Intersection of All Subrings Containing Subset is Smallest
(Redirected from Intersection of Subrings Containing Subset is Smallest)
Jump to navigation
Jump to search
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 subrings of $R$ containing $S$.
Then $L$ is the smallest subring of $R$ containing $S$.
Proof
From Intersection of Subrings is Subring, $L$ is indeed a subring of $R$.
Let $T$ be a subring of $R$ containing $S$.
Let $x, y \in L$.
By the Subring Test, we have that:
\(\ds x - y\) | \(\in\) | \(\ds L\) | ||||||||||||
\(\ds x \circ y\) | \(\in\) | \(\ds L\) |
By Intersection is Largest Subset, it follows that $x, y \in T$.
But $T$ is also a subring of $R$.
So, by the Subring Test again, we have that:
\(\ds x - y\) | \(\in\) | \(\ds T\) | ||||||||||||
\(\ds x \circ y\) | \(\in\) | \(\ds T\) |
So by definition of subset, $L \subseteq T$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$