Minimal polynomial of integral elements 
Let $R$ be an integrally closed domain and let $K$ be its fraction field. Let
  $L\supseteq K$ be a field. If $\alpha\in L$ is integral over $R$ (i.e.
  if it satisfies a monic polynomial in $R[x]$), does its minimal
  polynomial over $K$ lie in $R[x]$?

CONTEXT:
I'm trying to prove that the trace $t_{L/K}$ of integral elements lie in $R$ (provided that the extension $L/K$ is finite). I'm trying to use the fact that the trace of $\alpha$ is an integer multiple of certain coefficient of its minimal polynomial, so this trace lies in $R$ if such coefficient does. 
Since $\alpha$ satisfies an integral relation $p(\alpha)=0$ over $R$, it's minimal polynomial $q$ over $K$ exists and it divides $p$, does it imply that $q\in R[x]$? if so, then I'd be done. 
It's clear that the result is true if $R$ is a UFD. In such a case, it's only a matter of looking at the unique factorization of the polynomial in $R[x]$ and apply Gauss' lemma. However, I don't see a straight proof nor counter example in the general case.
 A: It is true for an integrally closed domain (Matsumura, Theorem 9.2), hence for a UFD, but not in general.
If $R$ is not integrally closed take $L=K$. Then   any element  $q\in K\setminus R$ integral over $R$ furnishes a counterexample: its minimal polynomial over $K$ is $X-q\notin R[X]$, whereas any monic relation of integral dependence for $q$ with coefficients in $R$ will have degree $\geq 2$.
Edit
The OP made the legitimate request that I provide a free source for users not having access to Matsumura.
Nothing beats our friend Pete Clark's splendid lecture notes, freely available here.
The relevant result is Theorem 14.18 b), page 226.
A: HINT:
To prove in the case of the ring integrally closed, show the following:
Assume $A$ commutative with $1$ and in $A[X]$ we have the equality between monic polynomials $f= g\cdot h$, where $f = X^m + a_1 X^{m-1} + \cdots + a_m$, $g = X^p + b_1 X^{p-1} + \cdots + b_p$, $h =X^q + c_1 X^{q-1} + \cdot + c_m$. Then all the coefficients of $g$ and $h$ are integral over the subring $\mathbb{Z}[a_i] $  of  $A$ (that is, the above equality $f= g\cdot h$ implies a series of integral dependences for all the $b_j$, $c_k$).
Let's state the following simple but important lemma:
Let $A$ be a ring and $P$ a monic polynomial in $A[X]$. There exists an extension of ring $A \subset S$ such that the polynomial $P$ splits completely in $S[X]$. The proof is similar to the analogous fact for fields.
Consider now $S$ an extension of $A$ in which $g$ splits completely. Write
$g(X) = (X-\beta_1)\cdot \ldots \cdot (X-\beta_p)$. Since the equality $f= g h$ also holds in $S[X]$ we have $f(\beta_j) = g(\beta_j) \cdot h(\beta_j) = 0$. Therefore, all the $\beta_j$ are integral over $\mathbb{Z}[a_i] $ and therefore, so are the symmetric functions in $\beta_j$'s and so the coefficients of $g$. Since the equality of integral dependence holds in $S$, it will also hold in $A$, since $A\subset S$.
$\bf{Added:}$ The $A$ in this proof is a general commutative, $1$, ring. For the purposes of the proof of the OP statement, one should take $A = K$, $f$ a monic polynomial in $R[X] \subset K[X]$ and $f = g h$ in $K[X]$.
