Ideal Contains Extension of Contraction
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Theorem
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b \subseteq B$ be an ideal.
Then $\mathfrak b$ contains the extension of its contraction by $f$:
- $\mathfrak b^{ce} \subseteq \mathfrak b$
Proof
\(\ds f \sqbrk {\mathfrak b ^c}\) | \(=\) | \(\ds f \sqbrk {f^{-1} \sqbrk {\mathfrak b} }\) | Definition of Contraction of Ideal | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \mathfrak b\) | Image of Preimage under Mapping |
Since $\mathfrak b^{ce}$ is generated by $f \sqbrk {\mathfrak b ^c}$ according to Definition of Extension of Ideal, it is contained in $\mathfrak b$
$\blacksquare$