This is an update of an [older question](https://mathoverflow.net/questions/276997/coloring-the-ideals-of-a-ring), suggested in [this comment](https://mathoverflow.net/questions/276997/coloring-the-ideals-of-a-ring?noredirect=1#comment682204_276997) by Zach Teitler.


Let $R$ be a ring with more than 1 element, and let $A$ be a non-empty set. We call a map $c:R\to A$ an *ideal coloring* if for every nonempty ideal $I$ with more than $2$ elements $c|\_{I\setminus\{0\}}$ is not constant (that is, the non-null elements of every ideal with sufficiently many points receive at least 2 "colors").


Note that the map $c:\mathbb{Z}\to\{0,1\}$ such that $c(z) = 1$ if $z = y^2$ for some $y\in\mathbb{Z}$, and $c(z) = 0$ otherwise is an ideal coloring. (Thanks to Zach Teitler for [pointing this out](https://mathoverflow.net/questions/276997/coloring-the-ideals-of-a-ring?noredirect=1#comment682100_276997).)


Is there a ring that can be colored with $3$ colors, but not with $2$ colors?



Ideal colorings of rings
There are such rings. I will give a nonunital example
first, then explain how to modify it to a unital example.


Given a colored ring, I call an ideal *monochromatic*
if all nonzero elements have the same color. 
I call a coloring of a ring *valid* if no ideal
of more than $2$-elements is monochromatic.


**First example.** (A nonunital ring with a valid 3-coloring, but no valid 2-coloring.)


Consider $\mathbb Z\_2^3$ considered as a ring
with zero multiplication. Suppose that its nonzero elements 
are colored red or blue. I first argue that
there must be three distinct elements $x, y, z$ of the same color
such that $x+y=z$.


Let $R$ be the subset of
red elements, and $B$ be the subset of blue elements.
We may assume that $|R|\geq |B|$. Since $R\cup B\cup \{0\}$
has $8$ elements, $|R|\geq |B|$ implies that
$|R|\geq 4$. Choose $r\in R$, and apply the function
$x\mapsto x+r$ to the subset $R$. If there do not exist
distinct red $x, y, z$ such that $x+y=z$, then
$R+r\subseteq B\cup \{0\}$. Since $x\mapsto x+r$ is a permutation
of order $2$,
$R+r = B\cup \{0\}$, and $(B\cup\{0\})+r=R$.
The choice $r\in R$ was arbitrary,
so $R+R=B\cup\{0\}$, $R+B\subseteq R$, and so $B+B\subseteq B\cup\{0\}$.
Thus, if $x, y\in B-\{0\}$ are distinct
and $z=x+y$, then $x, y, z$ are distinct and all blue.


Now the ideals of $\mathbb Z\_2^3$ are the same as its subgroups,
because we chose the multiplication to be trivial.
Choose any $2$-coloring of $\mathbb Z\_2^3$, then find
distinct $x, y, z$ of the same color such that $x+y=z$.
$I=\{0,x, y, z\}$ is a monochromatic ideal,
so the $2$-coloring is not valid. 


The above argument shows
that no $2$-coloring of $\mathbb Z\_2^3$ is valid.
It is easy to product a valid $3$-coloring:
$R=\{(1,0,0),(0,1,0),(0,0,1)\},
B=\{(1,1,0),(1,0,1)\}, G=\{(0,1,1),(1,1,1)\}$.
(Each color colors an independent set.)


**Second example.** (A unital ring with a valid 3-coloring, but no valid 2-coloring.)


I formally adjoin a unit to the first example.
The ring $\mathcal R$ is $\mathbb Z\oplus \mathbb Z\_2^3$.


All subgroups of $\mathbb Z\_2^3$ are again ideals, so the argument
for the nonunital example above works here to show that there
is no valid $2$-coloring of $\mathcal R$.
To produce a valid $3$-coloring,
use the $3$-coloring of the previous example to $3$-color the elements of
$\mathbb Z\_2^3\subseteq \mathcal R$. 
Extend this coloring to all of $\mathcal R$
by coloring $m+z\in\mathcal R\;(=\mathbb Z\oplus \mathbb Z\_2^3)$ red if
$m$ is positive and blue if $m$ is negative. \\\


