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Let $\mathcal{N}{oetherian}\mathcal{C}\mathcal{R}{ng} \overset{def}{=} {\left\lbrace{ R \in \mathcal{C}\mathcal{R}{ng} \wedge R \,\text{is}\, \mathcal{N}{oetherian} }\right\rbrace}$.

I define the property of the existence of maximal ideals for each non-invertible element as a class.

Let $R \in \mathcal{C}\mathcal{R}{ng}$ , I define a class of commutative rings which agrees with the wanted property:

$${\mathcal{C}\mathcal{R}{ng}\mathcal{W}{ith}\mathcal{M}{aximals}} {\overset{def}{=}}$$ $${\overset{def}{=}} {\left\lbrace{ R \in \mathcal{C}\mathcal{R}{ng}\,\mid\,\mathcal{M}{ax}{(R)}\neq\emptyset \wedge \forall x \in {R\!\setminus\!{R^{\times}}} \, \exists \mathfrak{m}\in \mathcal{M}{ax}(R) \, x\in\mathfrak{m}}\right\rbrace}$$

It is well known by all of us that $$ ZFC \vdash {\mathcal{C}\mathcal{R}{ng}\mathcal{W}{ith}\mathcal{M}{aximals}}\equiv \mathcal{C}\mathcal{R}{ng} $$

Too is well known by all of us that $$ ZF\neg{AC} \vdash {\mathcal{N}{oether}\mathcal{C}\mathcal{R}{ng}} \implies {\mathcal{C}\mathcal{R}{ng}\mathcal{W}{ith}\mathcal{M}{aximals}} $$

Then the class ${\mathcal{C}\mathcal{R}{ng}\mathcal{W}{ith}\mathcal{M}{aximals}}\neq\emptyset$ into $ZF\neg AC$.

Which I want find is a class in $ZF \neg AC$ such that ${\mathcal{X}\mathcal{C}{lass}} \implies {\mathcal{C}\mathcal{R}{ng}\mathcal{W}{ith}\mathcal{M}{aximals}}$ where ${\mathcal{N}{oetherian}\mathcal{C}\mathcal{R}{ng}}\implies{\mathcal{X}\mathcal{C}{lass}}$ and $\neg\left({\mathcal{X}\mathcal{C}{lass}}\implies{\mathcal{N}{oetherian}\mathcal{C}\mathcal{R}{ng}}\right)$.

Now, our $\mathcal{X}\mathcal{C}{lass}$ is the subject of the question.

$\mathcal{X}\mathcal{C}{lass}$ is non-trivial. Let $R\in\mathcal{N}{oetherian}\mathcal{C}\mathcal{R}{ng}$, and let $\mathcal{I}$ any set of indices. I claim that $R\left[{\left\lbrace{X_i}\right\rbrace}_{i \in \mathcal{I}}\right]$ is of $\mathcal{X}\mathcal{C}{lass}$ and it is non-Noetherian (this last note is an evidence).

Trying to prove:

Let $x\in R\left[{\left\lbrace{X_i}\right\rbrace}_{i \in \mathcal{I}}\right]$ such that is irreducible and is non-invertible. Then $Indeterminates(x)$ $=$ ${\lbrace{X_i}\rbrace}_{i\in F}$ where $F\subset\mathcal{I}$ and $\left\vert {F} \right\vert \in\mathbb{N}$.

Notation ${\boldsymbol{\mathcal{A}}}{\overset{def}{=}}{R{\left[{\left\lbrace{X_i}\right\rbrace}_{i \in \mathcal{I}}\right]}}$

And now we have another set of indices $\mathcal{F} := \mathcal{I}\!\setminus\!F$.

There are three possibilities for the polynomial $x$ that we can study.

  1. $x \in R\!\setminus\!R^{\times}$. Then , $\forall \mathfrak{m} \in \mathcal{M}{ax}(R)$ such that $x \in \mathfrak{m}$ we get that ${\left\langle\mathfrak{m}\right\rangle}_{\boldsymbol{\mathcal{A}}}+{\left\langle{\lbrace{X_i}\rbrace}_{i \in \mathcal{I}}\right\rangle}_{\boldsymbol{\mathcal{A}}}$ is a maximal ideal of the generated polynomial ring.

  2. $coefficient[0](x) \in \mathfrak(m) \in \mathcal{M}\!{ax}(R)$, then ${\left\langle\mathfrak{m}\right\rangle}_{\boldsymbol{\mathcal{A}}}+{\left\langle{\lbrace{X_i}\rbrace}_{i \in \mathcal{I}}\right\rangle}_{\boldsymbol{\mathcal{A}}}$ is maximal as the anterior case.

