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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?

Tom Holloway
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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 $n\Bbb Z$.

David C. Ullrich
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  • My question related specifically to ideals of the ring of integers. Ireland and Rosen, 2nd ed. pp 176,178 prove that any non-zero ideal of such rings is contained in a finite series of ascending ideals. I believe that I understand that proof. They use that to then claim that any element of the ring is contained in finitely many ideals period. That is then used to prove a finite ideal class number. My question concerns the claim that the fact of any ideal contained in the finite ascending series imples that it is containied in finitely many ideals period. – Tom Holloway May 19 '16 at 20:46
  • @TomHolloway I'm not sure what your point is. It's easy to see that any ideal other than ${0}$ in $\Bbb Z$ is contained in only finitely many ideals, just because $n$ has only finitely many factors. But that's not true for ideals in Noetherian rings in general. (And for rings other than $\Bbb Z$, ${0}$ is not the only counterexample. Consider ${0}\times\Bbb R$ in $\Bbb R^2$.) – David C. Ullrich May 19 '16 at 21:01
  • The ring of integers is the intersection between the set of algebraic integers and the field extension Q(a)/Q where a is complex, Q rational. Ireland and Rosen's proofs that any ideal of this ring is contained in a finite number of other ideals is, I believe, as I stated. My question still stands. – Tom Holloway May 23 '16 at 17:10