```
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}
```

# Tag: commutative

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

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## 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?

This question was asked at Mathematics Stackexchange, but not answered.

(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 example*injective* 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.

The busy reader is asked to skip the sequel.

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)?