## R a commutative ring and A an ideal in R with $A=m_1···m_r=n_1···n_s$ with $m_i$ distinct maximal ideals and all the $n_j$ distinct maximal ideals.

Let R be a commutative ring and A be an ideal in R satisfying $$A=m_1···m_r =n_1···n_s$$ with all the $$m_i$$ distinct maximal ideals and all the $$n_j$$ distinct maximal ideals. Show that $$r = s$$ and there exists a $$σ ∈ Sr$$ satisfying $$m_i = n_σ(i)$$ for all i.

I know that maximal ideal implies it being prime, and the product is contained in $$m_i$$ and $$n_j$$ for all $$i$$ and $$j$$, but I’m not sure how to proceed further.

## Commutative Convolution. Problem 26 Royden 2 ed.

Let $$f$$ and $$g$$ be functions in $$L^1(—\infty,\infty)$$, and define $$f\ast g$$ to be the 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 $$\int f(y-x)g(x)dx=\int f(z)g(y-z)(-dz)=-g\ast f$$

## Graded analogues of theorems in commutative algebra

Many theorems in commutative algebra hold true in a ($$\mathbb{Z}$$-)graded context. More precisely, we can take any theorem in commutative algebra and replace every occurrence of the word

• commutative ring by commutative graded ring (without the sign for commutativity)
• element by homogeneous element
• ideal by homogeneous ideal (i.e. ideal generated by homogeneous elements)

This results in further substitutions, e.g. a $$\ast$$local ring is a graded ring with a unique maximal homogeneous ideal, we get a notion of graded depth etc. After all these substitutions we can ask whether the theorem is still true.

One book that does some steps in this direction is Cohen-Macaulay rings by Bruns and Herzog, especially Section 1.5. For example, they have as Exercise 1.5.24 the following graded analogue of the Nakayama lemma:

Let $$(R,\mathfrak{m})$$ be a $$\ast$$local ring, $$M$$ be a finitely generated graded $$R$$-module and $$N$$ a graded submodule. Assume $$M = N + \mathfrak{m}M$$. Then $$M = N$$.

Moreover, a student of mine recently showed that the graded analogue of Lazard’s theorem (a module is flat if and only if it is a filtered colimit of free modules) is also true.

Usually one proves this kind of theorems essentially by a combination of two techniques:

1. Copy the ungraded proof and just substitute ungraded for graded concepts in the manner sketched above.
2. If one is annoyed by the length of the resulting argument, use some shortcuts by some translations between ungraded and graded. (E.g. a noetherian graded ring that is graded Cohen-Macaulay is also ungraded Cohen-Macaulay.)

Sometimes one can also be lucky and the statement is suitably algebro-geometry that one can argue geometrically with the stack $$[Spec R/\mathbb{G}_m]$$ for a graded ring $$R$$, using that a $$\mathbb{Z}$$-grading corresponds to a $$\mathbb{G}_m$$-action.

In any case, my question is the following:

Is there any class of statements, where one knows automatically that the graded analogue is true if the original statement in ungraded commutative algebra is true, without going through the whole proof?

I am not sure whether one can hope here for a model-theoretic approach as I know almost nothing about model theory, but any such statement could save a lot of work in proving graded analogues of known theorems.

## Propostion 1.6 on Atiyah’s commutative algebra text

Let $$A$$ be a ring (commutative with 1) and $$M$$ a maximal ideal of $$A$$. Let $$x \in A-M$$. Since $$M$$ is maximal, the ideal generated by $$x$$ and $$M$$ is $$(1)$$, i.e. the entire ring.

I don’t understand how that is true. If $$x$$ were a unit, certainly. However, $$x$$ could be contained in some other maximal ideal and is not guaranteed to be a unit. So how can we be sure that $$x$$ and $$M$$ can generate $$A$$?

## Weakly dense C*-algebra in a commutative von Neumann algebra and order convergence

Let $$H$$ be a Hilbert space and $$\mathscr{A}$$ a commutative norm-closed $$*$$-subalgebra of $$\mathcal{B}(H)$$. Let $$\mathscr{M}$$ be the weak operator closure of $$\mathscr{A}$$.

