Let $R$ be an integral domain, and let $K$ be its field of fractions. Let $K\subseteq L$ a field extension. Is it true that, if $x\in L$ is integral over $R$, all the coefficients of the minimal polynomial of $x$ over $K$ are integral over $R$?
Since one of the hypothesis regards minimality, I thought that it should be proved by absurd starting with a polynomial, and not using the theorems that link finiteness and integrality.
Take the minimal polynomial $p(X)\in K[X]$ of $x$ (over $K$); I should prove that if one of its coefficients is not integral over $R$, there is a polynomial of lesser degree that vanishes in $x$. In general, I could prove that if a polynomial $f(X)\in K[X]$, such that $f(x)=0$, has a coefficient not integral over $R$, then exist a $g(X)\in K[X]$ (of lesser degree than $f(X)$) such that $g(x)=0$. Here I'm stuck; is this really the most natural way of solving the exercise, or am I complicating things? Thanks for any hint
What you really want to use with the minimal polynomial $m_x(t)$ is the fact that for any polynomial $p(t)$ in $K[t]$ vanishing on $x$, the minimal polynomial will divide $p(t)$: in particular, if $p(t)\in R[t]$ is a monic polynomial vanishing on $x$, then via the inclusion $R[t]\hookrightarrow K[t]$ we see that $m_x(t)$ divides $p(t)$. In particular, every root of $m_x$ satisfies $p(t)$, so every root of $m_x$ is integral. Now from Vieta's formulas and the fact that sums and products of integral elements are again integral, it is direct to conclude that all the coefficients of $m_x(t)$ are integral over $R$.
This trick is occasionally useful in algebraic geometry, especially when $R$ is integrally closed in $K$ which implies $m_x(t)\in R[t]$.