## Ag.algebraic geometry – is noether after the commutative ring or not?

documentclass{article}
usepackage{amsmath,amssymb,amsthm}
$$begin{document} % Following commutative ring,is Noether or not? From Nagata cite{NagataLocalRings}203page. ( R= K(X_1,cdots,X_n,cdots)_S ,) where  K;field ,quad  K(X_1,cdots,X_n,cdots)  is polynomial ring over K, countably infinite indeterminate {X_j}_j, mathfrak{p}_i are genarated by {X_j}_{m_j}^{m_{j+1}}, S=cap_i(complementquad ofquad mathfrak{p}_i) \ This ring R is the polynomial ring of quotient by S. \ Is R Noetherian or not? begin{thebibliography}{1} bibitem{NagataLocalRings} Nagata,M. newblock {em Local rings}. newblock R. E. Krieger Pub. Co., Huntington, N.Y., United States, 1975, c1962 1975. end{thebibliography} end{document}$$

## Measure theory – Why is the commutative integral with the limit important in real analysis?

But avoid

• Make statements based on opinions; Provide them with references or personal experience.

Use MathJax to format equations. Mathjax reference.

## ac.commutative algebra – Can each locally free module be inverted via a commutative ring?

Since this topic is full of misunderstandings (see here, here, here and here), let's set out an exact terminology.
To let $$A$$ be a commutative ring and $$P$$ on $$A$$-Module.

I) We will say that $$P$$ is a local free module of the first rank or is reversible if $$P$$ is finally generated, projective and ranked first in the sense that for every prime ideal $$mathfrak p$$ of $$A$$ the localized $$A_ mathfrak p$$– module $$P_ mathfrak p$$ (which is free through projectivity) is of dimension $$1$$,
These modules correspond bijectively to a known result from Serre in FAC locally free sheaves $$tilde P$$ of rank $$1$$ on $$operatorname {Spec} A$$, also known as reversible rope sheaves. This is a motivation for the terminology above.
Another justification for the terminology "invertible" is that these modules are exactly those for which the canonical evaluation card is available $$P ^ * otimes_AP to A$$ is an isomorphism.

II) If $$B supset A$$ is an over ring of $$A$$ and $$P subset B$$ is a $$A$$Module, we will say that it is specifically reversible in memory of $$B$$ if $$P. (A: P) _B = A$$,
(By default $$(A: P) _B$$ denotes the set of elements $$b in B$$ so that $$bP subset A$$)
It is known that these specifically invertible modules are invertible. What about the reversal?

Question:
Is an invertible $$A$$-Module $$P$$ isomorphic as a $$A$$Module to a concrete invertible module $$P & # 39; subset B$$ in relation to a suitable overring $$B supset A$$?

Remarks
a) The answer is "yes" if $$A$$ is an integral domain. We can take $$P & # 39;$$ sit inside $$B = operatorname {Frac} A$$,
b) The answer is trivial yes if $$A$$ has been semi-local ever since $$P$$ is rank-free $$1$$: look here.
c) According to Eisenbud's commutative algebra, Theorem 11.6 b., every reversible module $$P$$ about a Noetherian ring $$A$$ is isomorphic to a concrete invertible submodule $$P & # 39; subset operatorname {Quot} A$$ of his entire quotient ring $$operatorname {Quot} A$$ obtained by inverting the regular (= not zero divisors) of $$A$$
d) However, it is not generally true that we can always find the ones we need $$P & # 39;$$ within the entire quotient ring $$B = operatorname {Quot} A$$,
Lam gives a counterexample in his lectures on modules and rings, example (2.22) (A), page 37.

## Commutative algebra – dimension of a given finite generated quotient module over a local ring.

I have dealt with the following question from the theory of dimension in commutative algebra.

To let $$(A, m)$$ be a local ring and $$M$$ a finally generated $$A$$-Module.

given $$x_1, …, x_r in M ​​$$, Prove that $$dim ( frac {M} {(x_1, …, x_r) M}) geq dim (M) – r$$with equality, if and only if {$$x_1, …, x_r$$} is part of a parameter system for $$M$$,

Now I can show that, though $$A$$ is a $$mathbf {regular}$$ local ring so $$frac {A} {(x_1, …, x_r) A}$$ is a regular local ring with dimension $$dim (A) – r$$ then and only if {$$x_1, …, x_r$$} is part of a parameter system for $$A$$, But I do not know how to show that for the given case. I also can not show the inequality. I could not find any proof, so I would be grateful for any help!

