# Primitive element for a number field, and ramification

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 $\theta$?

• The answer to your first question is no (if you mean for $\theta$ to be an algebraic integer). Dedekind discovered the first example of this: if $a^3 - a^2 - 2a - 8 = 0$ and $K = \mathbf Q(a)$ then $2$ splits completely in $K$ (in particular, 2 is unramified in $K$), but $2 \mid [\mathcal O_K:\mathbf Z[b]]$ for all $b \in \mathcal O_K - \mathbf Z$ and therefore $2$ divides the discriminant of the minimal polynomial of $b$, for all such $b$. The term to google is "essential discriminant divisor" (or, amazingly, also "inessential discriminant divisor," a term that never made any sense to me). – KConrad May 22 '16 at 3:20
• The answer to your second question is also no. The (unique) cubic field lying inside the 31st cyclotomic field is $F = \mathbf Q(c)$ where $c^3 + c^2 - 10c - 8 = 0$ (I used PARI to find this). The field $F$ is a Galois extension of $\mathbf Q$ (since it's inside a cyclotomic extension) and once again it turns out that 2 splits completely in $F$ but for every $b \in \mathcal O_F - \mathbf Z$ we have $2 \mid [\mathcal O_F:\mathbf Z[b]]$, so the minimal polynomial of $b$ has discriminant divisible by 2. – KConrad May 22 '16 at 3:33
• The lesson is that you need to give up on the idea of trying to understand everything in a ring of integers of a number field by relying on the crutch of subrings of the form $\mathbf Z[b]$. In fact it was really only after Dedekind discovered the cubic example above that he was compelled to give the full ring of integers of a number field the due respect it deserved, even if it didn't have the form $\mathbf Z[b]$. – KConrad May 22 '16 at 3:37
• @KConrad: Thanks for providing reasons for studying the full ring of integers of the number field. Your comments have answered the question completely. – P Vanchinathan May 22 '16 at 5:29