## NP-completeness and commutative property

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.

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

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.

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

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!

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

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

(The $$x_i$$ are undetermined.)

Question 5 Are 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.

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

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 $$X$$,

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

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.

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

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.

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

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?

## Commutative convolution. Problem 26 Royden 2 ed.

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 $$int f (y – x) g (x) dx = int f (z) g (y – z) (- dz) = – g ast f$$

## Graduated analogues of theorems in commutative algebra

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)
• 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:

1. Copy the ungraded proof and replace graded terms in the manner outlined above.
2. 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.