  3. $coefficient[0](x) \in R^{\times}$. Then we can suppose that the form of $x$ is $x=1+f$ where $f\in {\left\langle{\left\lbrace{X_i}\right\rbrace}_{i \in F}\right\rangle}_{\boldsymbol{\mathcal{A}}}$, and add a very restrictive hypothesis: $\left\vert \mathcal{I} \right\vert {\cong}_{\textbf{Sets}} {\mathbb{N}}$ and there exists ${\lbrace{r_i}\rbrace}_{i\in\mathbb{N}}$ that it generates $R$. We find a very, very large ideal, but before start to build it (by recursion) let us make some definitions : $$F\overset{\triangle}{\boldsymbol{:}}\boldsymbol{\mathcal{A}} \to \mathbb {\overset{\triangle}{\boldsymbol{:}\boldsymbol{:}}} y \mapsto IndicesOfIndeterminates(y)\overset{\triangle}{\equiv} F(y)$$ $$F_1 \overset{def}{=} F(x)$$ $${\mathbf{N}}_{1} = \mathbb{N}\!\setminus\!F_1$$ $$A_1 \overset{def}{=} {\left\lbrace{ x }\right\rbrace}$$ $$\mathfrak{A}_1\overset{def}{=}{\left\langle A_1 \right\rangle}_{\boldsymbol{\mathcal{A}}}$$ $$\exists y \in \boldsymbol{\mathcal{A}} \quad 1\notin{ \left\langle { A_{k-1}\cup\left\lbrace { y } \right\rbrace } \right\rangle }_{\boldsymbol{\mathcal{A}}}$$ $$F_k \overset{def}{=} F_k \cup F(y)$$ $${\mathbf{N}}_{k} = {\mathbf{N}}_{k-1}\!\setminus\!F(y)$$ $$A_{k} \overset{def}{=} A_{k-1}\cup{\left\lbrace{ y }\right\rbrace}$$ $$\mathfrak{A}_k\overset{def}{=}{\left\langle A_k \right\rangle}_{\boldsymbol{\mathcal{A}}}$$ The construction is: $$F_{\aleph_0}\overset{def}{=}\underset{\nu \in \Bbb{N}}{\bigcup}{F_\nu}$$ $$A_{\aleph_0}\overset{def}{=}\underset{\nu \in \Bbb{N}}{\bigcup}{A_\nu}$$ $$\boldsymbol{\mathfrak{M}}\overset{def}{=}{\left\langle A_{\aleph_0} \right\rangle}_{\boldsymbol{\mathcal{A}}}$$ The process has to be infinite of the length of $\Bbb{N}$ because the infinite set of indeterminates and the the finite character of $F(y)$ for any $y\in\boldsymbol{\mathcal{A}}$. On the other hand the process cover all the indeterminates in $\boldsymbol{\mathcal{A}}$. $1\notin\boldsymbol{\mathfrak{M}}$ by construction. If $1\in\boldsymbol{\mathfrak{M}}$ then there exists a finite set of elements $\lbrace x_1,x_2,\ldots,x_n \rbrace \subsetneq \boldsymbol{\mathfrak{M}}$ and a set of "scalars" $\lbrace \alpha_1,\alpha_2,\ldots,\alpha_n \rbrace \subsetneq \boldsymbol{\mathcal{A}}$ such that $\overset{n}{\underset{k=1}{\sum}}{\alpha_k x_k} = 1$. But these elements of the found ideal are found in the process in a finitely many steps against the defined development of the process. The ideal is maximal. Let I suppose that $$\boldsymbol{\mathfrak{M}} \subsetneq \boldsymbol{\mathfrak{M}}\boldsymbol{'} \subsetneq \boldsymbol{\mathcal{A}}$$, then the elements $A \overset{def}{=} \boldsymbol{\mathfrak{M}}\boldsymbol{'} \!\setminus\! \boldsymbol{\mathfrak{M}}$ are such that $A$ is agree with the set of steps of the algorithm to achieve a step to achieve a new index of ${A_k}$. Each of them can be inserted in the process altering the order of the index. Including $A$ was an infinite set of generators, $A$ is countable and we can reorder the index of the process to introduce all the elements of $A$.

$\neg$ Q.E.D. but quasi-Q.E.D.

The quasi-theorem is proved for $R$ countably many generated and for a set of indices of the indeterminates such that $\mathcal{I}=\mathbb{N}$. I do not know if it is true for the complete class $\mathcal{X}\mathcal{C}{lass}$. Has anyone an idea to enlarge that it has been shown?.

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I suppose you're asking about a (definable) class $\mathcal C$ of commutative rings for which we have $ZF\vdash$“For any $R\in \mathcal C$, every non-invertible element of $R$ is a member of a maximal ideal in $R$.”? What about the class of rings with the property that every non-invertible element is a member of a maximal ideal? ;) You need to word your question a little more carefully. Also, adding some line breaks wouldn't hurt. –  tomasz Apr 12 at 12:35
    
In a few moments I'll remake the ask, following your suggestions. But there is a problem, the ask is not well defined because therea are two questions in my ask, and because I do not want the only side of the question was those of the foundations or the logical side. –  Eärendil Elrond Arwen Apr 12 at 13:43
    
I have tried making a better question in the sense you are say me. Thanks. I hope you say me every problem with the question and with the result. –  Eärendil Elrond Arwen Apr 12 at 16:12
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Well, I'd rewrite your question to be more formally and logically sound and easier to read, but I'm kind of busy doing the same with my thesis right now... For what it's worth, I think your idea should work, but I'm pretty sure you need to be a little more careful choosing $m$, and you also need to explain why the ideal you get is maximal (and proper)... –  tomasz Apr 12 at 16:59
    
I have cut some afirmations to countable sets of generators, and I believe that I have found the way to prove this. For my purposes are sufficient but it was very nice to find a class more large of rings. –  Eärendil Elrond Arwen Apr 14 at 16:57

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