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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.

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    $\begingroup$ $9+4\sqrt{5}$ is not the fundamental unit of $K=\mathbb{Q}(\sqrt{5})$ - in fact it is the square of $2+\sqrt{5}\in K$, which in turn is the cube of a fundamental unit. Could you please double-check whether you're asking what you really intended to ask? $\endgroup$
    – GNiklasch
    Commented May 27, 2019 at 11:50
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    $\begingroup$ Sorry I meant its totally positive fundamental unit. $\endgroup$
    – A. Bailleul
    Commented May 27, 2019 at 15:25
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    $\begingroup$ Sorry, I think still not right. I would have thought that $\frac{3+\sqrt5}2$ was the totally positive fundamental unit. $\endgroup$
    – Lubin
    Commented May 27, 2019 at 16:08
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    $\begingroup$ The fundamental totally positive unit is $\frac{3+\sqrt{5}}{2}$. $9+4\sqrt{5}$ is its cube, and it's the generator of the group of totally positive units in $\mathbb Z[\sqrt{5}]$, which is a proper subring of the ring of integers of $K$. $\endgroup$
    – Wojowu
    Commented May 27, 2019 at 16:10
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    $\begingroup$ Thanks for the correction, I really didn't pay attention to the exact value of this fundamental unit, which is a bit irrelevant to my problem. $\endgroup$
    – A. Bailleul
    Commented May 27, 2019 at 18:35

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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.

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    $\begingroup$ Thank you very much. I believe the same argument shows that $d_n \leq C4^{n-1}$, where $C$ is $[L_1 : \mathbb Q]$. I think this may be close to an equality, as adjoining a square root of $\sqrt[5 \times 2^k]{\varepsilon}$ certainely won't provide a square root of $\zeta_{5 \times 2^k}$, but I hope I'm wrong because I would have loved to have something like $d_n = o(4^n)$. $\endgroup$
    – A. Bailleul
    Commented May 29, 2019 at 8:31
  • $\begingroup$ I admit I haven't thought carefully about what happens to the degree beyond $L_4$ - indeed I expect that it'll remain as large as possible, in other words that adjoining another $2$-power root of unity will never turn one of our iterated roots of $\varepsilon$ into a square when it wasn't one before. I don't have a quick argument, though. One would need to study the Galois action carefully, or to show explicitly that the $\sqrt{\eta}$ steps will necessarily turn some previously prime ideal above $2$ into a square of a prime ideal. $\endgroup$
    – GNiklasch
    Commented May 29, 2019 at 9:32

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