Recent MathOverflow Questions

Multiple mirrors phenomenon from SYZ and HMS perspective

Math Overflow Recent Questions - Fri, 02/15/2019 - 03:09

There is a set of ideas called mirror symmetry which, roughly speaking, relates symplectic and complex geometry of Calabi--Yau manifolds. There are also extensions to Fano and general type varieties but let's forget about them for now.

There are several more or less precise incarnations of mirror symmetry, including so-called SYZ and HMS programs. Each gives us a certain recipe to construct the manifold mirror to a given Calabi--Yau.

In SYZ, the expectation is that to construct the mirror manifold we need to put our Calabi--Yau in a family of Calabi--Yau's over the punctured disc so that the monodromy operator is maximally unipotent. Different maximally unipotent degenerations can lead to different mirror manifolds.

In HMS, let's assume that we have finally found the right definition of Fukaya category. After certain algebraic manipulations, we can cook up a triangulated category which should be equivalent to the derived category of coherent sheaves on the mirror manifold. If we pick a t-structure in that category, we can Gabriel-style reconstruct the mirror manifold itself from the heart of t-structure. However, if I understand correctly, there is no natural choice of t-structure so we may get several mirror manifolds.

The question is: are these two ambiguities (choosing a maximally unipotent degeneration and choosing a heart of t-structure) in some sort of bijective correspondence? Are there some references where this is studied?

P.S.: to avoid certain subtleties in hyperkaehler geometry, let's assume that the holonomy group of Calabi--Yau's is equal the special unitary group.

EDIT: on second thought, it feels like this should be wrong (Calabi--Yau's can be derived equivalent for reasons that have nothing to do with mirror symmetry). Is there some way to single out the mirrors arising from SYZ inside the set of mirrors arising from HMS?

Is it true that $\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k}$?

Math Overflow Recent Questions - Fri, 02/15/2019 - 02:07

On Jan. 27, 2012, I conjectured the identity $$\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^\infty\frac{\binom{2k}k^2H_{2k}}{(2k+1)16^k},\tag{$*$}$$ where $H_n$ denotes the harmonic number $\sum_{k=1}^n\frac1k$. As the two series converge slowly, I lack convincing numerical data to support $(*)$.

Question. Is the identity $(*)$ true? Can one check it further? If it is true, how to prove it?

Your comments are welcome!

Motivation. $(*)$ was motivated by my following conjecture on congruences.

CONJECTURE (Jan 26, 2012). For any prime $p>3$, we have

$$\sum_{k=1}^{(p-1)/2}\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k) \equiv -\frac 73pB_{p-3}\pmod{p^2}\tag{1}$$


$$\sum_{k=1}^{(p-3)/2}\frac{\binom{2k}k^2}{(2k+1)16^k}H_{2k} \equiv -2\left(\frac{-1}p\right)E_{p-3} \pmod p,\tag{2}$$

where $(\frac{\cdot}p)$ is the Legendre symbol, $B_0,B_1,\ldots$ are the Bernoulli numbers and $E_0,E_1,\ldots$ are the Euler numbers.

I noted in 2012 that (2) is equivalent to (1) modulo $p$. See also Conjecture 1.1 of my 2014 JNT paper.

Normal subgroups of $p$-groups

Math Overflow Recent Questions - Fri, 02/15/2019 - 01:48

I was reading Professor Yukov Berkovich' paper "On Subgroups of Finite $p$-groups" when I stumled upon the following theorem:

Let $G$ be a nonabelian $p$-group with cyclic Frattini subgroup, $|\Phi(G)|>p$ and let $\Phi_0$ be the subgroup of order $p$ in $\Phi(G)$. Let $Z$ be cyclic subgroup of maximal order in $G$ containing $\Phi(G)$; then $|Z|=p|\Phi(G)|$. Set $\Delta_1:=$ {$H<G|[H:Z]=p$}. Suppose that every $H\in \Delta_1$ contains a normal abelian subgroup of type $(p,p)$. Then $\Phi(G)\leq Z(G)$ and $G=(A_1***A_s)Z(G)$, where $A_i$ are minimal nonabelian and $\Phi_0=G'=A_i'$ for all $i$. If $|A_i|>p^3$, then $A_i$ has a cyclic subgroup of index $p$. Moreover, $G=AZ$, where $A$ is generated by all normal subgroups of $G$ of type $(p,p)$ containing $\Phi_0$.

