I am looking at the theorem that says the following:
Let $R$ be a UFD. Then $k$ is integral over $R$ iff $k$ is algebraic over the fraction field of $R$, $\mathrm{Frac}(R)$ and the minimal monic polynomial in $\mathrm{Frac}(R)[X]$ satisfied by $k$ is also in $R[X]$.
For my example of $F[X]$, let $k= \frac{1}{X}$. The fraction field of $F[X]$ is just all ratios of polynomials in $F[X]$. Now $\frac{1}{X}$ is clearly algebraic over the fraction field of $F[X]$, and the minimal monic polynomial in the fraction field satisfied by $\frac{1}{X}$ is the polynomial $T-\frac{1}{X}$ (since its monic and of degree $1$). However this polynomial is clearly not in $R[X]$.
Could someone clear things up for me? Where am I going wrong? Thanks.