Closed affine sets are quasi-compact

Question

I'd like to prove the following (from the beginning of Shafarevich's "Basic Algebraic Geometry"):

Zariski-closed subsets $X \subseteq \Bbb A ^n$ are quasi-compact.

I am able of developing the proof up to close to the final, where I encounter a difficulty.

Let $X \subseteq \bigcup _{i \in I} U_i$ be an open cover. Then

$$\Bbb A^n \setminus X \supseteq \bigcap _{i \in I} (\Bbb A^n \setminus U_i) .$$

Since $\Bbb A^n \setminus U_i = Z(F_{i,1}, \dots, F_{i,j_i})$, it follows that

$$\bigcap _{i \in I} (\Bbb A^n \setminus U_i) = Z(F_{i,j} \mid i \in I, 1 \le j \le j_i) .$$

Since the ideal $I \big( \bigcap _{i \in I} (\Bbb A^n \setminus U_i) \big)$ is finitely generated, it follows that there exist polynomials $G_1, \dots, G_d$ such that

$$Z(F_{i,j} \mid i \in I, 1 \le j \le j_i) = Z(G_1, \dots, G_d) = \bigcap _{k=1} ^d Z(G_k)$$

whence it follows that

$$X \subseteq \bigcup _{k=1} ^d \big( \Bbb A^n \setminus Z(G_k) \big) .$$

The problem is that I do not know how to relate the $G_k$ to the $F_{i,j}$. My finitely many open subsets $\Bbb A^n \setminus Z(G_k)$ have no connection whatsoever with the open subsets $U_i$. I have seen some proofs based upon the concept of "Noetherian topological space", but since this concept is not introduced prior to this problem, I believe that this is supposed to be solved using a different approach.

Answer 1

If $Z_i$ is a collection of closed sets of $\mathbb{A}^n$ whose interesection is empty, you need to show that there is a finite subcollection whose intersection is also nonempty. Let $$I_i = \mathcal I(Z_i) = \{ f \in k[T_1, ... , T_n] : f(x) = 0 \textrm{ for all } x \in Z_i\}$$ be the ideal corresponding to $Z_i$. Since $Z_i$ is closed, you can recover it as

$$\mathcal V(I_i) = Z_i = \{ x \in \mathbb{A}^n : f(x) = 0 \textrm{ for all } f \in I_i\}$$

In general, if $E_i$ are closed sets, then $\mathcal I(\bigcap\limits_i E_i) = \sum\limits_i \mathcal I(E_i) $, that is, the ideal generated by all the ideals $\mathcal I(E_i)$. So

$$k[X_1, ... , X_n] = \mathcal I(\emptyset) = \mathcal I(\bigcap\limits_i Z_i) = \sum\limits_i I_i$$

So there exists a finite subcollection, say $I_1, ... , I_t$, and elements $f_1, ... , f_t \in k[X_1, ... , X_n]$, such that $1 = f_1 + \cdots f_k$. Then $\sum\limits_{i=1}^t I_i = k[X_1, ... , X_n]$, and so

$$\emptyset = \mathcal V(k[X_1, ... , X_n]) = \mathcal V(\sum\limits_{i=1}^t I_i) = \mathcal V(\mathcal I(\bigcap\limits_{i=1}^t Z_i)) = \bigcap\limits_{i=1}^t Z_i $$