Here $"*"$ denotes the central product. In his paper, Yukov does not give a reference, and when I contacted him by mail he was still unable to do so. I think the theorem can play a vital role in the bit of research I am doing in character theory, so I would very much like to see the proof.

Can somebody give me a reference for the above theorem (or perhaps a similar statement)?

On spectral multiplicity of left shift operators

Math Overflow Recent Questions - Fri, 02/15/2019 - 01:45

Let $U$ be an operator defined on $l^{2}(\mathbb{Z})$ by $U(e_{n})=e_{n-1}$, where $e_{n}$ is an orthonormal basis of $l^{2}(\mathbb{Z})$. $U$ is a left shift operator. Since $U$ is unitary operator so spectrum is on $S^{1}$. What is spectral measure of $U$? What is its spectral decomposition with respect to multiplicity?

Motivation behind Analytic Number Theory

Math Overflow Recent Questions - Thu, 02/14/2019 - 16:25

I am an undergraduate student of mathematics and recently took an introductory course in analytic number theory, where the instructor roughly followed Apostol's first text on the subject. I have now started reading Davenport's 'Multiplicative Number Theory'. Without going into too much detail, I found that, while the results used some common techniques in their proofs, they were otherwise quite independent. Since I am relatively inexperienced, I found that quite strange, considering that most subjects I have read so far (real and complex analysis, abstract algebra, measure theory, functional analysis, algebraic topology) seem to have a coherent development, rather than simply be a collection of problems that have been solved using roughly similar machinery.

I was wondering if there is an overarching idea behind the study of analytic number theory (classical, as well as sieve methods), any specific open problems that motivated past research in the field, and if there is any textbook that treats it from that perspective, rather than just a collection of interesting problems. For instance, although I know very little about it, my professor once told me about the Langlands Program and said that the current goal of several mathematicians working in areas of algebraic number theory and automorphic forms to resolve the conjectures of that program.

Also, unlike many other areas, I couldn't effectively use the approach many of my professors seem to recommend, of reading the theorem and attempting to prove it by myself. I am inclined to believe that it is my own shortcomings that prevent this approach, but if it is a part of a wider trend, and I am reading it "wrong", for want of a better phrase, I would like to know the same, and would like to know how exactly such a subject is to be most efficiently learnt.

I haven't quite managed to phrase the question as well as I had hoped to, so, if you have any replies to the title in itself, any motivation towards a coherent understanding of the subject, then, I would be much obliged to you for your input/advice.

I wasn't quite sure what tag to use, and so, chose what I thought was most appropriate. I hope this will not be an issue.

What do the eigenvectors of the $n$th roots of $I_n$ look like?

Math Overflow Recent Questions - Thu, 02/14/2019 - 14:23

This was asked at math stackexchange a long time ago with no answers but some upvotes.

Let $A^n=I,$ where $A$ is $n\times n,$ and assume that $A^k\neq I,$ for all $1\leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues,

Thus there are a full set of linearly independent eigenvectors.

  • What do they look like?
  • If we assume $A$ is orthogonal, what do they look like?
  • If we assume $A$ is real, can one say anything more?

Commutative rings : Topoi = Fields :?

Math Overflow Recent Questions - Thu, 02/14/2019 - 13:56

The following is probably a bad question, but hopefully, it might have a very good answer.

In category theory there is a quite famous analogy between topoi and commutative rings, I was never convinced by this analogy, but the best way to see how far an analogy can be pushed is to challenge it. Clicking on the link that I provided above you can have an extensive presentation of the analogy, the general motto can be grasped by the following table.

That corresponds to Rem in my version of HTT by Lurie.

Q. According to this analogy, what should be a field?

Maybe I should say why this might be a stupid question or even a stupid challenge for the analogy. In fact it might be the case that:

  1. The notion of field is interesting only in low dimension.
  2. The correct generalization of the notion of field looks very different in categories and trivializes for sets because of their intrinsic rigidity.

Decompose geometric mean [on hold]

Math Overflow Recent Questions - Thu, 02/14/2019 - 13:27

Sorry if this is a super-dumb question. I have three data series: organic growth, volume growth and price growth. Organic growth = volume growth + price growth. I have converted them into indices and calculated a ten year geometric average for each series. The ten year organic growth geometric average does not equal the sum of the ten year volume and price growth geometric averages

Taylor Expansion on a Riemannian Manifold in Normal Coordinates

Math Overflow Recent Questions - Thu, 02/14/2019 - 13:15

Let $\phi: (M,g)\hookrightarrow (N,\tilde{g})$ be an isometric embedding of a Riemannian manifold $M$ of dimension $m$ into a Riemannian manifold $N$ of dimension $n$. I am interested in trying to do a Taylor expansion of this mapping in Riemannian Normal coordinates.

