This question comes from an example on page 2 of Lang's Introduction to Algebraic and Abelian Functions.
Let $k$ be an algebraically closed field, $x$ a transcendental element over $k$, and $\mathfrak{o}$ a discrete valuation ring in $k(x)$ containing $k$. Without loss of generality, we assume that $x \in \mathfrak{o}$. Lang then says that $\mathfrak{p} \cap k[x] \neq 0$, where (I assume) $\mathfrak{p}$ is the maximal ideal of $\mathfrak{o}$.
I think I am missing something obvious here, but why is $\mathfrak{p} \cap k[x]$ necessarily non-empty? Clearly $\mathfrak{o} \cap k[x]$ is non-empty since $x \in \mathfrak{o} \cap k[x]$, but why can't every element of $\mathfrak{o} \cap k[x]$ be a unit in $\mathfrak{o}$?
Let $\phi$ be a generator of $\mathfrak{p}$, then $\phi=f/g$ for some polynomials $f,g \in k[x]$. Since $x\in\mathfrak{o}$, we have $g\in\mathfrak{o}$. Since $\mathfrak{p}$ is stable under multiplication by elements in $\mathfrak{o}$, we must have $f = g\phi \in \mathfrak{p}$.
Just to clarify what this proof really "means". The fact that $x\in\mathfrak{o}$ says $x$ is a regular function at the point corresponding to the dvr $\mathfrak{o}$. Hence this point can be any "finite" point on the projective line over $k$ with affine coordinate $x$, say $x=\alpha$. Then a polynomial $f$ is contained in $\mathfrak{p}$ iff $\alpha$ is a zero of $f$. In particular it is clear that such $f$ exist.