Intersection of Ring Ideals is Ideal

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Theorem

Let $\struct {R, +, \circ}$ be a ring

Let $\mathbb L$ be a non-empty set of ideals of $R$.


Then the intersection $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an ideal of $R$.


Proof

Let $L = \bigcap \mathbb L$.

From Intersection of Subrings is Subring, we have that $L$ is a subring of $R$.


Let $x \in L$ and $y \in R$.

Then:

$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$

as every element of $\bigcap \mathbb L$, including $T$, is an ideal of $R$.


If $y \in R$, then $x \circ y$ and $y \circ x$ are in every element of $\mathbb L$, and hence in $L$.

So $L$ is an ideal of $R$.

$\blacksquare$


Sources

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