Question: For given a projection $$P\in\mathscr{M}$$, is the following true? $$P=\inf\left\{A\in\mathscr{A}:P\leq A\leq 1\right\}$$

It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $$P$$. Also, if the above is true, what happens if $$\mathscr{A}$$ is non-commutative?

## What happens if I replace a *unique natural number* that form a commutative Monoid with *the set of integers* Z that form a commutative Ring?

In mathematical logic, a Gödel numbering is a function that assigns to each symbol and well-formed formula of some formal language a unique natural number, called its Gödel number. The concept was used by Kurt Gödel for the proof of his incompleteness theorems. (Gödel 1931)

The natural numbers, N, form a commutative monoid under addition (identity element zero), or multiplication (identity element one). A submonoid of N under addition is called a numerical monoid. The positive integers, N ∖ {0}, form a commutative monoid under multiplication (identity element one).

The set of integers Z forms a commutative ring under addition and multiplication

What happens if I replace a unique natural number that form a commutative Monoid with the set of integers Z that form a commutative Ring ?

The more specific question should be: What happens if I use (not if I replace) a unique natural number that form a commutative Monoid as (not with) the set of integers Z that form a commutative Ring ?

Monoid $$\Rightarrow$$ Ring

P.S: I don’t want replace formally but I do not want to really change, but to make it believe that it is as if I had replaced.

## On existence of finitely generated projective generator with commutative endomorphism ring in ${}_R Mod$

Let $$R$$ be a ring with unity (not necessarily commutative). Let $${}_R Mod$$ be the category of left $$R$$-modules. If $${}_R Mod$$ has a projective generator with commutative endomorphism ring , then does $${}_R Mod$$ have a finitely generated projective generator with commutative endomorphism ring ?

## Have you ever seen this bizarre commutative algebra?

I have encountered very strange commutative nonassociative algebras without unit, over a characteristic zero field, and I cannot figure out where do they belong. Has anybody seen these animals in any context?

For each natural $$n>1$$, the $$n$$-dimensional algebra $$A_n$$ with the basis $$x_1$$, …, $$x_n$$ has the multiplication table $$x_i^2=x_i$$, $$i=1,…,n$$, and $$x_ix_j=x_jx_i=-\frac1{n-1}(x_i+x_j)$$ (for $$i\ne j$$).

The only thing I know about this algebra is its automorphism group, which is the symmetric group $$\Sigma_{n+1}$$. This can be seen from how I obtained the algebra in the first place.

Let $$I$$ be the linear embedding of $$A_n$$ onto the subspace of the $$(n+1)$$-dimensional space $$E_{n+1}$$ of vectors with zero coefficient sums in the standard basis, given by $$I(x_i)=\frac{n+1}{n-1}e_i-\frac1{n-1}\sum_{j=0}^ne_j,\quad i=1,…,n$$ and retract $$E_{n+1}$$ back to $$A_n$$ via the linear surjection $$P$$ given by $$P(e_0)=-\frac{n-1}{n+1}(x_1+…+x_n)$$ and $$P(e_i)=\frac{n-1}{n+1}x_i,\quad i=1,…,n.$$ Then the multiplication in $$A_n$$ is given by $$ab=P(I(a)I(b)),$$ where the multiplication in $$E_{n+1}$$ is just that of the product of $$n+1$$ copies of the base field; more precisely, it has the multiplication table $$e_ie_j=\delta_{ij}e_i,\quad i,j=0,…,n$$ (where $$\delta$$ is the Kronecker symbol).

$$A_n$$ is not Jordan either, in fact even $$(x^2)^2\ne(x^2x)x$$ in general.

Really don’t know what to make of it.

## Exercises for commutative algebra

I’m currently studying for my final exam of commutative algebra. In the class we covered Artinian rings, Dedekind domains, Integral closures, Grobner basis, (disctrete )Valuation rings… I’m particulary interested in exercises involving the exactecness of $$\operatorname{Hom}(\cdot,N)$$ and $$\operatorname{Hom}(N,\cdot)$$. Thought that you guys could provide some exercises. Thank you all in advance

## Is the diagram commutative

In the following diagram $$f,g$$ are isomorphisms and $$\pi, \tau$$ are canonical projections. Is the diagram then commutative ?


($$A, B$$ are groups) What if $$\pi$$ and $$\tau$$ are just surjective ?