Properties of a ring $R$ guaranteed by an equivalence relation and certain operations

Question

Let $R$ be a ring and let $S$ be a nonempty subset of $R$.

1) Consider the relation ~ on $R$ given by $a$~$b\Leftrightarrow a-b\in S$. Which properties must $S$ have so that ~ is an equivalence relation?

2) Suppose you have $S$ such that ~ is an equivalence relation. Consider the set $R/ S:= \{[a] :a\in R\}$ of equivalence classes of $R$ under ~. What additional properties must $S$ have so that the operations $[a]+[b]=[a+b]$ and $[a][b]=[ab]$ are well-defined for all $a,b\in R$?

Now, before you answer this question, please give a yes or no response to my answers below.

1) $S$ must have an additive identity (reflexive property of ~), an additive inverse (reflexive property of ~), commutativity of addition (since if $a+b=a-(-b)\in S$, then $b+a=-((-b)-a)=-(-(a+b))=a+b\in S$)), and associativity of addition (since if $(a+b)+c=(a-(-b))-(-c)\in S$, then $a+(b+c)=a-(-(b-(-c)))=(a+b)+c\in S$). I'm really unsure about my reasoning for the last two properties (it's most likely not enough or entirely incorrect).

2) I'm stuck here.

Edit: sorry about the last paragraph. Shouldn’t ‘ve been so negative

Answer 1

1) The answer to be angled at is

$(S,+)$ must be a subgroup of $(R,+)$

Each of the following points corresponds to an axiom of equivalence relations:

1. $0\in S$
2. For all $a\in S$, $-a\in S$.
3. For all $a, b\in S$, $a-b\in S$.

You may recognize these conditions as the subgroup test.

You explained correctly that the first is due to the reflexive property, but I wouldn't say that inverses are due to reflexivity. Try to sort out which ones the last two correspond to.

You don't have to worry about commutativity or associativity since these are inherited from the group $(R,+)$. The operation used in $S$ is the same $+$, so it's obvious a subset of $R$ can't violate the commutativity or associativity of $+$ in $R$.

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2) The answer to be angled at here is

$S$ must be an ideal (two-sided) of $R$.

There are already good explanations as to how the axioms of an ideal justify that the "obvious" multiplication operation $[a][b]=[ab]$ is well-defined. If $S$ is not an ideal, then this proposed operation doesn't work because it is not a well-defined map. (The addition operation $[a]+[b]=[a+b]$ is already well defined since $S$ is a normal subgroup of $R$.) Finally, the multiplication distributes over addition so that $R/S$ really is a ring.

What you wrote for this second point has virtually nothing to comment on. It looks like a confused mess of guesses.