Let $K=\mathbb Q(\theta)$ be a number field with integral primitive element $\theta$, and let $f(x)$ be the minimal polynomial of $\theta$. Let $p$ be a rational prime. It's well known that if $p$ does not divide the discriminant of $f(x)$, then $p$ is unramified in $K$.

I am interested in knowing whether a sort of converse for the above statement holds. Suppose a rational prime $p$ is unramified in a number field $K$. Is it always possible to find a primitive element $\theta$ for $K$ such that $p$ does not divide the discriminant of the minimal polynomial of $\theta$?

I have a second question in the case where $K$ is a splitting field. Suppose that $K$ is the splitting field of a polynomial $F(x)\in\mathbb Q[x]$, and let $x_1,\ldots, x_n$ be the roots of $F(x)$ in $K$. By the Primitive Element Theorem, there exists a primitive element for $K$ of the form $a_1x_1+\cdots+a_nx_n$, where $a_i\in\mathbb Z$. Suppose that a rational prime $p$ is unramified in $K$. Is it possible to find a primitive element for $K$ having the above form *and* such that $p$ does not divide the discriminant of $\theta$?

fullring of integers of a number field the due respect it deserved, even if it didn't have the form $\mathbf Z[b]$. $\endgroup$ – KConrad May 22 '16 at 3:37