# In noetherian domains, do we have $x \in P^{n-1} \setminus P^n,xy \in P^n \implies y \in P$?

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!

• I'd try the example of prime ideal whose square is not primary given here. – user26857 Nov 6 '16 at 10:22
• As a side comment, the conclusion is true if $P^n$ is $P$-primary. This holds if e.g. $P$ is maximal, explaining why it works for Dedekind domains. – user26857 Nov 6 '16 at 14:56

$$R = k[X,Y,Z]/(XY-Z^2),\quad P = (x,z)$$ 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.
• $Y \not \in (X,Z,XY-Z^2)$ can be shown just by evaluating at $(0,1,0)$. – Watson Jan 23 '17 at 18:48