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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.

$$A=Bbb ZsubsetBbb R=B;,;;I:=3Bbb Zimplies IB=IBbb R=Bbb R;,;;textso...$$

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 $Ito Acap IB$ is surjective if and only if for all primes $newcommandppmathfrakpppinnewcommandSpecoperatornameSpecSpec A$ the localized map $I_ppto (Acap IB)_pp=A_ppcap (IB)_pp=A_ppcap I_pp B_pp$ 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 $S^-1IM subseteq (S^-1I)(S^-1M)$, and conversely if $sum_i fraca_is_ifracm_it_i in (S^-1I)(S^-1M)$, multiplying by $prod_i s_it_i$ gives an element of $IM$, so the original element of $(S^-1I)(S^-1M)$ must have been an element of $S^-1IM$.

Hence to show that $I=Acap IB$ 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 $I_ppsubset A_pp$ we have $I_pp=A_ppcap I_pp B_pp$.

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 $pi^n-1in(pi^n)B_pp$, we would have $pi^n-1=pi^n b$ for some $bin B_pp$, or $1=pi b$, but $frac1pi$ cannot possibly be integral over $A_pp$, since then $frac1pi^m+fraca_m-1pi^m-1+cdots+a_0=0$, and multiplying by $pi^m$, we have $pi mid 1$ in $A_pp$, which is a contradiction. Hence $(pi^n)=A_ppcap (pi^n)B_pp$. 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.