I have shown previously that for any real-valued function $f: M\rightarrow \mathbb{R}$, the covariant $k$-th derivative $$ \nabla^{k+1} f [X_1,...,X_{k+1}]= X_{k+1}(\nabla^{k}[X_1,...,X_k])-\sum_{i=1}^k\nabla^k f[X_1,...,X_{i-1},\nabla_{X_{k+1}}X_i,...,X_k] $$

written Riemannian normal coordinates (or in in any coordinates where the Christoffel symbols vanish) is simply the expression $$ \sum_{I}X^I \frac{\partial^I f}{\partial X^I} $$ Where $I$ is a multiindex of size $k$ varying over all choices of $m$ integers from the set $\{1,...,m\}$ (This looks like the $k$-th derivative.)

This suggests that when I choose normal coordinates in $M$, and perform a Taylor expansion of $f$, I am actually computing the coordinate expression of $$ \nabla f[X] + \nabla^2 f[X,X] +\nabla^3f [X,X,X] + ... $$

If I wish to more generally understand the taylor expansion of a mapping between manifolds, $\phi: (M,g)\hookrightarrow (N,\tilde{g})$ as above, I can choose a coordinate mapping $\eta:U\rightarrow \mathbb{R}^n$ in $N$ and regard the $i$-th component function of the mapping $\eta\circ\phi$.

This suggests to me that there might be a way to Taylor expand $\phi$ in terms of it's covariant derivatives that is independent of choosing a particular chart $\eta$ in $N$. Is this the case? I suppose that the question more generally is is there a way to view a linear connection $\nabla_X Y$ as some combination of covariant derivatives on scalar functions?

How to find the peaks in a curve [on hold]

Math Overflow Recent Questions - Thu, 02/14/2019 - 12:13

I'd like to find the peaks in a curve using derivatives. Attached is a snapshot of the curve. enter image description here I'd also like to filter notches in the curve corresponding to points which are not clearly an optimum but represent a flat response surface. Is their a way to find all the peaks in the curve incrementally traversing through the points?

Does every uncountably categorical theory have a $\varnothing$-definable strongly minimal imaginary?

Math Overflow Recent Questions - Thu, 02/14/2019 - 12:07

An uncountably categorical theory always has a strongly minimal set definable over its prime model, but sometimes this set needs parameters to define.

By a $\varnothing$-definable imaginary I mean the quotient of some finite power of the home sort by a $\varnothing$-definable equivalence relation. No added generality is gained by considering quotients of definable sets, since we can always just extend the equivalence relation and send everything in the complement to a single point.

Every example of an uncountably categorical theory I can think of has a $\varnothing$-definable strongly minimal imaginary. So the question is:

Does every uncountably categorical theory have a $\varnothing$-definable strongly minimal imaginary?

The best I've been able to come up with is the theory of $\mathbb{Z}$ with a symmetric successor relation and a predicate for $x\equiv y\text{ (mod 2})$. By a direct argument you can show that no $\varnothing$-definable equivalence relation on elements gives a strongly minimal imaginary, but I believe there is a strongly minimal quotient of 2-tuples, specifically unordered adjacent pairs.

A few easier questions that would be relevant would be the same but allowing $\text{acl}^\text{eq}(\varnothing)$-definable imaginaries or only requiring that the imaginary have Morley rank $1$, but allowing it to have Morley degree greater than 1.

Decidability: Presentations vs. Groups

Math Overflow Recent Questions - Thu, 02/14/2019 - 11:59

This question is just a curiosity for me as a non-expert. Quite often we ask about decidability of various properties in a group. Often the answer is that the property is undecidable in general. However, what it usually means, as far as I know, that there is a presentation in which this property is undecidable. Are there examples, not-necessarily explicit, such that for all presentations of the group the problem is undecidable?

For instance, is there a finitely presented group such that for all its presentations we cannot decide whether it is trivial?

Are framed manifolds cubulatable?

