TITLE: Ring of integers of a degree $5$ extension
QUESTION [5 upvotes]: Consider the polynomial $P(X) = X^5 - X + 1 \in \mathbb{Q}[X]$, and let $x \in \mathbb{C}$ be a root of $P(X)$. Let $K = \mathbb{Q}(x)$. 
How can you prove that the ring of integers $\mathcal{O}_K$ is a principal ideal domain?

REPLY [4 votes]: First, $P(X)$ is irreducible with discriminant $2869=19\cdot 151$, which is squarefree.
Hence the ring of integers of $K$ is $\mathbb{Z}[x]$. Using the Minkowski bound, we see that
every class of ideals for K contains an integral ideal $\frak{a}$ with norm
$$
N(\frak{a})<0.062\cdot \sqrt{2869}<4.
$$
Now we have that there are no prime ideals $\frak{p}$ with norm $2$ or $3$ (because then the residue field would be $\mathbb{F}_2$ or $\mathbb{F}_3$, so that $P(X)$ would have a root modulo $2$ (or $3$), which is not the case). From this we see that the ring of integers is a PID, i.e., the class number of $K$ is $1$. The Galois group of the spiltting field of $P(X)$ is the full symmetric group $S_5$.