All Questions
613 questions
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}: ...
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, "...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 $...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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)...
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}...
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 $M$ ...
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 $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$. ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 $\mathfrak{S}_n$ ...
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 ...
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-...
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 ...
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
...
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, ...
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 ...
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 ...
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 ...
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 $\mathcal{O}_p(X)$ ...
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 ...
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 ...