Intersection of All Ring Ideals Containing Subset is Smallest

Theorem

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

Let $S \subseteq R$ be a subset of $R$.

Let $L$ be the intersection of the set of all ideals of $R$ containing $S$.

Then $L$ is the smallest ideal of $R$ containing $S$.

Proof

From Intersection of All Subrings Containing Subset is Smallest, $L$ is the smallest subring of $R$ containing $S$.

From Intersection of Ring Ideals is Ideal‎, $L$ is an ideal of $R$.

As $L$ is the smallest subring of $R$ containing $S$, and it is an ideal of $R$, there can be no smaller ideal of $R$ containing $S$ as it would then not be a subring.

So the intersection of the set of all ideals of $R$ containing $S$ is the smallest ideals of $R$ containing $S$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$