If X is NP-complete and for some $ Y, X\leq_p Y\cap Y\leq_p X$ what can we say about Y?

My intuition says that this is only the case when $ X==Y$ but I’m not sure how to justify this.

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# Tag: commutative

## NP-Completeness and commutative property

## Is it possible to apply a symmetric transformation (commutative) to two vectors of angles, but where the original values can be retrieved?

## $R$ commutative ring with 1 and not every ideal is principal. Prove $R$ has ideal that is not principal.

## A example of commutative ring which has conditions regarding Jacobson radical

## R a commutative ring and A an ideal in R with $A=m_1···m_r=n_1···n_s$ with $m_i$ distinct maximal ideals and all the $n_j$ distinct maximal ideals.

## Commutative Convolution. Problem 26 Royden 2 ed.

## Graded analogues of theorems in commutative algebra

## Propostion 1.6 on Atiyah’s commutative algebra text

## Weakly dense C*-algebra in a commutative von Neumann algebra and order convergence

## What happens if I replace a *unique natural number* that form a commutative Monoid with *the set of integers* Z that form a commutative Ring?

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If X is NP-complete and for some $ Y, X\leq_p Y\cap Y\leq_p X$ what can we say about Y?

My intuition says that this is only the case when $ X==Y$ but I’m not sure how to justify this.

I have a problem where I have two vectors a and b representing a list of angles.

I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)

For example: a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify

a.b doesn’t qualify because I don’t have a metric to compare square angles.

a-b doesn’t qualify because it is not commutative

Do you think this is possible?

What I have tried that didn’t work:

T(a,b) = {cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)}

And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.

This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn’t get back my original vectors.

Thank you!

I am wondering how to go about proving this,

Let $ R$ be a commutative ring with identity such that not every ideal of $ R$ is principal.

A) Use Zorn’s lemma to show that $ R$ has an ideal $ J$ such that (i) $ J$ is not principal ideal (ii) $ J$ is not properly contained in any non-principal ideal.

B) Show that $ R/J$ is a principal ideal ring (where $ J$ is the ideal from part (a)).

First of all , I define notations for ring $ R$ .

$ J(R)$ is Jacobson radical of $ R$ .

$ N(R)$ is nilradical of $ R$ .

Next, I say my question.

I want to get a commutative ring $ R$ which has following $ 2$ conditions.

$ (1)$ $ J(R) \neq N(R)$

$ (2)$ Krull dimension of $ A/J(R)$ is NOT $ 0$

Can you construct a commutative ring $ R$ satisfying (1) and (2) $ ??$

Let R be a commutative ring and A be an ideal in R satisfying $ A=m_1···m_r =n_1···n_s$ with all the $ m_i$ distinct maximal ideals and all the $ n_j$ distinct maximal ideals. Show that $ r = s$ and there exists a $ σ ∈ Sr$ satisfying $ m_i = n_σ(i)$ for all i.

I know that maximal ideal implies it being prime, and the product is contained in $ m_i$ and $ n_j$ for all $ i$ and $ j$ , but I’m not sure how to proceed further.

Let $ f$ and $ g$ be functions in $ L^1(—\infty,\infty)$ , and define $ f\ast g$ to be the function $ h$ defined by $ h(y) = \int f(y — x)g(x) dx$ .

Why $ f\ast g=g\ast f$ ?

I have this: If $ y-x=z$ then $ \int f(y-x)g(x)dx=\int f(z)g(y-z)(-dz)=-g\ast f$

Many theorems in commutative algebra hold true in a ($ \mathbb{Z}$ -)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word

- commutative ring by commutative graded ring (without the sign for commutativity)
- module by graded module
- element by homogeneous element
- ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)

This results in further substitutions, e.g. a $ \ast$ local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.

One book that does some steps in this direction is *Cohen-Macaulay rings* by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:

Let $ (R,\mathfrak{m})$ be a $ \ast$ local ring, $ M$ be a finitely generated graded $ R$ -module and $ N$ a graded submodule. Assume $ M = N + \mathfrak{m}M$ . Then $ M = N$ .

Moreover, a student of mine recently showed that the graded analogue of Lazard’s theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.

Usually one proves this kind of theorems essentially by a combination of two techniques:

- Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
- If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)

Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $ [Spec R/\mathbb{G}_m]$ for a graded ring $ R$ , using that a $ \mathbb{Z}$ -grading corresponds to a $ \mathbb{G}_m$ -action.

In any case, my question is the following:

Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?

I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.

Let $ A$ be a ring (commutative with 1) and $ M$ a maximal ideal of $ A$ . Let $ x \in A-M $ . Since $ M$ is maximal, the ideal generated by $ x$ and $ M$ is $ (1)$ , i.e. the entire ring.

I don’t understand how that is true. If $ x$ were a unit, certainly. However, $ x$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $ x$ and $ M$ can generate $ A$ ?

Let $ H$ be a Hilbert space and $ \mathscr{A}$ a commutative norm-closed $ *$ -subalgebra of $ \mathcal{B}(H)$ . Let $ \mathscr{M}$ be the weak operator closure of $ \mathscr{A}$ .

**Question**: For given a projection $ P\in\mathscr{M}$ , is the following true? $ $ P=\inf\left\{A\in\mathscr{A}:P\leq A\leq 1\right\}$ $

It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $ P$ . Also, if the above is true, what happens if $ \mathscr{A}$ is non-commutative?

In mathematical logic, a Gödel numbering is a function that assigns to each symbol and well-formed formula of some formal language

a unique, called its Gödel number. The concept was used by Kurt Gödel for the proof of his incompleteness theorems. (Gödel 1931)natural numberThe natural numbers, N, form a commutative

monoidunder addition (identity element zero), or multiplication (identity element one). A submonoid of N under addition is called a numerical monoid. The positive integers, N ∖ {0}, form a commutative monoid under multiplication (identity element one).The set of integers Z forms a commutative

ringunder addition and multiplication

What happens if I replace a *unique natural number* that form a commutative Monoid with *the set of integers* Z that form a commutative Ring ?

The more specific question should be: What happens **if I use** (not if I replace) a unique natural number that form a commutative Monoid **as** (not *with*) the set of integers Z that form a commutative Ring ?

**Monoid $ \Rightarrow$ Ring**

**P.S:** I don’t want replace formally but I do not want to really change, but to make it believe that it is as if I had replaced.

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