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 that.

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# Tag: commutative

## NP-completeness and commutative property

## Commutative Algebra – Characterization of Non-Integral Domains?

## algebraic manipulation – defining a differential operator that does not commutative

## ct.category theory – Is every commutative ring a bound of noether rings?

## Gr.group Theory – A Common Name for a Functional Construction of Commutative Algebra?

## Commutative Algebra – Does the category of artistic rings allow finite limits?

## Commutative Algebra – Quotient of the Notherian (Artinian) ring is also Notherian (Artinian)

## Commutative Algebra – Generalization of the Atiyah-Macdonald Proposal 5.7

## Commutative convolution. Problem 26 Royden 2 ed.

## Graduated analogues of theorems in commutative algebra

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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 that.

In Miles Reid's *Diplomated commutative algebra* he proves the following sentence:

If $ R $ is then a ring with zero divisors $ R $ has an element that is not zero, or multiple minimum prime prime values.

I would like to know how this is related to the non-trivial *Idempotents*, Is it true that a ring with more than a minimum prime ideal has a non-trivial idempotent? So is the following statement correct?

If $ R $ is then a ring with zero divisors $ R $ has a nonzero zero potential or a nontrivial idempotent.

On a non-commutative basis $ {x_0, x_1, x_2} $ I want to define a difference operator that works that way

$$ Delta_i ( sum_ {n = 0} ^ infty c_n x_i ^ n) = sum_ {n = 1} ^ infty c_n x_i ^ {n-1}, quad Delta_i x_j = 0 quad text {if} quad i neq j $$

So if this operator was a "derivative" in terms $ x_0 $when operating on a train in $ x_0 $ it will only reduce the power of $ x_0 $ individually. Also, I want to define it to be specific from the left, so if I had a string

$$ Delta_0 (x_0 x_0 x_0 x_1 + x_1 x_0 x_2 x_0 + x_2 x_2 x_1) = x_0 x_0 x_1 $$

so if a $ Delta_0 $ meets everything that is not $ x_0 $ it will destroy it on the left side.

To create non-commutativity, I understand that I can easily write the functions I want to work with between primitives. However, I have difficulty writing down an operator that meets the characteristics described above.

Any help would be appreciated!

I asked the question about Mathematics Stackexchange, but got no answer.

To let $ mathsf {Noeth} $ the category of noetheric rings, which is considered as a complete subcategory of the category $ mathsf {CRing} $ of commutative rings with one.

To let $ A $ in his $ mathsf {CRing} $,

Question 1.Is there a functor from a small category too $ mathsf {Noeth} $ their border in $ mathsf {CRing} $ is $ A $?

To let $ f: A to B $ be a morphism in $ mathsf {CRing} $ so the card

$$

circ f: text {Hom} _ { mathsf {CRing}} (B, C) to text {Hom} _ { mathsf {CRing}} (A, C)

$$

Send $ g $ to $ g circ f $ is bijective for everyone $ C $ in the $ mathsf {Noeth} $,

Question 2.Does that mean that? $ f $ is an isomorphism?

Yes to question 1, yes would mean question 2.

Question 3.Does the inclusion functor work? $ iota: mathsf {Noeth} an mathsf {CRing} $ commute with colimits? If so $ A in mathsf {Noeth} $ is the colimit of a functor $ alpha $ from a small category to $ mathsf {Noeth} $is $ A $ of course isomorphic to the colimit of $ iota circ alpha $?

Yes to question 2, yes would mean question 3, and yes to question 3 would mean that many colimits, and in particular many binary coproducts, do so *Not* exists in $ mathsf {Noeth} $: see this answer from Martin Brandenburg.

Here are two special cases of the above questions:

Question 4is $ mathbb Z[x_1,x_2,dots]$ a limit for noetheric rings?

(The $ x_i $ are undetermined.)

Question 5Are there binary coproducts in $ mathsf {Noeth} $?

You can try to attack the first question as follows:

To let $ A $ in his $ mathsf {CRing} $ and $ I $ the set of these ideals $ mathfrak a $ from $ A $ so that $ A / mathfrak a $ is noetherian. Then $ I $ is an ordered set and can therefore be considered a category. We can see the limit of $ A / mathfrak a $ With $ mathfrak a in I $and we have a natural morphism $ A $ up to this limit. I would like to know if this morphism is bijective.

I'm interested in whether the following construction, which occurs naturally in Commutative Algebra, has some familiar and accepted names.

Given a commutative monoid $ (M, +) $ and a set $ X $Consider the family $ F (X, M) $ of functions $ varphi: X to M $ that has finite support $ supp ( varphi): = {x in X: varphi (x) ne 0 } $ from where $ 0 $ is the neutral element of $ M $ in terms of the commuative operation $ + $,

The sentence $ F (X, M) $ has an obvious structure of a commutative monoid (actually a submonoid of power) $ M ^ X $).

Every function $ f: X to Y $ between sentences induces a monoidic homomorphism $ Ff: F (X, M) to F (Y, M) $ this is for everyone $ varphi in F (X, M) $ the function $ psi: Y to M $, $ psi: y mapsto sum_ {x in f ^ {- 1} (y)} varphi (x) $ (The latter sum is well defined, since it contains only finitely many non-zero terms).

The construction $ F (X, M) $ determines a functor $ F: mathbf {set} on mathbf {Mon} $ from the category $ mathbf {set} $ from sets to the category $ mathbf {Mon} $ of commutative monoids.

If I'm interested, if the radio operator $ F $ has a known reserved name.

