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If we have a quadratic number field $K=\mathbb{Q}(\sqrt{d})$ then its discriminant is $$\Delta_K = \begin{cases} d & d\equiv 1 \pmod 4\\ 4d & d \equiv 2,3 \pmod 4 \end{cases}.$$ In particular, $\Delta_K$ in this case is completely determined by $d$.

In a more general setting, say if $F$ is any number field or a function field (some field not $\mathbb{Q}$) and we have a quadratic extension of $F$: $K=F(\sqrt{\alpha})$ for some $\alpha \in F\setminus F^2$ how much about the relative discriminant $\Delta_{K/F}$ is determined by $\alpha?$

That is, given $\alpha$, can one determine something about $\Delta_{K/F}$?

user10039910
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  • It probably depends on your particular field; showing $\Delta_K$ satisfies the above is just a calculation. – Dzoooks Sep 02 '19 at 02:38
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    @reuns the discriminant of an extension of number fields $K/F$ is an ideal in $\mathcal O_F$ and it can be defined in several ways (the equivalence of these different ways not necessarily being obvious). Just google "discriminant ideal" or "relative discriminant". The prime ideal factors of $\Delta_{K/F}$ turn out to be the prime ideals in $\mathcal O_F$ that are ramified in $\mathcal O_K$, so it shares that important property with the "absolute discriminant" (the usual discriminant, with base field $\mathbf Q$, where a positive integer $d$ can be viewed as a principal ideal $d\mathbf Z$). – KCd Sep 02 '19 at 03:18
  • Watch out: the discriminant formula for quadratic fields needs squarefree $d$: $\mathbf Q(\sqrt{18})$ has discriminant 8 since it is $\mathbf Q(\sqrt{2})$ and $2$ is squarefree. In any case, $\alpha$ alone doesn't determine anything: you need to know the base field $F$ (consider $\mathbf Q(\sqrt{2})$ and $(\mathbf Q(\sqrt{5}))(\sqrt{2})$). Since $\alpha$ and $F$ completely determine $K = F(\sqrt{\alpha})$, the answer to your question is in principle "everything". Maybe you mean to ask "can we compute $\Delta_{F(\sqrt{\alpha})/F}$ from the prime ideal factorization of $\alpha\mathcal O_F$?" – KCd Sep 02 '19 at 03:24
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    Some quadratic extensions of number fields are unratified at all prime ideals, so the discriminant ideal is the unit ideal (see what I wrote above about its prime ideal factors). This happens when $F = \mathbf Q(\sqrt{-5})$ and $K = F(i)$ ($K$ is the Hilbert class field of $F$, where $F$ has class number $2$) and also when $F = \mathbf Q(\sqrt{3})$ and $K = F(i) = F(\sqrt{-3}) = F(\zeta_3)$ ($K$ is the narrow Hilbert class field of $F$: $F$ has class number $1$ but narrow class number $2$ and $K/F$ is unratified at all prime ideals but ramifies at $\infty$). – KCd Sep 02 '19 at 03:35
  • For elementary considerations, a number field $F,\alpha \in O_F,\not \in F^2,K=F(\sqrt{\alpha})$, $Disc(u) =\prod_{i \ne j}(\sigma_i(u)-\sigma_j(u))= -(u-\sigma(u))^2$, the discriminant ideal $\Delta=(Disc(O_K))$, looking at the ideal $I=(4 (\alpha F^2 \cap O_F))=(Disc( \sqrt{\alpha} F\cap O_K))$ tells the ramification away from the primes dividing $2$. For those prime ideals $P | I$ whenever $v_P(\alpha)$ is odd, $P^2 \nmid I$ and the same holds for $\Delta$. If $P^2=(b)$ is principal but $P$ is not principal then $F(\sqrt{b})$ is unramified away from $2$. – reuns Sep 02 '19 at 05:21

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