## real analysis – Prove that the finite difference operator is commutative

I try to prove the symmetry of mixed partial derivatives in which the following lemma is addressed.

$$textbf {Lemma:}$$ To let $$X$$ be open in $$mathbb R ^ n$$. $$f: X to F$$, and $$m in mathbb N$$, We $$h in mathbb R$$ and $$j in {1, ldots, n }$$let's define a map $$Delta_j ^ h f$$ by $$Delta_j ^ h f: X to F, quad x mapsto f (x + he_j) -f (x)$$

Accept $$j_1, j_2, ldots, j_m in {1, ldots, n }$$ and $$sigma$$ is a permutation of $${1, ldots, m }$$, Then $$Delta_ {j_1} ^ {h_1} Delta_ {j_2} ^ {h_2} cdots Delta_ {j_m} ^ {h_m} f (a) = Delta_ {j _ { sigma (1)}} ^ {h_ { sigma (1)} delta_ {j _ { sigma (2)}} ^ {h_ { sigma (2)}} cdots delta_ {j _ { sigma (m)}} ^ {h_ { sigma (m)}} f (a)$$

Could you please check if my proof looks good or contains logical gaps / mistakes? Thank you for your help!

$$textbf {my try:}$$

Since every permutation is a product of neighboring transpositions, it is sufficient to prove this $$Delta_ {j_1} ^ {h_1} Delta_ {j_2} ^ {h_2} f (a) = Delta_ {j_2} ^ {h_2} Delta_ {j_1} ^ {h_1} f (a)$$

We have begin {align} delta_ {j_1} ^ {h_1} delta_ {j_2} ^ {h_2} f (a) & = delta_ {j_1} ^ {h_1} left (f (a + h_2 e_ { j_2}) – f (a) right) \ & = Delta_ {j_1} ^ {h_1} f (a + h_2 e_ {j_2}) – Delta_ {j_1} ^ {h_1} f (a) \ & = (f (a + h_2e_ {j_2} + h_1e_ {j_1}) – f (a + h_2e_ {j_2}) – (f (a + h_1e_ {j_1}) – f (a)) & = f ( a + h_2e_ {j_2} + h_1e_ {j_1}) – f (a + h_2e_ {j_2}) – f (a + h_1e_ {j_1}) + f (a) end {aligned} and begin {align} delta_ {j_2} ^ {h_2} delta_ {j_1} ^ {h_1} f (a) & = delta_ {j_2} ^ {h_2} left (f (a + h_1 e_ { j_1}) – f (a) right) \ & = Delta_ {j_2} ^ {h_2} f (a + h_1 e_ {j_1}) – Delta_ {j_2} ^ {h_2} f (a) \ & = (f (a + h_1e_ {j_1} + h_2e_ {j_2}) – f (a + h_1e_ {j_1}) – (f (a + h_2e_ {j_2}) – f (a)) \ & = f ( a + h_1 e_ {j_1} + h_2 e_ {j_2}) – f (a + h_1 e_ {j_1}) – f (a + h_2 e_ {j_2}) + f (a) end {aligned}

Hence the extensions of $$Delta_ {j_1} ^ {h_1} Delta_ {j_2} ^ {h_2} f (a)$$ and $$Delta_ {j_2} ^ {h_2} Delta_ {j_1} ^ {h_1} f (a)$$ are identical. Consequently $$Delta_ {j_1} ^ {h_1} Delta_ {j_2} ^ {h_2} f (a) = Delta_ {j_2} ^ {h_2} Delta_ {j_1} ^ {h_1} f (a)$$, This concludes the proof.

## Why is the multiplication between two Markov chain transition matrices commutative?

Can someone intuitively prove the following sentence?

For any two Markov chain transition matrices $$A$$. $$B$$ (n times n matrix, which has every row sum to 1), we have $$A * B = B * A$$,

## ct.category theory – Are epimorphic endomorphisms of noether commutative rings always injective?

(In this post, "ring" means "commutative ring with a".)

To let $$A$$ be a noetherian ring, and let $$f: A to A$$ be an endomorphism, which is also an epimorphism.

is $$f$$ necessarily injective?

Eric Wofsey gave an example of an epimorphic endomorphism of a noether ring that is not surjective. (It is known and easy to prove that surjektiv Endomorphisms of noether rings are automatically bijective.)

(By definition, a morphism of rings $$f: A to B$$ is a epimorphism if for all pairs of morphisms $$(g, h): B right arrows C$$ the equality $$g circ f = h circ f$$ implied $$g = h$$, Surjective morphisms are epimorphic, but the reverse is not always true: for example, inclusion $$mathbb Z to mathbb Q$$ is an epimorphism.)

For more details on epimorphisms, see

$$bullet$$ MathOverflow Thread What are Epimorphisms of (Commutative) Rings?

$$bullet$$ Stacks Project Section Epimorphisms of rings.

$$bullet$$ Samuel seminar.

