D+F 7.3.18

(a) If $I$ and $J$ are ideals of $R$, prove that their intersection $I \cap J$ is also an ideal of $R$.
(b) Prove that the intersection of an arbitrary nonempty collection of ideals is again an ideal (do not assume that the collection is countable)

$\textit{Proof}$ : For (a), we wish to show that $I \cap J$ is a nonempty subring of $R$ which is closed under left and right multiplication. With $I$ and $J$ being ideals of $R$, $0\in{I}$ and $0\in{J}$ so, $0\in{I \cap J}$ so $I \cap J \neq \emptyset$.
By exercise 7.1.4, we know that the intersection of subrings is a subring. We just need to show that $I \cap J$ is closed under left and right multiplication. Let $x\in{I \cap J}$ and $r\in{R}$. We want to show that $rx,xr\in{J} \quad \forall r\in{R}$.
$$x\in{I \cap J} \Longleftrightarrow x\in{I} \wedge x\in{J}$$ $$x\in{I} \Longrightarrow rx,xr\in{I}$$ $$x\in{J} \Longrightarrow rx,xr \in{J}$$ Thus, $xr,rx \in{I \cap J}$.
$\therefore \quad I \cap J$ is an ideal of $R$

For (b) let $$\bigcap_{k\in{\Lambda}}I_k$$ be the intersection of ideals in $R$ where $\Lambda$ is an arbitrary index of ideals.
$$0\in{I_k} \quad \forall k \in{\Lambda} \Longrightarrow 0\in{\bigcap_{k\in{\Lambda}}I_k} \Longrightarrow \bigcap_{k\in{\Lambda}}I_k \neq \emptyset$$ Let $$x\in{\bigcap_{k\in{\Lambda}}I_k}$$ and $r\in{R}$. Then, $rx,xr\in{I_k}$ $\forall k\in{\Lambda}$ since each $I_k$ is an ideal. This implies that $$rx,xr \in{\bigcap_{k\in{\Lambda}}I_k}$$ $$\Longrightarrow \bigcap_{k\in{\Lambda}}I_k$$ is an ideal of $R \quad \blacksquare$