong R[X,Y]$ and $R \not\cong R[X]$? This is a ring-theoretic analog of my previous question about abelian groups: In fact, in any algebraic category we may ...
26
votes
6answers
6k views

What is the universal property of normalization?

What is the universal property of normalization? I'm looking for an answer something like If X is a scheme and Y→X is its normalization, then the morphism Y→X has property P and any ...
36
votes
5answers
2k views

Explicit elements of $K((x))((y)) \setminus K((x,y))$

In an answer to the popular question on common false beliefs in mathematics Examples of common false beliefs in mathematics I mentioned that many people conflate the two different kinds of formal ...
32
votes
2answers
2k 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 ...
25
votes
5answers
6k views

Local complete intersections which are not complete intersections

The following definitions are standard: An affine variety $V$ in $A^n$ is a complete intersection (c.i.) if its vanishing ideal can be generated by ($n - \dim V$) polynomials in $k[X_1,\ldots, X_n]$. ...
16
votes
5answers
5k views

When a formal power series is a rational function in disguise

Given a formal power series $f \in k[[X]]$, where $k$ is a commutative field, is there any good way to tell whether or not $f\in k(X)$? Edit: To clarify, "good way to tell" means "computable ...
22
votes
2answers
2k views

Criteria for irreducibility of polynomial

If $f, g\in \mathbb C[a,b]$ are polynomials in two variables, are there easy criteria that allow to see if $f(x,y)-g(t,z)\in \mathbb C[x,y,t,z]$ is irreducible? Thank you very much, best
14
votes
5answers
4k 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 ...

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