example for $(\mathfrak{a}:\mathfrak{b})^e\subsetneq (\mathfrak{a}^e:\mathfrak{b}^e)$? showing that the inclusion holds is an easy exercise, but can someone give an example where the inclusion $(\mathfrak{a}:\mathfrak{b})^e\subseteq (\mathfrak{a}^e:\mathfrak{b}^e)$ is strict?
 A: I assume that $R$ is a commutative ring, $\mathfrak{a},\mathfrak{b}$ are ideals of $R$, $(\mathfrak{a}:\mathfrak{b})=\{x \in R : x\mathfrak{b} \subseteq \mathfrak{a}\}$ is the ideal quotient, we have a homomorphism $f : R \to S$ of commutative rings, and $\mathfrak{a}^e = f(\mathfrak{a})S$ is the extended ideal. Please explain your notation text time so that we don't have to guess.
Instead of showing you some example directly, let me explain how to come up with one systematically (in fact, as I write these lines, I can't write down an example, yet).
We want $(\mathfrak{a}:\mathfrak{b})^e \neq (\mathfrak{a}^e:\mathfrak{b}^e)$. Notice that for $\mathfrak{a}=0$ this becomes: $\mathrm{Ann}(\mathfrak{b})^e \neq \mathrm{Ann}(\mathfrak{b}^e)$. It suffices to find an ideal $\mathfrak{b}$ with $\mathrm{Ann}(\mathfrak{b}) = 0$, but $\mathrm{Ann}(\mathfrak{b}^e) \neq 0$. It may happen that $\mathfrak{b}^e = 0$, for example when $f$ is $R \twoheadrightarrow R/\mathfrak{b}$. Then $\mathrm{Ann}(\mathfrak{b}^e)=R/\mathfrak{b} \neq 0$ when $\mathfrak{b}$ is proper, and we therefore only have to find some proper ideal of a ring with trivial annihilator. Of course such ideals exist. In fact, we can take any integral domain $R$ which is not a field, and then take $\mathfrak{b}=(a)$ for any $a \in R \setminus (\{0\} \cup R^*)$.
