Polynomials with Unit Discriminant Let $f=x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0\in \mathbb{C}[x]$ be a monic polynomial with algebraic integer coefficients and $n>1$. Let $K$ be the number field $\mathbb{Q}(a_{n-1},\dots,a_0)$. Suppose that the discriminant of $f$, denoted $D(f)$, is a unit in $K$. In other words, $D(f)\in \mathcal{O}_K^*$, where $\mathcal{O}_K$ is the ring of integers of $K$. Is it possible for such an $f$ to be irreducible over $K$?
When $K=\mathbb{Q}$ this is not possible. I'm used to thinking about things over number fields, but I'm happy to hear about answers in different settings.
 A: Let $K$ be a real quadratic field and $u$ be a fundamental unit of $\mathcal O_K$.  Then $\sqrt{u} \not\in K$, so the field 
$K(\sqrt{u}) = K((1+\sqrt{u})/2)$ is a quadratic extension of $K$ and the minimal polynomial of $(1+\sqrt{u})/2$ over $K$ is 
$$
\left(x - \frac{1+\sqrt{u}}{2}\right)\left(x - \frac{1-\sqrt{u}}{2}\right) = 
x^2 - x + \frac{1-u}{4}, 
$$
so this is irreducible over $K$. 
We want this to have algebraic integer coefficients, so we need $u \equiv 1 \bmod 4\mathcal O_K$.  Assuming we can find such an example, this fits your desired conclusion since the discriminant of $x^2 - x + (1-u)/4$ is $u$.
To find examples of this, write $K = \mathbf Q(\sqrt{d})$ where $d$ is a squarefree integer. Then $\mathcal O_K$ has $\mathbf Z$-basis $\{1,\omega\}$,  where $\omega = \sqrt{d}$ if $d \not\equiv 1 \bmod 4$ and $\omega = (1+\sqrt{d})/2$ if $d \equiv 1 \bmod 4$. Writing $u = a + b\omega$ for integers $a$ and $b$, the constraint $u \equiv 1 \bmod 4\mathcal O_K$ is the same as $a \equiv 1 \bmod 4\mathbf Z$ and $b \equiv 0 \bmod 4\mathbf Z$. Searching through the table of real quadratic fields and their fundamental  units at the back of Borevich and Shafarevich's "Number Theory" (Table 1 on p. 422), three real quadratic fields where the fundamental unit is $1 \bmod 4\mathcal O_K$ are 
$$
\mathbf Q(\sqrt{39}), \ \ u = 25 + 4\sqrt{39},
$$
$$
\mathbf Q(\sqrt{55}), \ \ u = 89 + 12\sqrt{55},
$$
$$
\mathbf Q(\sqrt{66}), \ \ u = 65 + 8\sqrt{66}.
$$
A: The quadratic $x^2 + \sqrt{3} x + 1$ has discriminant $(\sqrt{3})^2  - 4 = -1$. On the other hand, this is certainly irreducible over $\mathbb{Q}(\sqrt{3})$, since it has roots
$$\frac{-\sqrt{3} \pm \sqrt{-1}}{2},$$
and $\sqrt{-1} \not\in \mathbf{Q}(\sqrt{3})$ (it is imaginary, for example).
