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The set of ideals of a ring is closed under arbitrary intersections


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.18

Solution:

(1) In Exercise 7.1.4, we showed that $I \cap J$ is a subring of $R$. Thus it suffices to show that $I \cap J$ absorbs $R$ on the right and the left. To that end, let $r \in R$ and $x \in I \cap J$. Now $x \in I$, so that $rx, xr \in I$. Likewise, $rx, xr \in J$. Thus $rx, xr \in I \cap J$, and $I \cap J \subseteq R$ is an ideal.

(2) Again, we showed in Exercise 7.1.4, that $\bigcap_A I_k$ is a subring of $R$, thus it suffices to show absorption. Let $r \in R$ and $x \in \bigcap_A I_k$. Now $x \in I_k$ for each $k \in A$, so that $rx, xr \in I_k$ for all $k \in A$. Thus $xr, rx \in \bigcap_A I_k$. Hence $\bigcap_A I_k$ is an ideal of $R$.


Linearity

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