## commutative algebra – special case of isomorphism \$ S otimes_R mathrm {Hom} _R (M, N) simeq mathrm {Hom} _S (S otimes_R M, S otimes_R N) \$

So we have the S-module isomorphism $$S otimes_R mathrm {Hom} _R (M, N) simeq mathrm {Hom} _S (S otimes_R M, S otimes_R N)$$ given that $$S$$ is flat over $$R$$ and $$M$$ will finally be presented. Even though $$S$$ is not flat or $$M$$ is finally presented, there is at least $$S$$Homomorphism between the modules.

What I am considering is a limitation of isomorphism. To let $$phi in mathrm {End} _R (M)$$, $$1 otimes_R phi$$ will designate endomorphism of $$S otimes_R M$$, $$S otimes_R R[phi] Simeq S[1otimes_R phi]$$
Is the above isomorphism without the condition that $$S$$ is flat over $$R$$ or that $$M$$ is finally presented?

In the face of that $$S$$ is flat over $$R$$ and $$M$$ is finally presented, $$S otimes_R R[phi]$$ is a module of $$S otimes_R mathrm {End} _R (M)$$ and the isomorphism in question follows easily.

It is obvious that the induced homomorphism is surjective, but I can not verify exactly if it is injective.

1. Does Isomophism have no flatness or finite presence?
2. If the isomorphism holds, what is this structural difference between $$R[phi]$$ and the whole $$mathrm {End} _R (M)$$ that makes isomorphism possible on one side and not on the other?

## Commutative Algebra – A property of Cohen-Macaulay modules.

To let $$(R, mathfrak {m})$$ Be a local ring of Cohen-Macaulay (CM) $$d> 0$$, Suppose that $$M$$ is a finely generated CM $$R$$Module of $$dim_R M = i ,

Prove that $$M$$ is the homomorphic image of a CM $$R$$Module of dimension $$i + 1$$,

My idea is to continue using the induction $$d & # 39; s 0$$, The case $$i = 0$$ is quite simple, but I stick to the inductive step (I tried to find) $$U$$ in the shape $$U & # 39; oplus M$$ but it seems to me that this does not work $$mathrm {depth} (U \ plus M) leq mathrm {depth} M$$). I do not know if there are ways to overcome this.

## Commutative Algebra – In a local noether ring, every ideal is a force of the Jacobson Radical

I want to prove that $$R$$ a commutative noetheric local ring if the Jacobson radical is $$J$$ is then nilpotent $$R$$ is Artinian. I want to use the fact that in a commutative noetherian ring $$bigcap_ {k = 1} ^ { infty} J ^ k = 0$$ and show that $$R, J, J ^ 2, Dots J ^ s = 0$$ are all ideals in $$R$$, But how can I show that every ideal is a power of the Jacobson Radical?

## 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?

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 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.