Minimal polynomial of integral elements

Question

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.

Answer 1

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.

Answer 2

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 $K[X]$.