Math Overflow Recent Questions - Thu, 02/14/2019 - 11:19

Let's say an $n$-manifold is cubulated if it is glued out of cubes $[0,1]^n$ in a way that looks locally like the standard cubulation of $\mathbb R^n$. For instance, the face $[0,1]^{k-1} \times \{1\} \times [0,1]^{n-k}$ of some cube must be glued to the face $[0,1]^{k-1} \times \{0\} \times [0,1]^{n-k}$ of some other cube, for the same value of $k$; the torus $(\mathbb R/\mathbb Z)^n$ is cubulated with just one cube (coming from the standard map $[0,1] \to \mathbb R/\mathbb Z$); the Klein bottle is not cubulated in my sense.

One can imagine an equivalence relation on cubulations of $M$ in which two cubulations are equivalent if they share a common refinement. But I haven't thought through the details of exactly what "refinement" should mean.

It's clear that every cubulated manifold is framed. (A framing of an $n$-manifold $M$ is a homotopy class of trivializations of the tangent bundle, i.e. a homotopy class of vector bundle isomorphisms $TM \cong \mathbb R^n \times M$.) Locally, a framing determines a cubulation (or rather an equivalence class of cubulations). But globally I'm not so sure. I'm worried about things like the irrational line on the torus, which could prevent a framing from arising from any finite cubulation.

Question: Does every framed manifold admit a framing-compatible cubulation?

Equality condition for Araki–Lieb–Thirring inequality

Math Overflow Recent Questions - Thu, 02/14/2019 - 09:22

I'd like to have the equality condition in the Araki–Lieb–Thirring inequality $$\operatorname{Tr} [(BAB)^r]\leq \operatorname{Tr} [(B^{r}A^{r}B^{r})],$$ valid for $A,B$ semidefinite positive and $r\geq1$ (I'm interested in the case $r=2$). Any clue? I didn't find the proof of it yet, I guess it would be a good start!

As explained in comment, finding the equality condition for the inequality $$\operatorname{Tr} [X^2]\leq \operatorname{Tr} [XX^\dagger]$$ is sufficient!

It is the inequality n°13 of Lieb and Thirring - Inequalities for the moments of the Eigenvalues of the Schrödinger Hamiltonian and their relation to Sobolev inequalities.

What is the centralizer of a Coxeter element?

Math Overflow Recent Questions - Thu, 02/14/2019 - 08:55

Let $(W,S)$ be a Coxeter system (of finite rank) and $c \in W$ a Coxeter element.

If $W$ is finite, then the centralizer $C_W(c)$ is the cyclic group generated by $c$ (e.g. see the book "Reflection Groups and Invariant Theory" of Kane).

If $(W,S)$ is irreducible and not finite or affine, then the cyclic group $\langle c \rangle$ is of finite index in $C_W(c)$ (see Theorem 2.6 in the paper "Irreducible Coxeter Groups" by Luis Paris).

Are there more results in this direction? For affine Coxeter groups? Is it conjectured that $C_W(c) = \langle c \rangle$ holds in general? Or is there a counterexample?

Property $\Gamma$ in terms of Correspondences

Math Overflow Recent Questions - Thu, 02/14/2019 - 06:58

A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset $\{ x_{1}, x_{2},..., x_{n} \} \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)

A correspondence of von Neumann algebras $N$ and $M$ is a binormal Hilbert $N$-$M$ bimodule. One such bimodule $H$ is weakly contained in another $K$ if the first is in the closure of a countable (Hilbert) direct sum of copies of $K$ in the Fell topology defined in the preprint referred to below. In Popa's preprint, he considered a stronger notion of weak subequivalence which omits the direct sum in the above closure condition. Popa calls the latter condition weak containment, but in the literature that follows the above distinction is made. Before reading the following paragraph, choose and stick with one of these two notions and let $H\prec K$ denote ``$H$ weakly contained in/subequivalent to $K$''. I am interested in an answer to my question using either interpretation.

It seems that in Popa's 1986 Correspondence Preprint (cf. p. 45 of the preprint section 3.3 on "asymptotic commutativity") the correspondence characterization of property $\Gamma$ of a type $II_{1}$-factor is not quite right. There it is claimed that such an $N$ has $\Gamma$ if $L^{2}(N)\oplus L^{2}(N)\prec L^{2}(N)$. I think what was intended was the characterization found here on p. 27 in Theorem 3.9 part (2). That condition is more like ``$L^{2}(N)\prec L^{2}(N)\ominus \mathbb{C}1$'' which is not grammatical since it isn't properly a property of correspondences since the orthocomplement of the scalars is not an $N$-bimodule.

