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?



Primary ideals in valuation rings
There are certainly valuation rings in which each non-maximal prime ideal has no primary ideal except itself. Any Noetherian valuation ring has this property for trivial reasons. Also, this is true for any valuation ring $R$ with value group of the form $G\_1 \oplus \cdots \oplus G\_n$ under lexicographic order, where each $G\_i$ is a *non-discrete* (i.e. dense) subgroup of $\mathbb{R}$.


To see this, note that we can parametrize the entire prime spectrum of $R$ from looking at the value group. Indeed, the $\operatorname{Spec} R$ looks like $0 \subset \mathfrak p\_1 \subset \mathfrak{p}\_2 \subset \cdots \subset \mathfrak{p}\_n$. To use the $G\_i$, note that the nonzero elements of $\mathfrak p\_1$ are the ones where the first coordinate of the value is positive; the elements of $\mathfrak p\_2 \setminus \mathfrak p\_1$ are those whose values' first coordinate is zero and second coordinate positive, those of $\mathfrak p\_3 \setminus \mathfrak p\_2$ have value with the first *two* coordinates $0$ and *third* coordinate positive, and so forth. The fact that $\mathfrak p\_i^2= \mathfrak p\_i$ comes directly from the fact that $G\_i$ is dense in $\mathbb R$.


However, you are right, in that any non-Noetherian valuation domain with finite rank and *discrete* value group will fail this property. That is, every nonzero non-maximal prime has nontrivial primary ideals. 


