3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Your proof looks good, although I have a few minor stylistic comments.

  1. You should be explicit in stating that $X = \operatorname{Spec}A$ is an affine scheme at the beginning of the statement, because there certainly exist schemes which are not quasi-compact.
  2. Depending on your intended audience, you might want to quickly state why it is sufficient to suppose your open cover consists of basic opens. You could also be more explicit about how you obtain your first contradiction ($J\supseteq (\{f_i\})$ implies that $J\in\bigcap_{I}V(f_i)$) and about why there necessarily exists a prime ideal $J$ containing the ideal generated by all the $f_i$. Of course, whether you include these details is a matter of style and depends on who the intended audience of this proof is.
  3. At one point you write the expression ``$\bigcup_{i = 1}^m f_i$," and it would probably be more accurate to write $\bigcup_{i = 1}^m\{f_i\}$, or even to say $$\bigcap_{i = 1}^m V(f_i) = V\left((f_1) + \dots + (f_m)\right) = V((f_1,\dots, f_m)),$$ as one does not typically take the union of elements, but rather of sets.
$\endgroup$
  • $\begingroup$ Thanks a lot with your feedback yeah I will add those details. I mainly want to solve the text for myself to get a better grip of commutative algebra, but it is always good to always be careful and include details, then things become more clear in your mind. $\endgroup$ – user111750 Jun 12 '17 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy