Fulton’s exercise 6.11: varieties are quasi-compact

Question

I am dealing with the following exercise from Fulton's Algebraic Curves, 2008.

I have read several posts such as:

Closed affine sets are quasi-compact

Show that affine varieties are quasi-compact.

But these topics do not handle the full generality of the proposition we want to prove. More precisely, here "variety" means an open subset of an irreducible algebraic set, with respect to Zariski topology.

I could not prove any of the items. These were my attempts: suppose $V$ is an irreducible algebraic set and let $X$ be an open subset of $V$.

For $(a)$, I tried this: Suppose we have $(F_n)_{n\in \mathbb{N}}$ a sequence of closed subsets such that $F_{n+1}\subseteq F_n$ for every $n\in \mathbb{N}$. We want to show that there exists $n_0\in \mathbb{N}$ such that $F_{n_0}=F_{n_0+k}$ for every $k\in \mathbb{N}$. Let $C_i$ be the $V$-closure of $F_i$. Then $F_i=C_i\cap X$ and $C_i=V(I_i)$ for ideals $I_i$ such that $I_i=I(C_i)$. Since $F_{i+1}\subseteq F_i$, then $V(I_{i+1})\cap X\subseteq V(I_i)\cap X$. If we were able to show that $V(I_{i+1})\subseteq V(I_i)$, then we would just make use of Hilbert's theorem to say that there exists $n_0\in \mathbb{N}$ such that $I_{n_0}=I_{n_0+k}$ for every $k\in \mathbb{N}$.

I am not succeeding in proving it. How would you proceed?

For $(b)$, I tried to show that every family of closed subsets of $X$ which has an empty intersection, has a finite subfamily which has also an empty intersection. I could not prove it, but I could prove the following:

Let $(F_i)_{i\in I}$ be a family of closed subsets of $X$. Let $C_i$ be the $V$-closure of $F_i$. Then there exists a finite subset $J$ of $I$ such that $\bigcap_{i\in I}C_i=\bigcap_{j\in J}C_j$ (1)

So, I think that we would be done if we were able to show this: if $(F_i)_{i\in I}$ is a family of closed subsets of $X$ with empty intersection, then $(C_i)_{i\in I}$ is a family of closed subsets of $V$ with empty intersection, where $C_i$ is the $V$-closure of $F_i$.

In fact, if we showed this, we would argue as follows: Let $(F_i)_{i\in I}$ be a family of closed subsets of $X$ with empty intersection, and let $C_i$ be the $V$-closure of $F_i$. Then, by (1), there exists a finite subset $J$ of $I$ such that $\bigcap_{j\in J}C_j=\emptyset$. Since $F_i\subseteq C_i$, then $\bigcap_{j\in J}F_j\subseteq \bigcap_{j\in J}C_j$.

Any help?

Answer 1

A noetherian topological space is a space such that one of the two (equivalent) properties below hold:

• If $ Y_1 \supseteq Y_2 \supseteq Y_3 \supseteq \cdots $ is an decending chain of closed sets, then there is some $ n \in N $ such that $ Y_n = Y_{n + 1} = \cdots $. We say the chain terminates.
• If $ \mathscr{A} $ is a non-empty set of closed sets of $ X $, then $ \mathscr{A} $ has a minimal element under inclusion.

(Note: if decending chains of closed sets terminate then so do ascending chains of open sets; and if sets of closed sets have minimal elements then sets of open sets have maximal elements.)

You can show that the Zariski topology is Noetherian, since there is a correspondence between closed sets and algebraic sets that reverses inclusion. (This is part (a) of the question.)

Now, we have the following result: Every subset of a Noetherian space is Noetherian if given the induced topology. I leave this to you - it follows immediately from the definition.

Finally, a Noetherian space (and so by the previous result every subset of a Noetherian space) is compact. A proof of this may be found in this question, but I encourage you to try it yourself. This solves part (b), since a variety is a subset of a Noetherian space.