Let $R$ be a noetherian integral domain, let $P$ be a non zero prime ideal of $R$ and let $x,y \in R$. Assume that for some $n \geq 1$, we have $x \in P^{n-1} \setminus P^n$ and $xy \in P^n$. Does it follow that $y \in P$ ?
This is true if $R$ is a Dedekind domain. Indeed, $x$ has $P$-valuation $n-1$ and $xy$ has $P$-valuation at least $n$, so that $$v_P(y) = v_P(xy)-v_P(x) \geq n-(n-1) = 1$$ and $y \in P$. I tried to do it by induction on $n$, but we don't have $P^{n} \setminus P^{n+1} \subset P^{n-1} \setminus P^{n}$... Otherwise we would have been done.
If $R$ is only an integral domain or if $P$ is not prime, this could fail, I believe. Anyway, any hint would be helpful.
Thank you!
Following the useful hint given by user26857 above, we can try the example 3, p. 51, in chap. 4 of Introduction to commutative algebra (Atiyah, MacDonald) :
$$ R = k[X,Y,Z]/(XY-Z^2),\quad P = (x,z)$$ where $x,y,z$ denote the equivalence classes of $X,Y,Z$ respectively. Notice that $P$ is a prime ideal of the noetherian integral domain $R$ (since $z^2-xy$ is Eisenstein at $p=x$ in $k[x,y][z]$).
We have $$x \in P \setminus P^2,\quad xy = z^2 \in P^2, \quad y \not \in P$$ because $Y \not \in (X,Z,XY-Z^2)$ for degree reasons and $X \not \in (X^2,XZ,Z^2,XY-Z^2)$ also for degree reasons. Therefore the intial claim is wrong in general.