Let $A$ be a ${\bf valuation}$ ${\bf ring}$ in the classical sense: $A$ is a domain with quotient field $K$ and for every non-zero $x\in K$ one has $x\in A$ or $x^{-1}\in A$.

Now ${\bf Bourbaki}$ (Commutative Algebra, Chapter VI, exercise 1 for §4) suggests that if $\mathfrak{p}$ is any non-maximal prime ideal of $A$, then $A$ does not possess any $\mathfrak{p}$-primary ideals other than $\mathfrak{p}$ itself.

But $\mathfrak{p} = \mathfrak{p}A_\mathfrak{p}$ (cf. ${\bf Matsumura}$, Commutative ring theory, Theorem 10.1), which is the maximal ideal of $A_\mathfrak{p}$ (another valuation ring of $K$). Hence $\mathfrak{p}^2$ is $\mathfrak{p}$-primary in $A_\mathfrak{p}$, and it follows that $\mathfrak{p}^2\cap A = \mathfrak{p}^2$ is $\mathfrak{p}\cap A$ - primary in $A$ - that is, $\mathfrak{p}$-primary.

And it is easy to find examples where $\mathfrak{p}^2 \ne \mathfrak{p}$.

What am I missing?

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