**Annotation.** For some special monoids $ M $ the functor $ F $ is known in algebra. For example,

$ bullet $ for the group $ mathbb Z $ of integers the monoid $ F (X, mathbb Z) $ can be identified with the free Abelian group of $ X $;

$ bullet $ for the cyclic group with 2 elements $ C_2 $the monoid $ F (X, C_2) $ can be identified with the free Boolean group of $ X $,

$ bullet $ for the cyclic group with n elements $ C_n $the monoid $ F (X, C_n) $ can be identified with the free Abelian group of $ X $ in the diversity of the abelian groups that satisfy the identity $ x ^ n = 1 $;

$ bullet $ for the 2-element half lattice $ 2 = {0,1 } $ with the operation $ max $the monoid $ F (X, 2) $ can be identified with the free half grid with the unit above $ X $,

To let $ mathsf {Artin} $ Be the category of artistic rings considered as a complete subcategory of the category $ mathsf {CRing} $ of rings. (Here "ring" means "commutative ring with one".)

Question 1.does $ mathsf {Artin} $ To admit finite limits?

As $ mathsf {Artin} $ has finite products, question 1 corresponds

Question 2.does $ mathsf {Artin} $ Add equalizer

A closely related question is

Question 3.To let $ A to B $ the equalizer in $ mathsf {CRing} $ from two morphisms $ B $ to $ C $, Accept that $ B $ and $ C $ are artinian. Does that mean that? $ A $ is artinian?

Yes to question 3 would mean yes to questions 1 and 2.

This answer from MooS implies that the category of *Noetherian* Makes rings *Not* Allow finite limits.

**Proposal 1:** To let $ 0 to M xrightarrow {right} M xrightarrow {g} M & # 39; & # 39; to 0 $ to be an exact sequence of $ A $Modules. Then:

I) $ M $ is noetherian **iff** $ M & # 39; $ and $ M & # 39; & # 39; $ are noetherians.

ii) $ M $ is Artinian **iff** $ M & # 39; $ and $ M & # 39; & # 39; $ are Artinian.

**Proposal 2:** To let $ A $ to be a Notherian (Artinian), $ I $ an ideal of $ A $, Then $ A / I $ is a Notherian (Artinian) ring.

My question is this: I know the proof of Theorem 1. How do I use it to prove Theorem 2?

I tried to construct some examples of the exact order, but I did not succeed.

The proposal is

To let $ A $ $ subseteq $ $ B $ be holistic domains, $ B $ all over $ A $, Then $ B $ is a field iff $ A $ is a field.

The proof is simple. I would like to generalize this proposal. I want to prove that

To let $ A $$ to $ $ B $ be holistic domains, $ B $ all over $ A $, Then $ B $ is a field iff $ A $ is a

Field.

One page is simple: Suppose $ A $ is a field, let $ y in B, y not = 0 $, Since $ B $ is all over $ A $, To let $ y ^ {n} + f (a_ {1}) y ^ {n-1} + … + f (a_ {n}) (a_ {i} in A) $ be an equation of integral dependence for y of the least possible degree. Then $ a_n not = 0 $, so $ y ^ {- 1} = – f (a_ {n} ^ {- 1}) (y ^ {n-1} + … + f (a_ {n-1})) in B $therefore $ B $ is a field. But I do not know how to prove the other side: if $ B $ So it's a field $ A $ is a field.

Can you tell me how to prove it?

To let $ f $ and $ g $ Functions $ L ^ 1 (- infty, infty) $and define $ f ast g $ to be

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 apply in one ($ mathbb {Z} $-) graduated context. More specifically, we can use any sentence in commutative algebra and replace any occurrence of the word

- Commutative ring by commutative grading ring (unsigned for commutativity)
- Module through graduated module
- Element by homogeneous element
- Ideal by homogeneous ideal (that is, ideally generated by homogeneous elements)

This leads to further substitutions, eg. on $ ast $Local ring is a graded ring with a unique maximum homogeneous ideal, we get an idea of graded depth, etc. After all these substitutions we can ask if the sentence is still true.

A book that takes a few steps in this direction is *Cohen-Macaulay rings* Bruns and Herzog, especially section 1.5. For example, in Exercise 1.5.24 they have the following graded analogue of the Nakayama lemma:

To let $ (R, mathfrak {m}) $ be a $ ast $local ring, $ M $ to be a finitely graded one $ R $Module and $ N $ a graduated submodule. Accept $ M = N + mathfrak {m} M $, Then $ M = N $,

A student of mine has recently shown that the graduated analogue of Lazard's theorem (a module is flat even if it is a filtered colimit of free modules) is also true.

Normally, this kind of theorem is essentially proved by a combination of two techniques:

- Copy the ungraded proof and replace graded terms in the manner outlined above.
- If you get annoyed about the length of the resulting argument, use some abbreviations for some translations between unrated and graded. (For example, a noetheric ring graded after Cohen-Macaulay is also unnamed Cohen-Macaulay.)

Sometimes you can be lucky, and the statement is suitably an algebra geometry that you can argue geometrically with the stack $[Spec R/mathbb{G}_m]$ for a stepped ring $ R $with this a $ mathbb {Z} $-grading corresponds to one $ mathbb {G} _m $-Action.

In any case, my question is this:

Is there a class of statements that automatically recognizes that the graded analog is true if the original statement in the unclassified commutative algebra is true without going through all the proof?

I am not sure if one can hope for a model-theoretical approach here, since I know almost nothing about model theory, but such a statement could save a lot of work in the detection of graded analogues of known theorems.

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