Let $ R $ be a commutative ring with unity (Like in Atiyah-Macdonald, from now "ring" would mean "commutative ring with unity"). We can ask ourselves :
$\textbf{Q}$) In how many ways can we put an equivalence relation $ \sim $ on $ R $ (i.e. partition set $ R $), and put binary operations $ \oplus,\odot $ on set $ R/{\sim} $, such that "Equations in $ R $ give corresponding equations between equivalence classes in $ R/{\sim} $" that is "($a+b=c$ in $R$ implies $[a] \oplus [b] = [c]$ in $ R/{\sim} $) and ($ab=c$ in $ R$ implies $ [a] \odot [b] = [c]$ in $R/{\sim} $)" ?
[Under such $ \sim, \oplus, \odot $, notice $ (R/{\sim}, \oplus, \odot) $ automatically becomes a ring]
$\underline{\textbf{Part-1}}$ (Looking at the potential candidates for $ \sim, \oplus, \odot$)
Let $ \sim, \oplus, \odot $ be as needed. Unwrapping the constraints one by one...
$$ (1) \text{ } \sim \text{ is an equivalence relation} $$
and
$$ (2) \text{ } [a]=[a'], [b]=[b'] \implies [a]\oplus [b] = [a'] \oplus [b'], [a]\odot[b] = [a']\odot [b'] $$
and
$$ (3) \text{ } [a]\oplus [b] = [a+b], [a]\odot [b] = [ab]. $$
Using (3), we see (2) modifies as $ a \sim a', b \sim b' \implies a+b \sim a' + b' , ab \sim a' b'.$
Notice $ a \sim b \iff a-b \in [0] $
($ \implies $: As $ a \sim b $ and $ (-b) \sim (-b) $, we have $ a + (-b) \sim b + (-b) = 0 $.
$ \impliedby $: As $ a - b \sim 0 $ and $ b \sim b $, we have $ (a-b)+b \sim 0+b $, that is $ a \sim b $ )
Using this, constraint (1) can be rewritten as {$a - a \in [0]$; $a - b \in [0] $ implies $b-a \in [0] $; $ a-b, b-c \in [0] $ implies $ a-c \in [0] $}, which just means "$ [0] $ is a subgroup of $ (R, +) $".
Similarly constraint (2) becomes "$a-a', b-b' \in [0] $ implies $ (a+b)-(a'+b') , ab-a'b' \in [0]$", that is "$ a - a', b - b' \in [0] $ implies $ ab - a'b' \in [0] $", that is "$ x,y \in [0] $ implies $ (a' + x)(b' + y) - a'b' \in [0] $", that is "$x,y \in [0] $ implies $ a' y + b' x + xy \in [0] $", that is "$ t \in [0] $ implies $ at \in [0]$".
Constraint (3) remains the same.
So to summarise, in any such $ \sim, \oplus, \odot $ we have :
- $ a \sim b \iff a - b \in I $, where $ I $ is a subgroup of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $
- [Above condition gives $ [a] = a+I $] The operations $ \oplus, \odot $ satisfy $ (a+I) \oplus (b+I) = (a+b+I) $ and $ (a+I) \odot (b+I) = (ab + I) $.
$ \underline{\textbf{Part-2}} $ (That all such $ \sim, \oplus, \odot $ work)
Let $ I $ be a subgroup of $ R $ with $ aI \subseteq I $ for all $ a \in R $. We can readily verify $ a \sim b \overset{\text{def}}{\iff} a - b \in I $, and $ (a+I) \oplus (b+I) \overset{\text{def}}{=} (a+b+I) $, $ (a+I) \odot (b+I) \overset{\text{def}}{=} (ab+I) $ satisfy the constraints in question.
To summarise the entire discussion, equivalence relations $ a \sim b \iff a - b \in I $, arising from subgroups $ I $ of $ (R, +) $ satisfying $ a I \subseteq I $ for all $ a \in R $, are precisely the ones we were looking for. The subgroups here are traditionally called "Ideals".