Suppose $A$ is a integral domain, and $L$ is its fraction field.
$A$ is integral closed in $L$, that is any element of $L$ that is integral over $A$ is actually in $A$.
Let $\alpha$ be an element that is integral over $A$, and $\alpha \not\in A$.Let $f_\alpha(x) \in A[x]$ be its minimal polynomial in $A$.
Since $f_\alpha[x]\in L[x]$, $\alpha$ is algebric over $L$ and let $m_\alpha(x) \in L[x]$ be its minimal polynomial in $L$, is it true that $m_\alpha(x) \in A[x]$? Thus by uniqueness $f_\alpha \equiv m_\alpha$.
Let $n$ be the degree of $f_\alpha$ and $m$ be the degree of $m_\alpha$. We have $n\geq m$ since $m_\alpha | f_\alpha $ in $L$. Now if $n>m$, when we clear the denominators of coefficients of $m_\alpha$, we will get a non monic polynomial $cm_\alpha(x)$ in $A[x]$ whose degree is strictly less than the minimal polynomial $f_\alpha$, then $f_\alpha | cm_\alpha$ (which is impossible?) so $m=n$, they must have the same degree. But we don't know if $c$ is a unit in $A$. And do we need the integral closed assumption?
We look at $$A\subset L \subset \overline L$$ where $\overline L$ is the algebraic closure of the field $L$.
$m_\alpha(x)$ factors completely in the algebraic closure $\overline L$ to $\Pi (x-\beta_i)$ where $\beta_i\in \overline L$.
We know $m_\alpha(x) | f_\alpha (x)$ in $L$, thus each $f_\alpha(\beta_i) = 0$, thus $\beta_i$ are all integral over $A$.
Since each coefficient in $m_\alpha(x)$ is a algebraic combination of $\beta_i$'s, we see the coefficient are integral over $A$, at the same time they are also in $L$, thus since $A$ is integral closed in $L$, thus the coefficients of $m_\alpha(x)$ are in $A$.