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 $\mathfrak b$

$\blacksquare$