13A15-ExtendedIdeal.tex

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Let $f: A \to B$ be a ring map. We can look at the ideal generated by the image of $\mathfrak{a}$, which is called an {\em extended ideal} and is denoted by $\mathfrak{a}^e$. \\
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It is not true in general that if $\mathfrak{a}$ is an ideal in $A$,  the image of $\mathfrak{a}$ under $f$ will be an ideal in $B$. (For example, consider the embedding $f: \mathbb{Z} \to \mathbb{Q}$. The image of the ideal $(2) \subset \mathbb{Z}$ is {\em not} an ideal in $\mathbb{Q}$, since the only ideals in $\mathbb{Q}$ are $\{0\}$ and all of $\mathbb{Q}$.)
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