## Commutative algebra – counterexample for tensor product of modules and completion

I'm reading Qing Liu's Algebraic Geometry Book, and here's a problem I can not find an answer to:

Find a ring $$A$$ and a $$A$$-Module $$M$$ so that $$M otimes_ {A} has {A} to {has} {M}$$ is not surjective where $$has {A} = lim A / I ^ {n}$$ and $$has {M} = lim M / I ^ {n} M$$ are $$I$$-adische graduations for some ideal $$I subset A$$,

I know that the card is surjective, though $$M$$ is finally over generated $$A$$, in order to $$M$$ should be generated infinitely over $$A$$ to be a counterexample. I tried $$A = mathbb {Z}, M = mathbb {Q}$$ and $$I = p mathbb {Z}$$that is not an exampleinjective do not fall surjektiv, Can someone give a hint to find such an example? Thank you in advance.

## ct.category theory – Does the Cantor Schröder-Bernstein theorem belong to the category of noetheric commutative rings?

I asked this question at Mathematics Stackexchange, but did not get an answer.

To let $$A$$ and $$B$$ be noetherian commutative rings with one, and let them $$f: A to B$$ and $$g: B to A$$ Be epimorphisms.

Are the rings $$A$$ and $$B$$ necessarily isomorphic?

(In this post, "ring" means "commutative ring with one", and morphisms are required for assignment $$1$$ to $$1$$, By definition, a morphism of rings $$f: A to B$$ is a epimorphism if for all pairs of morphisms $$(g, h): B right arrows C$$ the equality $$g circ f = h circ f$$ implied $$g = h$$, Surjective morphisms are epimorphic, but the reverse is not always true: for example, inclusion $$mathbb Z to mathbb Q$$ is an epimorphism.

For more details on epimorphisms, see

$$bullet$$ MathOverflow Thread What are Epimorphisms of (Commutative) Rings?

$$bullet$$ Stacks Project Section Epimorphisms of rings.

$$bullet$$ Samuel seminar. See in particular section 2 of Exposé No. 7 by Daniel Ferrand.

The answers to the following variants of the above question are known:

(1) If $$f: A to B$$ and $$g: B to A$$ are injecting morphisms of noether rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is no, as the following example shows, taken from a comment by Sam Lichtenstein on this question. To let $$K$$ be a field, $$x$$ an indefinite, $$f: K (x ^ 2, x ^ 3) to K (x)$$ the recording and $$g: K (x) to K (x ^ 2, x ^ 3)$$ the (clearly injective) morphism defined by $$g (p (x)) = p (x ^ 2)$$, Note that $$K (x ^ 2, x ^ 3)$$ is not isomorphic too $$K (x)$$ because that's ideal $$(x ^ 2, x ^ 3)$$ from $$K (x ^ 2, x ^ 3)$$ is not home.

(2) If $$f: A to B$$ and $$g: B to A$$ are surjective morphisms of rings, are $$A$$ and $$B$$ necessarily isomorphic? The answer is no, as the following example shows, taken from the same commentary by Sam Lichtenstein. to adjust
$$A: = mathbb Z / (4) times mathbb Z / (4) times cdots, quad B: = mathbb Z / (2) times A,$$
To let $$f: A to B$$ be defined by $$f (x_1, x_2, points) = (h (x_1), x_2, points)$$, Where $$h$$ is the unique ring morphism out $$mathbb Z / (4)$$ to $$mathbb Z / (2)$$, and let $$g: B to A$$ be defined by $$g (x_1, x_2, dots) = (x_2, x_3, dots)$$, The Rings $$A$$ and $$B$$ are not isomorphic because the equations $$2x = 0$$ and $$x ^ 2 = x$$ have no non-zero simultaneous solutions in $$A$$and such a solution in $$B$$ (namely $$x = (1,0, dots)$$).

(3) If $$f: A to B$$ and $$g: B to A$$ are surjective morphisms of noether rings, are $$A$$ and $$B$$ isomorphic? The answer is yes, because surjective endomorphisms of noether rings are isomorphisms. But epimorphic endomorphisms of noether rings are not always isomorphisms: see this answer by Eric Wofsey.

## Commutative algebra – motivation for Jordan Holder theorem in an abelian category

The Jordan Holder Theorem states that any chain of subobjects of a finite-length object can be refined into a series of compositions, and that each composition series has the same length.

This sentence applies to every abelian category, and a notable example is the case of modules. Although I do not need an example of the usefulness of the JH theorem in the context of modules, I would like to ask:

What are applications of the JH theorem in a general abelian category that is not a category of modules (or in a situation where it is not so trivial to prove that this category corresponds to a category of modules)?