$B/I$ and $B/J$ flat $A$-algebras; does $I=J$ hold? 
Let $A\to B$ be a ring homomorphism. Consider $I$ and $J$ ideals of $B$ such that $B/I$ and $B/J$ are flat $A$-algebras. We know furthermore that there exists a non zero-divisor $t\in A$ such that $(B/J)_t = (B/I)_t$. We want to prove $I=J$.

I have a hint: Use that $IB_t=JB_t$ (which should be implied by $(B/J)_t = (B/I)_t$  (?)). But I didn't find this very useful so far.
I've tried to find a useful exact sequence in order to apply flatness. For example: I would like to use $0\to A\to A_t$ to prove that $B/I=B/J$, tensoring with $B/I$ and $B/J$ themselves. But it didn't work since I couldn't find a way to apply 5-lemma as I wished.
Does anyone have suggestions? Am I making any mistakes? Is there a counterexample?
 A: First, the assumption $(B/I)_t = (B/J)_t$ has to be made precise: I guess it's meant to say that the kernels of the projections $B_t \to (B/I)_t$ and $B_t \to (B/J)_t$ are equal, i.e. $IB_t = JB_t$. 
Now, suppose $i\in I$. Considering it as an element of $IB_t=JB_t$, there exist $j\in J$ and $k> 0$ such that $i = j/t^k$, i.e. $t^{k+l} i = t^l j$ in $B$ for some $l>0$. In particular, $t^{k+l} i \in J$. Now, use the flatness of $B/J$ in considering the exact sequence $0\to A\xrightarrow{\cdot t} A\to A/(t)\to 0$ to conclude that you already have $i\in J$. 
Digression: Also, it is instructive to think about the problem geometrically: The morphism $A\to B$ gives rise to a map $Y:=\text{Spec}(B)\to X:=\text{Spec}(A)$, and $I,J$ correspond to closed subschemes $S,T$ of $Y$, while $t\in A$ describes an open, dense subscheme $D$ of $X$. The assumption $IB_t = JB_t$ means that over $D$, $S$ and $T$ agree, and the question is whether this suffices to see that the are already equal over the whole of $X$. 
Intuitively, the problem is that we don't know in general how $S$ and $T$ look like over the zero set of $t$: For example, you might imagine the projection ${\mathbb A}^2 \to {\mathbb A}$ of the plane to the $x$-axis, say, take for $S$ the $x$-axis, for $T$ the union of the $x$-axis and the $y$-axis, and for $t$ the identity. Then $S$ and $T$ agree over the complement of the origin, but they differ in general. Algebraically this example corresponds to taking $B = k[x,y]$, $A = k[t]$, $I=(y)$, $J=(xy)$. 
The flatness of $S$ and $T$ now means vaguely that the shape of $S$ and $T$ does not drastically change when you pass to fibers over the zero set of $t$.
