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Let $L$ be a number field, and let $K_1, \cdots, K_r$ be the maximal subfields of $L$ (that is, $K_j \subset L$ but for each $j$ there does not exist a proper subfield field $M_j$ of $L$ such that $K_j \subset M_j \subset L$). Let $\Delta(L)$ denote the discriminant of $L$ (respectively, $\Delta(K_j)$ is the discriminant of $K_j$). Define the essential discriminant of $L$ to be

$$\displaystyle \Delta^\ast(L) = \frac{\Delta(L)}{\prod_{j=1}^r \Delta(K_j)}.$$

Is $\Delta^\ast(L)$ always an integer? If so, what are the arithmetic/geometric interpretations for it?

Note that when $L$ is a primitive field, meaning it doesn't contain any proper subfields other than $\mathbb{Q}$, then $\Delta^\ast(L) = \Delta(L)$. As a trivial consequence, all cubic and quadratic fields have essential discriminants which are integers. One can also check that for any $D_4$ or $C_4$-quartic field the essential discriminant is always an integer.

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    $\begingroup$ I have no idea offhand about your original question, but what if you multiply by the discriminants of the maximal subfields of the K_j, then divide by the maximal subfields of those, etc. and get some sort of multiplicative inclusion-exclusion-type expression? $\endgroup$ – Liam Baker Nov 29 '17 at 22:43
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It is trivially false for $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$.

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    $\begingroup$ Or indeed any $({\bf Z}/2{\bf Z})^3$ extension. $\endgroup$ – Noam D. Elkies Dec 30 '17 at 1:27
  • $\begingroup$ In general the discriminant of a $({\bf Z}/2{\bf Z})^k$ extension is exactly equal to the product of the discriminants of the $2^k-1$ minimal subfields. (Arguably the easiest proof is to compare the functional equations of the zeta functions involved!) For $k=2$ these are also the maximal subfields, but for $k \geq 3$ the product over the $2^k-1$ maximal subfields is strictly larger. $\endgroup$ – Noam D. Elkies Dec 30 '17 at 1:29

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