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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 $I\neq\{0\}$ the restriction $c|_I$ is not constant (that is, the elements of every nontrivial ideal receive at least 2 "colors").

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

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  • $\begingroup$ Your example doesn't work, the function $c$ is constant on $2\Bbb{Z}$. $\endgroup$
    – abx
    Commented Jul 22, 2017 at 7:15
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    $\begingroup$ Isn't $c_{(2)}$ constant? How about the $2$-coloring of $\mathbb{Z}$ given by $c(n) = 0$ if $n$ is a square, $1$ otherwise. $\endgroup$
    – Zach Teitler
    Commented Jul 22, 2017 at 7:16
  • $\begingroup$ Right sorry will remove the example! $\endgroup$
    – Dominic van der Zypen
    Commented Jul 22, 2017 at 7:31
  • $\begingroup$ I'm curious where this definition came from. Could you sketch the motivation? Would it possibly be interesting to consider colorings of the nonzero elements of the ring? $\endgroup$
    – Zach Teitler
    Commented Jul 22, 2017 at 18:13
  • $\begingroup$ I was trying to see whether something like this was being done on rings: dominiczypen.wordpress.com/2017/07/17/… The question asked in the post was asked here but has not been answered: mathoverflow.net/questions/275489/… Your suggestion of the nonzero elements might be interesting! $\endgroup$
    – Dominic van der Zypen
    Commented Jul 23, 2017 at 7:23

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Every ring can be ideal-colored with $2$ colors. Let $0$ be colored by $0$, and every other element by $1$.

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    $\begingroup$ How foolish of me not to see this answer...! I apologize and thank you for writing it up. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 22, 2017 at 8:42
  • $\begingroup$ Oh, well, you're welcome. It was a coin toss, to write it here or in a comment. $\endgroup$
    – Wlod AA
    Commented Jul 22, 2017 at 8:59

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