Cutting down an Ideal Let $A \subset B$  two commutative rings and $I \subset A$ an Ideal. I shall write  $I B$ for the Ideal in $B$ generated by $I$.  Now, is it always true that $I = A \cap I B$ ?
We used something like this in our lecture (however we did not prove it and used it only for Dedekind-Domains) and I was wondering if the statement holds for every commutative rings. The inclusion "$\subset$" is trivial, but I don't  know about "$\supset$". Any help would be much appreciated.
 A: $$A=\Bbb Z\subset\Bbb R=B\;,\;\;I:=3\Bbb Z\implies IB=I\Bbb R=\Bbb R\;,\;\;\text{so...}$$
A: DonAntonio gave a good counterexample, I just thought I'd try to make the problem of which rings have this property a bit easier by reducing it to a local question.
The inclusion map $I\to A\cap IB$ is surjective if and only if for all primes $\newcommand\pp{\mathfrak{p}}\pp\in\newcommand\Spec{\operatorname{Spec}}\Spec A$ the localized map $I_\pp\to (A\cap IB)_\pp=A_\pp\cap (IB)_\pp=A_\pp\cap I_\pp B_\pp$ is surjective. The first equality is proved here, and the second is true, since if $I$ is an ideal in a commutative ring, $A$, $S$ is a multiplicative set, and $M$ is an $A$ module, it's fairly clear that $S^{-1}IM \subseteq (S^{-1}I)(S^{-1}M)$, and conversely if $\sum_i \frac{a_i}{s_i}\frac{m_i}{t_i} \in (S^{-1}I)(S^{-1}M)$, multiplying by $\prod_i s_it_i$ gives an element of $IM$, so the original element of $(S^{-1}I)(S^{-1}M)$ must have been an element of $S^{-1}IM$.
Hence to show that $I=A\cap IB$ for all ideals $I$ in $A$ it is sufficient to prove the property locally. I.e., that for all points $\pp$ of $\Spec A$ and ideals $I_\pp\subset A_\pp$ we have $I_\pp=A_\pp\cap I_\pp B_\pp$.
In fact it suffices to just prove this for the maximal ideals. See page 7 here or any standard text on commutative algebra.
It's not too hard to show that this property holds for the maximal ideals in a Dedekind domain, (with $B$ integral over $A$ ofc) since $A_\pp$ is a DVR (when $\pp$ is maximal). Thus we have that the ideals of $A_\pp$ are $(\pi^n)$ for $\pi$ a uniformizer for $A_\pp$, and if $\pi^{n-1}\in(\pi^n)B_\pp$, we would have $\pi^{n-1}=\pi^n b$ for some $b\in B_\pp$, or $1=\pi b$, but $\frac{1}{\pi}$ cannot possibly be integral over $A_\pp$, since then $\frac{1}{\pi^m}+\frac{a_{m-1}}{\pi^{m-1}}+\cdots+a_0=0$, and multiplying by $\pi^m$, we have $\pi \mid 1$ in $A_\pp$, which is a contradiction. Hence $(\pi^n)=A_\pp\cap (\pi^n)B_\pp$. Note that we used that if $B$ is integral over $A$, then $B_\pp$ is integral over $A_\pp$. See pages 220-221 of Pete L. Clark's notes.
I'm not too sure personally which rings are the ones with this property, but hopefully this is helpful.
