primes splitting completely in cyclic extensions Let $K$ be a quadratic number field. It is a well known result that a prime $p$ splits completely in $K$ if and only if $\left(\frac{d_K}{p}\right)=1$.
What about cubic extensions? Can we find similar statements in terms of the discriminant? What about extensions of degree $n$ with cyclic galois group $\mathbb{Z}_n$?
 A: My understanding when $K$ is a quadratic field with discriminant $D$, $p$ splits completely into $K$ if and only if $D$ is a quadratic residue $\pmod p$. This results in a set of primes $p$, determined by some modular congruence. This is because all quadratic fields are cyclic fields (and therefore subfields of cyclotomic fields). 
The constraint for cubic fields can be split into two cases.
The first case is where $K=Q(r)$ is a cyclic cubic field, where $r$ is a root of the polynomial $P$. The discriminant $D$ of $P$ is a perfect square. Let $D=q^2$ where $q$ is a prime congruent to $1 \pmod 3$. The primes $p$, in particular, the set of primes $p$ which will split completely into $K$ are those which are a cubic residue $\pmod q$. Therefore, there is a single modular congruence constraint on these primes $p$.
The second case is a little tricky. If $K = Q(r)$ is a cubic field with $r$ a root of $P$ and the discriminant of $P$ is $D=q$, then the set of primes $p$ which completely split into $K$ are those of the form $p=x^2-qy^2$.
To apply this, we wish to find the set of primes $p$ which split completely into the field $K=Q(r)$ where $r$ is a root of $P=x^3+x^2-2x-1$. So $P$ has discriminant $D=49=7^2$ and therefore $K$ is a cyclic cubic field. Since $q=7$, if $p$ splits completely into $K$, then $p$ is a cubic residue $\pmod 7$. This leads to $p=±1\pmod 7$ which are the set of primes that split completely into the field $K$.
Another example involving the second case, we wish to find the set of primes $p$ which split completely into the field $K=Q(r)$ where $r$ is a root of $P=x^3-x^2-4x-4$. The discriminant of $P$ is $D=-464$, so if $p$ splits completely into $K$, then $p=x^2+464y^2$ for integers $x$ and $y$. 
I hope this helps.
