Example of Galois, non-monogenic number field Question
I am after an example of $f\in \mathbb{Z}[x]$ with the following properties:


*

*monic, irreducible

*$L := \mathbb{Q}(\alpha) / \mathbb{Q}$ is Galois, where $\alpha$ is a root of $f$.

*The ring of integers of $L$ is not monogenic (not of the form $\mathbb{Z}[\beta]$)


If you are able to, I'd also love:


*The Galois group of $\mathbb{Q}(\alpha) / \mathbb{Q}$ is not abelian,


but this starts to force the degree to be larger.
It would be nice if as many of the properties as possible were provable by hand, without a computer algebra system, but beggars can't be choosers... (I don't actually know of computer algebra packages that can test whether a ring of integers is monogenic)
What I've tried
I've played with some cubic examples (which are nice because they are 'Galois' iff they have a square discriminant). The problem seems to be that I don't have many tools in my belt to see whether a field is monogenic (at least none that seem to fit with the conditions for Galoisness). E.g. Stein gives a short proof in Algebraic Number Theory, a Computational Approach example 4.3.2 that  $x^3 + x^2 − 2x + 8$ defines a non-monogenic number field... but the properties it uses don't seem to be blatantly related to whether or not it is Galois (it turns out not to be).
I've tried to locate a few papers in the literature which give necessary conditions for monogeneity, but haven't stumbled across any that lend themselves to easy construction of smallish examples. 
Any help would be very much appreciated!
 A: Here is one local obstruction to being monogenic: If $2$ splits completely in $F$ and $d = [F:\mathbf{Q}] > 2$, then $F$ is not monogenic. Otherwise $\mathcal{O}_F/2 = (\mathbf{F}_2)^d$ is a quotient of $\mathbf{Z}[x]$, but there are no such maps if $d > 2$ (note that the image of $x$ in any copy of $\mathbf{F}_2$ must satisfy $x^2 = x$, so the largest such quotient is $\mathbf{Z}[x](x^2-x) = \mathbf{F}^2_2$.
But now it is easy to produce examples.
The simplest one: Let $p$ and $q$ be two primes which are $1$ modulo $8$. Then take $F = \mathbf{Q}(\sqrt{p},\sqrt{q})$, and take
$$f(x) = (x^2 - p - q)^2 - 4pq,$$
with roots $\pm \sqrt{p} \pm \sqrt{q}$.
The more complicated field one writes down, the harder it is to write down a minimal polynomial explicitly.
Let $H = \mathbf{Q}(\zeta_{pq})$ with $p,q$ primes that are $1$ modulo $N$. Then $H$ contains a $(\mathbf{Z}/N \mathbf{Z})^2$ extension in which $2$ is unramified and thus a $(\mathbf{Z}/N \mathbf{Z})$ extension in which $2$ is totally split.
It's also easy to write down explicit polynomials whose splitting field $F$ has the property that the Galois group is $S_n$ and $2$ splits completely. However, it's not so easy to write down the minimal polynomial of the splitting field by hand. The degree would be at least $6$, for a start.
If you want to do something completely by hand, start by looking for polynomials of degree $3$ with small height which split completely over $\mathbf{Q}_2$ (which you can check using Hensel's Lemma by hand).
I found $x^3 + 4 x^2 - x + 4$ has this property. The discriminant of the polynomial is $-2^2 \cdot 431$, and the discriminant of the field is $-431$, because it is unramified at $2$. (The argument above shows that this cubic field is non-monogenic, of course!)
Now you want the minimal polynomial of the Galois closure. (If $2$ splits completely in a field $K$, it does so also in the Galois closure $F$.) If $\alpha$ is a root of the cubic above, you could take $\beta = \alpha + \sqrt{-431}$. It's easy to see this will be a primitive element of the Galois closure. The minimal polynomial of $\beta$ over $\mathbf{Q}(\sqrt{-431})$ is obviously
$$(z - \sqrt{-431})^3 + 4 (z -\sqrt{-431})^2 - (z - \sqrt{-431}) + 4.$$
So the minimal polynomial is the produce of this with $\sqrt{-431}$ replaced by $-\sqrt{-431}$. Not super fun to do by hand, but not impossible. The result is
$$z^6 + 8z^5 + 1307z^4 + 6896z^3 + 571108z^2 + 1472288z + 83393344.$$
