I'm very confused as to the use of Zorn's lemma in showing that the nilradical of a ring is the intersection of all the prime ideals.
Namely, we let $a \notin N$, where $N$ is the nilradical. Then we let $\Sigma$ be the set of ideals of our ring $A$ such that $a^n \notin \Sigma$ for any $n \in \mathbb{N}$. $\Sigma$ is non-empty as the zero ideal is in it.
Next, we order $\Sigma$ by inclusion, and in any chain we just take the union of all the ideals, which due to the inclusion is just the sum of the ideals in the chain. We then conclude that this ideal is an upper-bound of the chain, and apply Zorn's lemma to get that there is a maximal element in $\Sigma$.
First, for the sake of my sanity, the upper bound we got in this case also happens to be a maximal element, yes? It's an ideal in $\Sigma$ that dominates every other element in the chain by inclusion.
My main question is this: Can't this be done for any ring and for any set $\Sigma$ of ideals in the ring? We just order $\Sigma$ by inclusion and conclude that there's a maximal element by using the union in any chain as an upper bound. But then doesn't this imply every ring is Noetherian (which is obviously false)?
Thanks,
Garnet
Let's see what happens when we do this for a chain of ideals in an arbitrary commutative ring.
Given $I_0 \subset I_1 \subset I_2 \subset \cdots$, we form $I=\bigcup I_j$. This is an ideal, and an upper bound for the chain. But it need not belong to the chain, which is a requirement for verifying the noetherian condition.