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Let $A$ be a ring and $X=spec(A)$, the prime spectrum of $A$. Prove that $X$ is quasi-compact.

Definition of quasi compact: each open covering of $X$ has a finite subcovering of $X$.

It is enough to considering the covering in the basis $\{X_f|f\in A\}$.

Let the set of $\{X_{f_i}| f_i\in A , i\in I\}$. I is some index set. It is obvious that $f_i$ with $i\in I$ generates the unit ideal. But why there exists a finite subset $J$ of $I$ such that $f_i$ with $i\in J$ generates the unit ideal?

user26857
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claire
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1 Answers1

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Hint. $\langle f_i\rangle_{i\in I}=\langle1\rangle$ if and only if $1\in\langle f_i\rangle_{i\in I}$. What does an arbitrary element of $\langle f_i\rangle_{i\in I}$ look like and use the latter statement I gave.

Karl Kroningfeld
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  • An arbitrary element in $<f_i>{i\in I}$ looks like that $\sum{i\in I} f_ig_i$ where $g_i\in A$ – claire Sep 01 '14 at 00:28
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    @claire You need to be a little more careful, since you may notice that your description allows for infinite sums in $A$ (something that makes no sense in your random ring $A$). – Karl Kroningfeld Sep 01 '14 at 00:30
  • Infinite sum is not allowed to describe an element in $A$? – claire Sep 01 '14 at 00:34
  • @claire Yep, there is no such thing as an infinite sum. We sometimes abuse notation and write $\sum_{i\in I}f_ig_i$, $g_i\in A$ and all but finitely many of the $g_i$ are zero. – Karl Kroningfeld Sep 01 '14 at 00:36
  • Oh, I do not know that... Thank you for your help, Karl~ – claire Sep 01 '14 at 00:40