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The union of a chain of ideals is an ideal

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.19

Solution: In Exercise 2.1.15, we saw that $S \subseteq R$ is an additive subgroup. To show that $S$ is an ideal, it suffices to show that $S$ absorbs $R$ on both sides. To that end, let $r \in R$ and $s \in S$. Since $s \in S$, we have $s \in I_k$ for some $k \in K$. Since $I_k$ is an ideal, $rs, sr \in I_k$. Thus $rs, sr \in S$, and $S$ absorbs $R$.

Thus $\bigcup_{K} I_k \subseteq R$ is an ideal.


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