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I'm doing exercise 17 in Atiyah-MacDonald, where I am proving that the prime spectrum of a ring is (quasi-)compact. I have shown that an open covering of sets $X_{f_i}$ for $i\in I$ for some index set $I$ implies that $f_i$ for $i\in I$ generates the unit ideal $(1) = A$. In the hint it is claimed that this means there is a finite combination of the generators $f_i$ which is equal to $1$, and I am having trouble proving this. Even on the Columbia stacks, this step is stated without much reason. Maybe I am just overlooking something simple.

Here is my attempt at reasoning. We can create a chain of ideals by adding more and more of the $f_i$. Supposing that no finite combination of the $f_i$ is equal to $1$, we can create an infinite chain of ideals. Now, I imagine Zorn's lemma might come in handy here? But again, I would like to think there is a simpler way to resolve this problem.

Andrea B.
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1 Answers1

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The ideal generated by $S = \{f_i : i \in I\}$ is the intersection of all ideals containing $S$. But you can also show that this intersection is just the set of all finite linear combinations from $S$:

$$ (S) = \bigcap\{J \trianglelefteq A : S \subseteq J\} = \left\{ \sum_{r = 1}^n a_r f_r : a_r \in A, f_r \in S, n \in \mathbf{Z}^+ \right\}. $$

Because that set of finite linear combinations is:

  1. an ideal containing $S$
  2. contained in any ideal which contains $S$
Trevor Gunn
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