# A ring with infinitely many maximal ideals

I am looking for a commutative ring with 1 and infinitely many maximal ideals such that for any infinite family $\{m_i\}_{i\in I}$ of maximal ideals there exists $j \in I$ such that $m_j\subseteq \cup_{j \not= i\in I}m_i$.

Let $R$ be the direct sum of countably infinitely many copies of the field $\mathbb{Z}_2.$ Each member of $R$ is a sequence $f\colon \mathbb{N}\to\lbrace 0,1 \rbrace$ which takes on the value $1$ only finitely often; addition and multiplication are pointwise.

Let $\delta_n \in R$ be defined by setting $\delta_n(k)=\begin{cases}1,&\text{ if }k = n, \\0,&\text{ if }k \ne n. \end{cases}$

For any $n\in \mathbb{N},$ let $I_n = \lbrace f\in R \mid f(n)=0\rbrace.$ Clearly each $I_n$ is an ideal in $R.$

1. Any proper ideal $J$ of $R$ is contained in some $I_n.$

Proof: If $J$ is an ideal not contained in any $I_n,$ then for each $n\in \mathbb{N},$ there exists $f_n \in J$ such that $f_n(n)=1.$ Since $J$ is an ideal, each $\delta_n = \delta_n \cdot f_n \in J.$ But every member of $R$ can be written as a finite sum of $\delta_n\text{'s},$ and $J$ is closed under addition, so $J=R.$

1. $I_m\subseteq I_n$ implies $m=n.$

Proof: If $m \ne n,$ then $\delta_n \in I_m\setminus I_n,$ so $I_m \not \subseteq I_n.$

1. Each $I_n$ is a maximal ideal in $R.$

Proof: Suppose $J$ is an ideal and $I_n \subseteq J.$ By 1, $J$ is contained in some $I_m.$ Then $I_n \subseteq J \subseteq I_m,$ so, by 2, $m=n,$ and we must have $J=I_n.$

1. The maximal ideals in $R$ are precisely the $I_n$ for $n \in \mathbb{N}.$

Proof: Let $J$ be any maximal ideal in $R.$ By 1, $J$ is contained in some $I_n.$ By maximality of $J$ and the fact that $I_n$ is an ideal, we must have $J=I_n.$

1. Each $f \in R$ belongs to all but finitely many of the maximal ideals of $R.$

Proof: $f(n) = 0$ for all but finitely many $n.$ For any such $n,$ $f$ belongs to $I_n.$



Now we can prove the required statement. Let $\mathscr{S}$ be an infinite family of maximal ideals in $R.$ Let $I$ be any member of $\mathscr{S}.$ We just need to show that $I \subseteq \bigcup_{J\in \mathscr{S}\setminus\lbrace{I\rbrace}}J,$ but you can see that as follows:

If $f$ is any member of $I,$ we know that $f$ belongs to all but finitely many of the maximal ideals of $R.$ Since $\mathscr{S}$ is infinite, $f$ must therefore belong to some member of $\mathscr{S}\setminus\lbrace I \rbrace.$ (If not, there would be infinitely many maximal ideals that $f$ didn't belong to.)

• It seems that your $R$ is not unitary. Commented Sep 24, 2016 at 13:57
• Whoops -- you're right. It was too late last night and I missed that. I'll think about it, but it doesn't look fixable unfortunately. Commented Sep 24, 2016 at 14:05