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It's (probably) a fairly basic result that the only equivalence relations on $\mathbb{Z}$ that are compatible with the ring operations are congruences modulo $n$ for $n \in \mathbb{Z}$... as it's usually stated without proof in more advanced works on algebra.

(This is a "completeness result"--if you'd like to call it that--of the more elementary fact (in the opposite direction) that congruences modulo $n$ are compatible with the ring operations on $\mathbb{Z}$. The only slightly tricky part of the reverse implication is to prove it for multiplication: $a_1 - a_2 = xn$ and $b_1 - b_2 = yn$ imply that $a_1 b_1 - a_2 b_2 = a_1 (b_1 - b_2) + b_2(a_1-a_2) = (a_1 y + b_2x)n$.)

But to restate the problem at hand here, let $E$ be an equivalence relation on $\mathbb{Z}$ compatible with the ring operations, i.e E is reflexive, transitive, symmetric, and additionally that $a_1Ea_2$ and $b_1Eb_2$ imply that $(a_1+a_2)E(b_1+b_2)$ and $(a_1a_2)E(b_1b_2)$. Prove that there exist $n \in \mathbb{Z}$ such that for all integers $a,b \in \mathbb{Z}$, $aEb$ implies that $a-b = kn$ for some $k \in \mathbb{Z}$.

the gods from engineering
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2 Answers2

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Consider the set $Z:=\{a\in \mathbb{Z}:aE0\}$. The positive integers in this set have a minimum, call it $n$. (If the only element is $0$, then $E$ is the equality relation.)

One can then prove the following statements:

  1. $Z=n\mathbb{Z}$ (by using the division algorithm).
  2. $aEb\iff (a-b)E(b-b)\iff (a-b)E0\iff a-b=kn$.
Chrystomath
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  • I guess your point (2) is that this problem is equivalent to showing that the only ideals of $\mathbb{Z}$ are $n\mathbb{Z}$. – the gods from engineering Mar 30 '21 at 09:43
  • One point that's unclear to me (from this sketch) is whether you have use the fact that E is compatible with multiplication or does the compatibility with addition (explicit in 2) suffice? – the gods from engineering Mar 30 '21 at 13:42
  • https://math.stackexchange.com/a/1301510/173347 is related to your first para, that $n\mathbb{Z}$ are the only subgroups of $\mathbb{Z}$ so there can't be any ideals besides those (and in this case all subgroups are ideals too.) – the gods from engineering Mar 30 '21 at 15:08
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    @Fizz Although the fact that E is compatible with multiplication is not directly used in the proof, this is misleading because in $\mathbb{Z}$ subgroups are automatically ideals. For any $a$, $na=0$, so if $a_1\sim a_2$ and $b_1\sim b_2$ then $a_1b_1=a_1+\cdots+a_1\sim a_2+\cdots a_2=a_2b_1= \cdots \sim a_2b_2$. – Chrystomath Mar 31 '21 at 06:54
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To collect here a couple of my comments that maybe someone else finds useful on this matter:

  • The problem is reduced to that of determining (all the) ideals of $\mathbb{Z}$ because (as Howie phrased this):

in a ring a congruence is determined if we know the ideal which is the congruence class containing the zero.

  • There can be no more ideals than there are (additive) subgroups and $n\mathbb{Z}$ are all subgroups of $\mathbb{Z}$.

  • More generally, every integral & Euclidean domain is a principal ideal domain, or in Dummit and Foote's words:

The first implication of a Division Algorithm for the integral domain $R$ is that it forces every ideal of $R$ to be principal. Every ideal in a Euclidean Domain is principal. More precisely, if $I$ is any nonzero ideal in the Euclidean Domain $R$ then $I$ = ($d$), where $d$ is any nonzero element of $I$ of minimum norm.

(For $\mathbb{Z}$ the absolute of an element value gives the norm.)

the gods from engineering
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