My question is the following:

Question: Can the Murray-von Neumann Property $\Gamma$ be characterized as Popa tried to via weak containment/weak subequivalence of certain correspondences?

Incidentally, this obviously would indirectly answer the question here pointed out to me by Yemon Choi.

Injective Change of Rings

Math Overflow Recent Questions - Thu, 02/14/2019 - 05:05

Sorry if this is too elementary, but when I was going to ask this question on math.stackexchange, I saw the same question with three up-votes and no answer. So I decided to post it here.

I am doing the following problem:

Let $R$ be a ring, $x\in R$ a central non-unit non-zerodivisor. If $A\neq 0$ is an $R/xR$ module with $id_{R/xR}A$ finite, then $$id_R(A)=1+id_{R/xR}A$$ where $id_{R}(id_{R/xR})$ means injective dimension as a module over $R(R/xR)$

This is Exercise 4.3.3 of Weibel's Introduction to homological algebra. I am trying to mimic the proof of the corresponding theorem for projective dimension (theorem 4.3.3 in the book), which uses induction on $n=pd_{R/x}A$. The problem is that I cannot prove the base case, which says that:

If $A$ is an injective $R/xR$ module, then $id_A=1$.

I can prove that $A$ is not an injective $R$ module, so $id_RA\geq1$, but I cannot prove the other inequality. Can anyone give some hints to me? Thanks in Advance.

I don't know if it's against the rules to copy-paste another question, if so, please tell me.

Equal principal minors of matrix plus rank-1 and inverse

Math Overflow Recent Questions - Thu, 02/14/2019 - 02:58

Given an invertible real matrix $A$ and real column vectors $b$ and $c$.

For which $A$,$b$ and $c$ are all corresponding principal minors of $B = A-bc^T$ and $A^{-1}$ equal?

According to a result by Loewy, this is true if $B$ and $A^{-1}$ are diagonally similar with transpose (plus some extra conditions). This work by Engel and Schneider seems more promising, but still I'm stuck. We can assume that both $A$ and $A^{-1}$ are adjacency matrices of fully connected graphs, i.e., all entries are non-zero.

My main interests are:

  1. For which matrices $A$ is the problem solvable?
  2. Given a matrix $A$, how attain $b$ and $c$ numerically (best possible if no exact solution exist)?
  3. As a general characterization of $A$ might be difficult, I am particularly interested in a solution of the form $A = G O G$, with diagonal matrix $G \neq I$ and orthogonal $O$. Can you think of a class of matrices $O$ which simplifies this problem?

Attempt Following the work of Engel and Schneider and assuming fully connectedness: Let $H = A \div B$, where $\div$ is element-wise. Then for $A$ and $B$ to be diagonally similar, we need $H_{ii} = 1$ for all $i$. Hence, $$ b_i = \frac{ A_{ii} - (A^{-1})_{ii} }{c_{i}}. $$ Note, $(A^{-1})_{ii}$ can be expressed via Jacobi's identity in terms of $A$. Remains, to determine $c$. In Corrolary 3.11., it can be easily seen (from the fully connectedness) that $H$ is diagonally similar to a matrix of only 1s.

A simple case is when assuming that $A$ and $bc^T$ are symmetric. Then $A-dd^T$ and $A^{-1}$ are in canonical form and therefore $dd^T = A - A^{-1}$. This is only true if all eigenvalues of $A$ are 1 except one.

Motivation: The problem arises in control theory, where transfer function from a state-space formulation is:

$$ H(z) = \frac{\det(A) \det(D(z) - (A - bc^T))}{\det(D(z) - A)}, $$ where $D(z) = diag([z^{m_1},\dots,z^{m_n}])$ for integer $m_i$. The goal is now to choose $A$, $b$ and $c$ such that $|H(z)|=1$ for all $z$. This is true if the numerator and denominator of $H(z)$ are "flipped", i.e.,

$$ flip(\det(D(z) - A)) = \det(A) \det(D(z) - A^{-1}). $$

Thus, for any $m_i$, we need: $$ \det(D(z) - A^{-1}) = \det(D(z) - (A - bc^T)), $$ which is true if all principal minors are equivalent.

