Seeking counterexample to the modulo $p$ reduction method in finding Galois group. Consider the following result of Dedekind:

For any polynomial $f\left(x\right) \in \mathbb{Z}\left[x\right]$ and any prime $q$ ${\textbf{ not dividing}}$ the discriminant of $f\left(x\right)$, if $f\left(x\right)$ factors modulo $q$ into a product of irreducible polynomials with degrees $d_1, \ldots, d_s$, then the Galois group $\text{Gal}\left(f\left(x\right)/\mathbb{Q}\right)$ contains a permutation with cycle structure $\left(d_1, \ldots, d_s\right)$.

I'm trying to find a counter-example to this result for a prime $q$ ${\textbf{dividing}}$ the discriminant of $f\left(x\right)$. More precisely, for example, I want to find $f\left(x\right)\in\mathbb{Z}\left[x\right]$ and a prime integer $q$ such that they have the following properties:


*

*$f\left(x\right)$ is an irreducible quatric polynomial, 

*$q$ divide the the discriminant of $f\left(x\right)$,

*the Galois group ${\text{Gal}}\left(f\left(x\right)/\mathbb{Q}\right)$ is $V_{4}$ (klein four group, which contains no four cycle), and

*$f\left(x\right)$ is irreducible in $\mathbb{F}_{q}\left[x\right]$, where $\mathbb{F}_{q}$ is the finite field with $q$ elements.


I have tried lots of examples but in vain. Any suggestion? Thanks in advance.
 A: One reason you may have had trouble: what you ask for is not possible unless you allow the degree to drop mod $q$.
The issue is that (2) and (4) are incompatible. For a fixed degree $n$, the discriminant is some expression in terms of the coefficients which in particular reduces mod $p$. So when the degree doesn't drop, the discriminant of $\bar f$ (reduction mod $q$) is the discriminant of $f$ reduced mod $q$. But if $\bar f$ is irreducible, then it is separable (working over a perfect field) and so its discriminant is nonzero (mod $q$), which means that $q$ did not divide the discriminant of $f$ in the first place.
Here is an example if you are ok with the degree dropping: $f(x) = 25x^4 - 48x^2 + 4$ with $q=5$ (it is a general fact that $ax^4 + bx^2 + c$ has Galois group $V_4$ if and only if it is irreducible and $c/a$ is a square, so I just ran a couple through Sage until I found one that worked). I assume there are more careful ways of finding an example, especially if you keep the above issue with the degree in mind.
More generally, if $f(x) \in \mathbb Z [x]$ is monic and irreducible modulo any $q$, then its Galois group has an $n$-cycle. Or even more generally, if $f(x)$ is $q$-integral for any prime $q$ and irreducible mod $q$, then its galois group has an $n$-cycle. This can be proven by looking at the decomposition group of a prime over $q$.
