# Zorn's Lemma applied to Commutative Algebra

I am trying to use Zorn's lemma to prove that the set of all prime ideals in a commutative ring with unity has a minimal element. I have already seen the other posts on the site asking for help proving the same theorem - so please finish reading before marking my question as a duplicate. My question here is more about how to use Zorn's lemma. I have a specific step that seems to be tripping me up.

So heres what I got: 1. Define the set of all prime ideals of some non-zero commutative ring with unity, then order this set with respect to reverse inclusion.

1. Since I have at my disposal theorems that 1) every such ring has a maximal ideal, and 2) every maximal ideal is a prime ideal, I know that my set of prime ideals is non empty. Further I have proven that reverse inclusion is a partial order on my non empty set of prime ideals.

2. We have gathered all the ingredients we need to apply Zorn's lemma - except the chain condition, and this is where I have a problem. If I let $T$ be an arbitrary chain, I can use intersections of prime ideals to produce my maximal element (in my case the minimal). But I have a problem, What if $T = \emptyset$. It appears since I am letting $T$ be an arbitrary chain, and $\emptyset$ appears to vacuously be an acceptable chain, so whatever my proof does should hold. But then I need some prime ideal $P$ such that $P \subseteq \emptyset$? It seems like this would force $P$ to also be the empty set which wont work for me since the empty set cant be an ideal.

I've been known to big deal minor issues - thats probably what I am doing here, but I cant move on until I have some closure here. Thanks for any help.

## 1 Answer

An upper bound of a chain $T$ is an element $p$ of the partial order which is bigger than every element of $T$. If $T$ is empty, then every element of the partial order is an upper bound of $T$!

For example, in this specific case, the condition "$P$ is an upper bound of $T$" does not mean "$P\subseteq T$", but rather "$P\subseteq Q$ for every $Q\in T$. So we're not looking for a prime ideal which is contained in the emptyset, we're just looking for a prime ideal which is contained in every element of the emptyset. But, since the emptyset has no elements, every prime ideal is contained in every element of the emptyset, in the same way that every person is taller than every pink elephant in my house.

• My pink elephant is very tall. I never get to do that example :-( – Mariano Suárez-Álvarez Jan 13 '17 at 2:45
• Oh man. I entirely see my confusion. $T$ $\textit{is}$ empty. I was entirely mistaking the empty set as an $\textit{element}$ of $T$. Thanks for slapping me straight. – Prince M Jan 13 '17 at 2:56