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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).

Now I believe that this is a proof:

Let $\mathcal{I}$ a set into the $ZF¬AC$ axiomatic set theory.

Let $\boldsymbol{\mathcal{A}} \overset{def}{=} R\left[{\left\lbrace{X_i}\right\rbrace}_{i \in \mathcal{I}}\right]$. With this notation we denote with $0\in\mathcal{I}$ the indeterminate which represents the unital into $R$, $X_0 \equiv 1_R$

For each $j \in \mathcal{I}$ I define the subring ${\boldsymbol{\mathcal{B}}}_{j}(R) \overset{def}{=} R\left[{\left\lbrace{X_i}\right\rbrace}_{i\in{\mathcal{I}\setminus\lbrace j\rbrace}}\right]$ of $\boldsymbol{\mathcal{A}}{(R)}$.

Let ${j_0}\in\mathcal{I}$ an arbitrary fixed element, and let $\mathfrak{m}\in \mathcal{M}{ax}\mathcal{S}{pec}(R)$ another one. We shall define: ${\mathfrak{M}}_{\left(\mathfrak{m},{j_0}\right)}$ ${\overset{def}{=}}$ $\left({\sum_{i\in\mathcal{I}}^{i\neq{j_0}}} {\left\langle{X_i}\right\rangle}_{{\boldsymbol{\mathcal{B}}}_{j_0}}\right) {\bigcup} {\left\langle{ \mathfrak{m} {X_{j_0}} }\right\rangle}_{\boldsymbol{\mathcal{B}}_{j_0}}$. This union is a disjoint union.

Now we can define a subset of $\mathcal{M}ax\mathcal{S}pec(\boldsymbol{\mathcal{A}})$ (that I shall have to prove it): $$ \mathcal{MAX}\left(\boldsymbol{\mathcal{A}}\right) \overset{def}{=} {\left\lbrace{\mathfrak{M}_{\left(\mathfrak{m}_0,j_0\right)}\,\mid\,\mathfrak{m}\in\mathcal{M}{ax}\mathcal{S}{pec}(R)\,\wedge\, j_0\in\mathcal{I}}\right\rbrace}$$

To prove: $\forall \mathfrak{m} \in \mathcal{M}{ax}\mathcal{S}{pec}\left(R\right) \, \forall j \in \mathcal{I} \quad \mathfrak{M}_{\left(\mathfrak{m},j\right)}$ is a maximal ideal of $\boldsymbol{\mathcal{A}}(R)$.

That the mentioned subset of $\boldsymbol{\mathcal{A}}$ is an ideal is easy to prove. If $p\,q\in\mathfrak{M}_{\left(\mathfrak{m},j\right)}$, then there exists only one index of $\mathcal{I}$ such that $p = x + {\sum_{k=1}^{k=\nu}} {z_k X_j^k}$ and $q = y + {\sum_{k=1}^{k=\mu}} {w_k X_j^k}$ where $\forall k \quad z_k \in {\left\langle{\mathfrak{m}{X_j}}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}} \wedge w_k \in {\left\langle{\mathfrak{m}{X_j}}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}}$ and $\forall j \in \mathcal{I} \quad x , y \in \sum_{i\in\mathcal{I}}^{i\neq j}{{\left\langle{X_i}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}}}$, now $x-y$ and $x\cdot y$ are polynomials in $\sum_{i\in\mathcal{I}}^{i\neq j} {{\left\langle{X_i}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}}}$, and the other part is into the ideal ${\left\langle{X_j}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}}$, ${{\sum_{k=1}^{k=\nu}} {q\cdot z_k X_j^k}\cdot{\sum_{k=1}^{k=\mu}} {p\cdot w_k X_j^k}}$.

The rest is to show that for any $\mathfrak{M}$ we have that is ideal maximal. We shall suppose that $p \in \mathfrak{M}_{(\mathfrak{m},j)}$ but $q \notin \mathfrak{M}_{(\mathfrak{m},j)} \cup {\boldsymbol{\mathcal{A}}}^{{}\times{}}$. Then, the piece of $q$ in ${\left\langle{X_j}\right\rangle}_{{\boldsymbol{\mathcal{A}}}{(R)}}$ has some coefficient that is not in ${\left\langle{\mathfrak{m}}\right\rangle}_{{\boldsymbol{\sum_{i\in\mathcal{I}}^{i\neq j}}}}$

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put on hold as unclear what you're asking by Eric Wofsey, Alex M., Najib Idrissi, Daniel W. Farlow, user26857 Nov 22 at 20:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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 '14 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 '14 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 '14 at 16:12
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 '14 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 '14 at 16:57