I have a very quick question about ideals.
From what I know: $I$ is an ideal of a ring $R$ if $I$ is a subring of $R$ and $rI \subseteq I$ for all $r \in R$ (assume $R$ is commutative).
If $I$ is an ideal of a ring $R$, is it the case that, for all $r,s \in R$, we have
$(r+I) \cdot (s+I) = rs +I$?
(Where, for two subsets $S,T \subseteq R$, $S \cdot T = \{st: s \in S, t \in T\}$.)
If this is true then I am having a difficult time proving it. It is not hard to show that $(r+I) \cdot (s+I) \subseteq rs+I$ for all $r,s \in R$. I am having trouble showing the other direction: $rs+I \subseteq (r+I) \cdot (s+I)$. Thank you in advance.
In group theory, it's assumed almost all the time that $S\cdot T = \{st\mid s\in S, t\in T\}$, no matter what type of subsets $S, T$ are. This is used in (to name a few)
This is not the case in ring theory. For example, $$(2+2\mathbb Z)(2+2\mathbb Z)=4+4\mathbb Z + 4\mathbb Z + 4\mathbb Z + 4\mathbb Z=4+4\mathbb Z\not=4+2\mathbb Z$$ refutes $(r+I)\cdot (s+I)=rs+I$, if $S\cdot T=\{st\mid s\in s, t\in T\}$ defined as above. In order to define quotient ring to make the first isomorphism theorem hold, it's necessary to make sure by definition $$(a+I)\cdot (b+I):=ab+I$$
What's required here is to show $ab+I$ doesn't depend on $a,b$ but only the cosets $a+I, b+I$.
The definition of $S\cdot T$ in ring theory very much depends on the context. For example, when $I, J$ are ideals, $IJ$ usually means the ideal generated by $\{rs\mid r\in I, s\in J\}$ (at least when the ring is commutative). This is different from the multiplication of additive cosets as above.