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Given a ring $A$ we can make $Spec(A)$ with the Zariski topology. A question in atiyah and macdonald asks you to prove that $Spec(A)$ is quasi-compact. It is clear that if $Spec(A)$ is covered by open sets $U_s, s\in S$, then it is covered by $X_{f_i}, i \in I$ where $X_{f_i}$ are the basic open sets of the Zariski topology. Then it is easy to show if the $X_{f_i}$ cover $Spec(A)$ then they generate the unit ideal. From this point I become very confused. This is because I do not see how the $X_{f_i}$ generating the unit ideal implies some finite subset $J$ of $I$ $X_{f_j}, j\in J$ generates the unit ideal.

edit: well now i understand why there must be a finite subset it comes directly from the definition of sum of ideals. but now i am really confused as to why the definition of sum of ideals is such that almost all of them are 0. i would appreciate some insight as to why this is.

MichaelMoore
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    If ${f_i}{i\in I}$ generates the unit ideal, the unit $1$ is actually generated by these $f_i$. That is $1=\sum{i\in J} a_i f_i$ for a finite subset $J\subset I$. – Roland Sep 26 '15 at 15:57
  • thanks a lot for the link martin, i now understand that the answer to my question lies in the definition of the sum of ideals. i must say that i am curious now why the sum of ideals is defined this way. – MichaelMoore Sep 26 '15 at 16:31

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