Ideal is Contained in Contraction of Extension
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Theorem
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in the contraction of its extension by $f$:
- $\mathfrak a \subseteq \mathfrak a^{ec}$
Proof
By definition of extension and generated ideal:
- $f \sqbrk {\mathfrak a} \subseteq \mathfrak a^e$
By Subset of Domain is Subset of Preimage of Image:
- $\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a}}$
By Preimage of Subset is Subset of Preimage:
- $\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a }} \subseteq f^{-1} \sqbrk {\mathfrak a^e}$
$\blacksquare$