If every ideal I in R is contained in a finite series of ascending ideals, prove that only finitely many ideals of any kind contain I.

Question

If a Noetherian ring is defined by the fact that all ideals are contained within a finite series of ascending ideals, how does this prove that the initial ideal is contained within finitely many ideals of any kind, for example ideals intersecting on the initial ideal but not containing each other?

Answer 1

Any ring satisfies the condition "all ideals are contained within a finite series of ascending ideals". The definition of Noetherian is that any ascending chain of ideals is finite.

Not to be pedantic, but given that you didn't state the definition of Noetherian quite correctly, it seems possible that the question you asked is also not quite what you meant. It's not true that every Noetherian ring has the property as stated. In fact $\Bbb Z$ is a counterexample. The ideal $I=\{0\}$ is contained in every one of the infinitely many ideals $n\Bbb Z$.