TITLE: Intuition behind "ideal"
QUESTION [59 upvotes]: To briefly put forward my question, can anyone beautifully explain me in your own view, what was the main intuition behind inventing the ideal of a ring? I want a clarified explanations in these points:

Why is the name "ideal" coined?. In English 'ideal' means "One that is regarded as a standard or model of perfection or excellence." Why did people gave the name of ideal to such group?
And why are the ideals not present in the case of groups?
And give me a very fantastic intuition and motivation behind the ideals and what are the roles served by them in advanced mathematics.

Thanks a lot, to every one.

REPLY [2 votes]: 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".