When is an integrally closed local domain a valuation ring?
A commutative ring $R$ is called coherent, if every finitely generated ideal $I$ is finitely presented, that is as an $R$-module $I$ is isomorphic to $R^n/J$ for some finitely generated $R$-submodule $J$ of $R^n$.
For two ideals $I,J$ of $R$ one defines the ideal $(I:J):=\{r\in R : rJ\subseteq I\}$.
Now the following is true: the local integrally closed domain $R$ is a valuation domain if and only if $R$ is coherent and there exist $r,s\in R$, $s\not\in rR$ such that the maximal ideal $M$ of $R$ is minimal among the prime ideals containing $(rR:sR)$.
This follows from results obtained by J. Mott and M. Zafrullah some decades ago.
References:
S. Glaz, Commutative coherent rings, Lecture notes in mathematics 1371, 1989. (general theory of coherence)
J. Mott, M. Zafrullah, On Prüfer -v-multiplication domains, Manuscripta Mathematica 35 (1981). (Theorem 3.2 is relevant)
M. Zafrullah, On finite conductor domains, Manuscripta Mathematica 24 (1978). (Theorem 2 is relevant)
Note the following statements.
I. A quasi local domain $(D,M)$ is a valuation domain if and only if $D$ is a Bezout domain (i.e. for every pair $a,b$ in $D,$ the ideal $(a,b)$ is principal or, equivalently, every finitely generated ideal of $D$ is principal).
If $D$ is a valuation domain then as for each pair $a,b$ in $D$ we have $a|b$ or $b|a$, giving $(a,b)=(a)$ or $(a,b)=(b)$, that is $D$ is Bezout.
Conversely, take $a,b$ in $D.$ If either of $a,b$ is zero or a unit $a|b$ or $b|a.$ So, let both $a,b$ be nonzero non units. Since $D$ is Bezout, $% (a,b)=(c)$ for some $c$ in $D.$ Clearly $c|a,b.$ Let $a=a^{\prime }c$ and $% b=b^{\prime }c.$ Substituting, we get $(a^{\prime }c,b^{\prime }c)=(c)$. Canceling $c$ from both sides we get $(a^{\prime },b^{\prime })=D.$ As in a quasi local domain nonzero non units generate a proper ideal, at least one of $a^{\prime },b^{\prime }$ is a unit. So, $a^{\prime }|b^{\prime }$ or $% b^{\prime }|a^{\prime }$ leading to $a^{\prime }c|b^{\prime }c$ or $% b^{\prime }c|a^{\prime }c$ and to $a|b$ or $b|a.$
II. A quasi local domain $(D,M)$ is a valuation domain if and only if $D$ is a Prufer domain (every two generated nonzero ideal is invertible or , equivalently, every finitely generated nonzero ideal is invertible).
Follows from I. once we note that in a quasi local domain each invertible ideal is principal.
Note that P: $D$ is Bezout or P: $D$ is Prufer both are non-trivial in that there are Bezout (resp., Prufer) domains that are not valuation domains. So perhaps that would suffice as an answer.
Now the above two results do not require the domain $(D,M)$ to be integrally closed and you are asking for a property P such that $(D,M)$ is a valuation domain. Here is the exact property P: Every finitely generated nonzero ideal of $D$ is a $v$-ideal (i.e. a divisorial ideal). So we have the statement.
III. An integrally closed quasi local domain $(D,M)$ is a valuation domain if and only if every nonzero finitely generated ideal of $D$ is a $v$-ideal.
For the proof look up Theorem 8, on pages 1710-1711, of an old paper of mine: [Z] The $v$-operation and intersections of quotient rings of integral domains, Comm. Algebra, 13 (8) (1985) 1699-1712.
The cited theorem says: An integrally closed fgv domain is a Prufer domain.
Now fgv domain is a fancy name for a domain whose nonzero finitely generated ideals are divisorial. Indeed as every invertible ideal is divisorial the converse of Theorem 8 of [Z] is also true. You can also get information on divisorial ideals (i.e. $v$-ideals) from [Z] or sources mentioned there.
Proof of III. Let $(D,M)$ be an integrally closed domain such that every nonzero finitely generated ideal of $D$ is divisorial. Then $D$ is a Prufer domain by Theorem 8 of [Z] and by II. above $D$ is a valuation domain. Conversely let $(D,M)$ be a valuation domain then every nonzero finitely generated ideal of $D$ is principal and so divisorial.
Note:If you would rather follow Hagen' suggestion here's how to go about it. Note that a nonzero ideal $A$ is a $t$-ideal if for each finitely generated nonzero ideal $I$ contained in $A$ the $v$-image (I sub v) is also contained in $A$. So Hagen wants you to use P: $(D,M)$ is such that $D$ satisfies FC and $M$ is a $t$-ideal. An easy way to see what Hagen means is look up Lemma 5 of the finite conductor domains paper mentioned above by him.