Talk:Proper Ideal iff Quotient Ring is Non-Null

The theorem is not true, because of the definitions here, concerning rings and trivial rings.

The theorem is true if nontrivial is replaced by non-null.

Then the theorem is a consequence of there existing more than one equivalence class iff the ideal is not the whole set.

The theorem is once again true if the rings are required to contain unity (in which case all non-null rings are nontrivial), but this form is therefore weaker than the one above.

These changes will not change the result in What links here: Proper Ideal of Ring is Contained in Maximal Ideal, in which Krull's Theorem is applied on a non-null ring with unity.

Counterexample 1: $S$ abelian group, equipped with trivial multiplication. $T$ a cyclic subgroup. Then $T$ is ideal of $S$ as ring.

If $T$ is proper then $S \ T$ is trivial because of the trivial multiplication.

Counterexample 2: $\set {0, 1} \times \set {0, 2}$, equipped with usual addition and multiplication in each coordinate: mod 2 in the 1st, mod 4 in the 2nd.

Then $\set {0, 1} \times \set 0$ is a proper ideal, and the quotient ring is trivial as $2 \times 2 = 0$.

The second example shows that the quotient ring (a non-trivial ring) / (proper ideal) can be trivial. RandomUndergrad (talk) 09:49, 30 April 2020 (EDT)

Please feel free to make whatever changes are needed here.

This page was added by someone who is no longer around. Much of the work he did was good, and contributed to the site, but he was not good at making sure his definitions were consistent and accurately linked. We had to block him from contributing after he decided to radically restructure large areas without consultation and without completing the work, leaving the site's integrity compromised. --prime mover (talk) 11:05, 30 April 2020 (EDT)

The statement and evolution of the result have had nontrivial ring replaced with non-null ring, which is probably what was meant in the first place. The moral of this story: whenever linking to a definition, always check that it says what you want it to say, or you may find your proof does not do what you want it to do. --prime mover (talk) 21:17, 29 May 2022 (UTC)

Yes, another possibility was to assume that the ring has a unity. But the statement with non-null is more useful.--Usagiop

Not all non-trivial rings have a unity. The ring of even integers under conventional addition and multiplication doesn't, for one. Does this theorem not apply to those too? --prime mover (talk) 21:39, 29 May 2022 (UTC)