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$\newcommand{\Spec}{\operatorname{Spec}}$

Let $A$ be an algebra and $\Spec(A)$ the scheme consisting of all prime ideals of $A$.

How to show that $\Spec(A)=\bigcup_{i=1}^n D(g_i)$ implies that $(g_1, \ldots, g_n)=A$? By definition, $D(g_i)=\{p \in \Spec(A): g_i \not\in p\}$. Let $a \in A$. Then $a \in p$ for some $p \in \Spec(A)$. Therefore $a \in p \in D(g_i)$ for some $i$. Hence $g_i \not\in p$. But how could we show that $a \in (g_1, \ldots, g_n)$? Thank you very much.

Michael Hardy
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LJR
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  • It is standard to use \cup in things like $A\cup B$ and $A_1\cup\cdots\cup A_n$ and \bigcup in things like $\bigcup_{i=1}^n A_i$. In a displayed, as opposed to inline, setting, the latter looks like this: $\displaystyle\bigcup_{i=1}^n A_i$. With \cup instead of \bigcup in an inline setting, you see $\cup_{i=1}^n A_i$ instead of $\bigcup_{i=1}^n A_i$. I changed it in the question. ${}\qquad{}$ – Michael Hardy Nov 26 '13 at 17:33
  • @MichaelHardy, thank you very much. – LJR Nov 26 '13 at 17:46

1 Answers1

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Hint: every proper ideal of $A$ is contained in a prime ideal.

Apply this to the ideal $I=\left<g_1, \dots, g_n\right>$. What does the condition that the $D(g_i)'s$ cover $\text{Spec }A$ tell you?

Bruno Joyal
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