Talk:Ideal induced by Congruence Relation defines that Congruence

Comment: "It is only shown that the relation is a congruence for $+$, not for $\circ$".

Well yes, that's all you need. The thing is:
 * We are given $\mathcal E$, a congruence relation.
 * From it we are given $J$, the ideal defined by $\mathcal E$.
 * We then demonstrate, solely by investigating the behaviour of $+$, that the relation induced by $J$ is $\mathcal E$ itself.
 * We already know that $\mathcal E$ is a congruence for $\circ$ because that's how we defined it.
 * Therefore we do not need to show it is a congruence for $\circ$.

--prime mover 18:17, 27 April 2012 (EDT)


 * Ah, I see. Thanks for elaborating. Some more issues come to mind:


 * The phrase 'ideal induced by $\mathcal E$' might justify a definition page; this will allow for the first two lines to be eliminated or shortly referred to
 * I feel that the point I made could be more stressed, or innocent wanderers may gloss over these details (this can eg. be done by including your explanation in some form onto the page, maybe using terminology like $+$-congruence or something like that)
 * In any case, I'm happy the proof still stands, but it needs to be clearer that it is a proof. --Lord_Farin 18:32, 27 April 2012 (EDT)
 * Added that bit of explanation.
 * I'm in two minds about defining that 'ideal induced by $\mathcal E$' as a definition. I'd rather not because it doesn't have a very wide scope: once it has been established that an ideal and a congruence are equivalent, all this tedious scaffolding is no longer needed. --prime mover 19:09, 27 April 2012 (EDT)