# Explanation for a step in a proof of Hartshorne 'Algebraic Geometry'

In Hartshorne's 'Algebraic Geometry', the following proposition is proved (proposition 1.5):

In a noetherian topological space $X$, every nonempty closed subset $Y$ can be expressed as a finite union $Y = Y_1 \cup \ldots \cup Y_r$ of irreducible closed subsets $Y_i$...

(since my question is about this part of the property, I did not include the uniqueness part of this statement). He proves this as follows: he defines the subset $\mathcal{S}$ of subsets $Y$ of $X$ which do not satisfy this statement and argues that if this subset $\mathcal{S}$ is nonempty, then because of $X$ being a noetherian topological space, the set $\mathcal{S}$ contains a minimal element, say $Y$.

I do not see why this is true. Moreover, only the definition of 'noetherian topological space' has been introduced, so it must be something basic I guess...

The set $S$ is specified as the set of those nonempty closed subsets of $X$ such that ...

Assume $S$ is nonempty.

Thus, let $W_1$ be an element of $S$.

If $W_1$ is minimal, you're done.

If not there must be an element $W_2 \in S$ such that $W_2$ is a nonempty proper subset of $W_1$.

As before, if $W_2$ is minimal, you're done.

Continuing in the same way, then, since $X$ is Noetherian, the process can't go on forever, hence must terminate. The last $W_k$ must be a minimal element of $S$.

• ah, I see! I have tried to do something similar, but then I got stuck on the fact that I could have that two sets are not necessarily contained in one another... but then the element would be minimal by definition. Thanks a lot! – Student Mar 18 '17 at 10:15

This is essentially equivalent to the following statement:

If $A$ is a ring TFAE:

1. Every ascending chain stabilizes.

2. Every nonempty collection of ideals contains a maximal element.

For example, the implication in question can be answered by considering the contrapositive. If there were no maximal element, then we could create an infinite chain that does not stabilize by taking the union of ideals.

To see the rest, let $Z(I_1) \supseteq Z(I_2),..$ must stabilize (just by inclusion reversal.) I've also answered a question here that shows that the Zariski topology is quasi-compact.)

The rest of the proof goes as follows: Let $\mathcal{A}$ be the collection of zariski closed sets that do not have a decomposition into irreducible ideals. Then there is a minimal element, say $Z$. If $Z$ were irreducible, then $Z \notin \mathcal{A}$ which is absurd. On the other hand, if $Z$ is irreducible, then it can be written as the disjoint union of two sets $Z_1, Z_2$, but by nimilaity, these are not in $\mathcal{A}$, so they have a decomposition and we are again done.

• Oh, okey, so I guess since the condition for a noetherian topological space uses descending chains, this is equivalent with every non empty collection having a minimal element then? – Student Mar 18 '17 at 10:13
• yes, :). I think it may be helpful to look at a proof for the zariski topology being "quasi-compact." I just linked it in my answer. – Andres Mejia Mar 18 '17 at 10:17
• Thanks for your answer :) the rest of the proof was clear to me and thanks for the extra link :) – Student Mar 18 '17 at 10:21
• No problem! Good luck :) – Andres Mejia Mar 18 '17 at 10:31