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42 votes
6 answers
7k 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}: ...
36min's user avatar
  • 3,806
120 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
19 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
  • 10.5k
38 votes
1 answer
10k 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
17 votes
1 answer
6k 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
5 votes
1 answer
630 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
  • 10.5k
78 votes
5 answers
14k 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
  • 14.3k
29 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
  • 5,052
78 votes
9 answers
26k 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
  • 30.5k
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
26 votes
2 answers
9k 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
  • 6,734
15 votes
3 answers
3k 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
10 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 $...
user avatar
8 votes
1 answer
362 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
  • 201
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
84 votes
31 answers
70k 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
4k 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 ...
user avatar
27 votes
13 answers
4k 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 ...
Chris Schommer-Pries's user avatar
22 votes
6 answers
8k 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
  • 14.3k
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
  • 1,256
16 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
  • 279
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
  • 1,788
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
  • 30.3k
132 votes
3 answers
21k 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 ...
111 votes
0 answers
17k 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
101 votes
31 answers
29k views

Errata for Atiyah–Macdonald

Is there a good list of errata for Atiyah–Macdonald available? A cursory Google search reveals a laughably short list here, with just a few typos. Is there any source available online which lists ...
65 votes
4 answers
22k 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
9 answers
13k views

Is there a preferable convention for defining the wedge product?

There are different conventions for defininig the wedge product $\wedge$. In Kobayashi-Nomizu, there is $\alpha\wedge\beta:=Alt(\alpha\otimes\beta)$, in Spivak, we find $\alpha\wedge\beta:=\frac{(k+l)...
agt's user avatar
  • 4,306
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
  • 1,742
52 votes
2 answers
4k 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
52 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
  • 5,407
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
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
  • 22.3k
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
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
33 votes
5 answers
4k views

(Short) Exact sequences with no commutative diagram between them

This question was asked by a student (in a slightly different form), and I was unable to answer it properly. I think it's quite interesting. The problem is to produce an example of the following ...
Pierre's user avatar
  • 2,287
32 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
  • 1,993
32 votes
6 answers
12k 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
  • 530
30 votes
6 answers
8k 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
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
1 answer
14k 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
  • 2,857
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 ...
24 votes
5 answers
6k views

To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field?

In his answer to a question about simple proofs of the Nullstellensatz (Elementary / Interesting proofs of the Nullstellensatz), Qiaochu Yuan referred to a really simple proof for the case of an ...
user2734's user avatar
  • 1,411
23 votes
6 answers
7k 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
  • 1,320
22 votes
4 answers
2k views

Two questions about finiteness of ideal classes in abstract number rings

Let us say that an abstract number ring is an integral domain $R$ which is not a field, and which has the "finite norms" property: for any nonzero ideal $I$ of $R$, the quotient $R/I$ is finite. (I ...
Pete L. Clark's user avatar
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
18 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
956 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
  • 8,597

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