Discriminant of a radical extension of a quadratic number field Let $K=\mathbb Q(\sqrt 5)$ and $\varepsilon = \frac{3 + \sqrt 5}{2}$ its totally positive fundamental unit (i.e. it generates the subgroup of totally positive units). For any $n \geq 3$, let $L_n = K(\zeta_{5 \times 2^n}, \sqrt[5\times 2^n]{\varepsilon})$, where $\zeta_{5 \times 2^n}$ is a primitive $5 \times 2^n$-th root of unity. In other words, $L_n$ is the splitting field of $X^{5 \times 2^n} - \varepsilon$ over $K$.
I would like to compute, or at least upper bound, the absolute discriminant of $L_n$. Thanks in advance.
 A: Write $D_n$ for the absolute discriminant $\left|\mathrm{disc}(L_n|\mathbb{Q})\right|$ of the field of interest and $d_n$ for its degree. Then the root discriminant is bracketed in the interval $$5^{31/20}2^{n-1} \le D_n^{1/d_n} \le 5^{31/20}2^{2(n-1)}.$$
The power of $5$ comes from $L_1 = \mathbb{Q}(\zeta_{10},\root{10}\of\varepsilon) = \mathbb{Q}\bigl(\zeta_5,\root{5}\of{(1+\sqrt{5})/2}\bigr)$ (by Kummer theory or a quick naive explicit computation).
Beyond $L_1$, we're stacking quadratic extensions, adjoining the square root of some algebraic unit $\eta$ on each of the $2(n-1)$ steps beginning with $n=2$. The relative discriminant at each step can be no worse than an ideal dividing the square of the principal ideal $(\sqrt{\eta}-(-\sqrt{\eta})) = (2)$, which implies the upper bound. Adjoining the successive square roots of $2$-power roots of unity first, each contributes exactly that, and thus contributes a factor of $2$ to the root discriminant; we do this $n-1$ times, whence the lower bound.
Running the first few examples through GP/PARI only takes a few minutes (but half a GB of RAM). It turns out that $L_2$ winds up right in the middle of the above interval at $D_2=2^{120}5^{124}$: passing from $\root 20\of\varepsilon$ to $\root 40\of \varepsilon$ only contributes another factor $2^{1/2}$ to the root discriminant. And this looks like the beginning of a trend:
$$D_3 = 2^{840} 5^{496}; D_3^{1/320}=2^{21/8}5^{31/20} = 2^{2+1/2+1/8}5^{31/20},$$
$$D_4 = 2^{4680}5^{2464}; D_4^{1/1280}=2^{117/32}5^{31/20} = 2^{3+1/2+1/8+1/32}5^{31/20}$$
where the first summand in the exponent of $2$ on the right comes from the $2$-power roots of unity and the remaining summands from the iterated square roots of the real unit.