Perform a univariate integral, involving a Gauss hypergeometric function

Math Overflow Recent Questions - Wed, 02/13/2019 - 12:02

This is a follow-up question to the one posed in Compute the two-fold partial integral, where the three-fold full integral is known . (I hope that doing so is viewed as a legitimate step. If not so, I can withdraw the question, and fall back on the original.)

The problem is to integrate over $p \in [0,1]$ the integrand, \begin{equation} -\frac{(p-1)^{2 b+1} \mu^b \Gamma (b+1)^2 \, _2F_1\left(b+1,b+1;2 (b+1);\frac{(p-1) \mu^2}{p}\right)}{p \Gamma (2 (b+1))}, \end{equation} where $\mu \in [0,1]$ and $b$ is a nonnegative integer.

The answer takes the form $v(b,\mu) + w(b,\mu) \log(\mu)$, where it is now known that \begin{equation} w(b,\mu)=\frac{\sqrt{\pi } 4^{-b} \mu^b \left(\mu^2-1\right)^{-2 b-1} \Gamma (b+1) \, _2F_1\left(-b,-b;1;\mu^2\right)}{\Gamma \left(b+\frac{3}{2}\right)}. \end{equation}

Additionally, \begin{equation} v(b,1)=\frac{\pi 4^{-2 b-1} \Gamma (b+1)^2}{\Gamma \left(b+\frac{3}{2}\right)^2}. \end{equation} So, a general formula for $v(b,\mu)$ is sought.

So, we would like the counterpart for $v(b,\mu)$ of the Rubey formula for $w(b,\mu)$, that is, \begin{equation} 4 u^b \left(u^2-1\right)^{-2 b-1} \frac{1}{4 \left(4 b^2-1\right)} \frac{b} {\binom{2 (b-1)}{b-1}} \Sigma_{k=0}^b u^{2 k} \binom{b}{k}^2, \end{equation} which he apparently obtained using the general purpose computer algebra system, FriCAS

Is every commutative ring a limit of noetherian rings?

Math Overflow Recent Questions - Wed, 02/13/2019 - 05:51

Edit of Feb. 14, 2019. After Laurent Moret-Bailly's accepted answer, only Questions 4 and 5 remain open. I don't care that much about Question 4, but I'm very curious about Question 5, which is

Do binary coproducts always exist in the category of noetherian commutative rings?

End of edit.

I asked the question on Mathematics Stackexchange but got no answer.

Let $\mathsf{Noeth}$ be the category of noetherian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of commutative rings with one.

Let $A$ be in $\mathsf{CRing}$.

Question 1. Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?

Let $f:A\to B$ be a morphism in $\mathsf{CRing}$ such that the map $$ \circ f:\text{Hom}_{\mathsf{CRing}}(B,C)\to\text{Hom}_{\mathsf{CRing}}(A,C) $$ sending $g$ to $g\circ f$ is bijective for all $C$ in $\mathsf{Noeth}$.

Question 2. Does this imply that $f$ is an isomorphism?

Yes to Question 1 would imply yes to Question 2.

Question 3. Does the inclusion functor $\iota:\mathsf{Noeth}\to\mathsf{CRing}$ commute with colimits? That is, if $A\in\mathsf{Noeth}$ is the colimit of a functor $\alpha$ from a small category to $\mathsf{Noeth}$, is $A$ naturally isomorphic to the colimit of $\iota\circ\alpha$?

Yes to Question 2 would imply yes to Question 3, and yes to Question 3 would imply that many colimits, and in particular many binary coproducts, do not exist in $\mathsf{Noeth}$: see this answer of Martin Brandenburg.

Here are two particular cases of the above questions:

Question 4. Is $\mathbb Z[x_1,x_2,\dots]$ a limit of noetherian rings?

(The $x_i$ are indeterminates.)

Question 5. Do binary coproducts exist in $\mathsf{Noeth}$?

One may try to attack the first question as follows:

Let $A$ be in $\mathsf{CRing}$ and $I$ the set of those ideals $\mathfrak a$ of $A$ such that $A/\mathfrak a$ is noetherian. Then $I$ is an ordered set, and thus can be viewed as a category. We can form the limit of the $A/\mathfrak a$ with $\mathfrak a\in I$, and we have a natural morphism from $A$ to this limit. I'd be interested in knowing if this morphism is bijective. [Edit. By a comment of Laurent Moret-Bailly the morphism in question is not always bijective.]


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