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I want to prove that any affine scheme $X = \operatorname{Spec} A$ is compact.

$\bigcup D(f_i)$ is an open cover of $X$ if and only if the sum ideal $\sum (f_i)$ contains $1$. That is, $ D(\sum f_i)=\bigcup D(f_i)$.

Why does a possible infinite sum of ideals which includes $1$ mean that there exists a finite subset of such ideal whose ideal sum will include $1$ ?

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If $\mathfrak{a}_i \subseteq A$ is an ideal for each $i \in Y$, $Y$ a set of any size, then $\sum_i \mathfrak{a}_i$ is (by definition) the set of all finite sums of elements of the $\mathfrak{a}_i$. So if $1 \in \sum_i \mathfrak{a}_i$, then there exist $i_1,\ldots,i_n$ and elements $f_{i_k} \in \mathfrak{a}_{i_k}$ such that $$f_{i_1}+\cdots+f_{i_n}=1.$$

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  • $\begingroup$ Oh ok. I didn't know that was the definition. Thanks $\endgroup$
    – user7090
    Commented Jul 26, 2016 at 22:16
  • $\begingroup$ Just FYI, the reason why that's the definition is because $\sum_i \mathfrak{a}_i$ is precisely the ideal generated by $\bigcup_i \mathfrak{a}_i$. $\endgroup$
    – Zach Blumenstein
    Commented Jul 26, 2016 at 22:19
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$1\in\sum(f_i)$ implies that $1=\sum_{j=1}^{j=n}a_{i_j}f_{i_j}$, thus $\bigcup_{j=1}^{j=n}D(f_{i_j})=Spec(A)$.

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  • $\begingroup$ I think your first implication is what I am asking about. Why is there necessarily such a finite sum? $\endgroup$
    – user7090
    Commented Jul 26, 2016 at 22:15
  • $\begingroup$ By definition, an element of $\sum(f_i)$ is a finite sum $\sum a_{i_j}f_{i_j}$ or a sum $\sum a_if_i$ such that only finitely many $a_i$ are not zero and the two assertions are equivalent. Thus $1\in \sum (f_i)$ implies that $1=\sum a_{i_j}f_{i_j}$ $\endgroup$
    – Tsemo Aristide
    Commented Jul 26, 2016 at 22:17

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