I am solving the following problem from AM commutative algebra. I am solving commutative algebra in my spare time to get good grip in it.
Prove that the Zariski topology is Quasi-compact. Here is my proof, I just want to make sure that I have everything correct.
We want to prove that X is quasi-compact. Suppose we have a covering of X by basic open sets $\{ X_{f_i} \}_{i \in \mathcal{I}}$, i.e we have that
$$X = \bigcup_{i \in \mathcal{I}}X_{f_i} \iff \bigcap_{i \in \mathcal{I}} V\big( (f_i) \big) = \emptyset.$$
If $\{(f_i)\}_{i \in \mathcal{I}}$ doesn't generate 1, then there exists a prime J such that J contains the ideal generated by all the $f_i$. Thus, $J \supseteq (f_i)$ for all $i \in \mathcal{I}$, which is a contradiction.
Thus, the ideal generated by all the $f_i$ agree with the whole ring, so in particular we must have:
$$1 = \Sigma_{i \in \mathcal{J}} g_i f_i \text{ } (g_i \in A)$$
Where $\mathcal{J}$ is some finite subset of I. To finish the proof we should show that:
$\{X_{f_i}\}_{i \in \mathcal{J}}$ cover X. That is we want to prove that
$$\bigcup_{i \in \mathcal{J}} X_{f_i} = X$$
Since $\mathcal{J}$ is a finite subset we can rewrite the above equality as, that is we want to show that:
$$\bigcup_{i = 1}^{i = m} X_{f_i} = X \iff \emptyset = \bigcap_{i = 1}^{i = m}V\big( (f_i) \big) = V(\bigcup_{i = 1}^{i = m} f_i) = V\big( (f_1,\ldots, f_m)\big)$$
Where the last equality is from exercise 1.15 part a.
Since we have that
$$1 = \Sigma_{i \in \mathcal{J}} g_i f_i \text{ } (g_i \in A)$$
This means that
$$(f_1,\ldots,f_m) = (1)$$
Thus the equality below is satisfied:
$$V\big( (f_1,\ldots, f_m)\big) = \emptyset = \bigcap_{i = 1}^{i = m}V\big( (f_i) \big)$$
And from the equivalence earlier we get our result.
Best Answer
Your proof looks good, although I have a few minor stylistic comments.