I am reading "Commutative Ring theory" written by Matsumura. I have a question about the proof of Kaplansky's theorem.
In Lemma 1, we construct a well-ordered family $\{F_{\alpha}\}$ such that (i)~(vi) hold.
Why does the well-ordered family $\{F_\alpha\}$ satisfy condition (ii), (i.e.) $\bigcup_{\alpha}F_\alpha= F$?
I appreciate your help.
Notice that in the successor step of the construction, it says to stop if $F=F_\beta$. This is the only way the construction ever stops, so if it does stop, the final $F_\beta$ will be equal to $F$ itself and so trivially $\bigcup_\alpha F_\alpha=F$.
If the construction never stops, then it continues and $F_\alpha$ is defined for all ordinals $\alpha$. Moreover, by construction $F_{\beta+1}$ is always strictly larger than $F_\beta$. But this would then imply that $F$ is a proper class, since it has elements in $F_{\beta+1}\setminus F_\beta$ for every ordinal $\beta$. So the construction must stop (and in fact it must stop at no later than $F_\kappa$ where $\kappa$ is the cardinality of $F$).