Let $k$ be a number field and let $\mathfrak{c}$ be an ideal of $k$ dividing $2\mathcal{O}_k$.
For an ideal $\mathfrak{a}$ of $k$, write $s(\mathfrak{a})$ for the squarefree part of $\mathfrak{a}$, i.e. $s(\mathfrak{a})$ is the unique squarefree ideal of $k$ such that there is some fractional ideal $I$ with $\mathfrak{a} = I^2s(\mathfrak{a})$. For $\alpha \in k^*$, write $s(\alpha)$ as shorthand for $s((\alpha))$. Let $Q(\mathfrak{c}^2)$ be the group of elements $\alpha \in k^*$ satisfying the following two conditions:
- The ideals $s(\alpha)$ and $\mathfrak{c}$ are coprime.
- There is some $x \in k^*$ such that $\alpha \equiv x^2 \pmod{\mathfrak{c}^2}$.
If $\mathfrak{c}_1 \mid \mathfrak{c}_2 \mid 2\mathcal{O}_k$, then $Q(\mathfrak{c}_2^2)\subseteq Q(\mathfrak{c}_1^2)$. Therefore, for $\alpha \in k^*$, we may define $\mathfrak{c}(\alpha)$ to be the smallest (with respect to containment) ideal of $k$ such that $\alpha \in Q(\mathfrak{c}(\alpha)^2)$. That is, if $\alpha \in Q(\mathfrak{c}^2)$ for any ideal $\mathfrak{c}$, then $\mathfrak{c}\mid \mathfrak{c}(\alpha)$.
Theorem 1.7 of the paper "Counting cyclic quartic extensions of a number field" by Cohen, Diaz y Diaz, and Olivier, says that for $\alpha \in k^*$, the discriminant ideal of $k(\sqrt{\alpha})/k$ is given by $$ \mathfrak{d}(k(\sqrt{\alpha})/k) = \frac{4s(\alpha)}{\mathfrak{c}(\alpha)^2}. $$ The authors say that this is a special case of a result of Hecke, which can be found in Chapter 10 of Cohen's book "Advanced Topics in Computational Number Theory". I do not have access to this book, but I would like to know what the general result is. Could somebody either tell me the general result or give me a freely accessible reference to it?
Section 10.2.3 is called "Hecke's Theorem". In it, Theorem 10.2.9 is about Galois radical extensions of prime degree $L/K$, where $L=K(\sqrt[\ell]{\alpha})$ for prime $\ell$ such that $\zeta_\ell \in K$ and $\alpha$ in $K^\times$ is not an $\ell$th power in $K^\times$ (so $x^\ell - \alpha$ is irreducible in $K[x]$). That makes $L/K$ a Kummer extension of degree $\ell$. Let $\mathfrak d(L/K)$ denote the discriminant ideal of $L/K$. You are dealing with the special case $\ell = 2$, where the hypotheses are automatic: every quadratic extension of a number field is a Kummer extension.
In quadratic extensions of $\mathbf Q$, a prime number that ramifies is totally ramified, and a prime number that is unramified is either inert or splits completely. The same applies to Galois extensions of $K$ of prime degree: a nonzero prime ideal in $\mathcal O_K$ that is ramified in the extension is totally ramified, and a nonzero prime ideal that is unramified in the extension is either inert or splits completely.
In the theorem below, a congruence $x^\ell \equiv \alpha \bmod \mathfrak p^m$ is called solvable in $K$ if there is $x \in K$ such that $v_\mathfrak p(x^\ell - \alpha) \geq m$.
Theorem 10.2.9. With notation as above and $\mathfrak p$ a prime ideal in $\mathcal O_K$, set $$ e(\mathfrak p|\ell) = v_\mathfrak p(\ell) = (\ell - 1)v_\mathfrak p(1-\zeta_\ell), $$ so this is the ramification index of $\mathfrak p$ over $\ell$ if $\mathfrak p \mid \ell$ and it is $0$ otherwise.
$(1)$ Suppose $\ell \nmid v_\mathfrak p(\alpha)$. Then $\mathfrak p$ is totally ramified in $L$ and $$ v_\mathfrak p(\mathfrak d(L/K)) = \ell-1+\ell e(\mathfrak p|\ell), $$ so $v_\mathfrak p(\mathfrak d(L/K)) = \ell-1$ if $\ell \nmid v_\mathfrak p(\alpha)$ and $\mathfrak p \nmid \ell$.
$(2)$ Suppose $\ell \mid v_\mathfrak p(\alpha)$ and $\mathfrak p \nmid \ell$. Then $\mathfrak p$ is unramified in $L$, so $v_\mathfrak p(\mathfrak d(L/K)) = 0$. In addition, $\mathfrak p$ splits completely in $L$ if and only if the congruence $$ x^\ell \equiv \alpha \bmod \mathfrak p^{1+v_\mathfrak p(\alpha)} $$ is solvable in $K$. Otherwise $\mathfrak p$ is inert in $L$.
$(3)$ Suppose $\ell \mid v_\mathfrak p(\alpha)$ and $\mathfrak p \mid \ell$. Set $$ z(\mathfrak p,\ell) = \ell\frac{e(\mathfrak p|\ell)}{\ell-1}+1 = {\ell}v_\mathfrak p(1-\zeta_\ell)+1. $$ Consider the congruences $$ x^\ell \equiv \alpha \bmod \mathfrak p^{k+v_\mathfrak p(\alpha)} $$ for $k \geq 0$. Let $a$ be the largest $k$ for which that congruence is solvable, where we use $a = \infty$ if that congruence is solvable for all $k$. Then
$(a)$ $\mathfrak p$ is split completely in $L$ if and only if $a \geq z(\mathfrak p,\ell)$, in which case we must have $a = \infty$;
$(b)$ $\mathfrak p$ is inert in $L$ if and only if $a = z(\mathfrak p,\ell)-1$;
$(c)$ $\mathfrak p$ is totally ramified in $L$ if and only if $a \leq z(\mathfrak p,\ell)-2$, in which case $a \geq 1$, $\ell \nmid a$, and $$ v_\mathfrak p(\mathfrak d(L/K)) = (\ell-1)(z(\mathfrak p,\ell)-a). $$
What is special about the integer $z(\mathfrak p,\ell)$ is that it detects when $\alpha$ is an $\ell$th power modulo all sufficently high powers of $\mathfrak p$:
Lemma 10.2.10. If $v_\mathfrak p(\alpha) = 0$ and $\mathfrak p\mid \ell$, then the congruence $x^\ell \equiv \alpha \bmod \mathfrak p^k$ is solvable in $K$ for $k - z(\mathfrak p,\ell)$ if and only if it is solvable for all $k \geq z(\mathfrak p,\ell)$.
The last result in this section is the following corollary of the theorem.
Corollary 10.2.12. With notation as above, we have the following.
$(1)$ The prime ideal $\mathfrak p$ is unramified in $L$ if and only if $\ell \mid v_\mathfrak p(\alpha)$ and either $(i)$ $\mathfrak p \nmid \ell$ or $(ii)$ $\mathfrak p \mid \ell$ and the congruence $$ x^\ell \equiv \alpha \bmod \mathfrak p^{z(\mathfrak p,\ell)-1+v_\mathfrak p(\alpha)} $$ is solvable in $K$.
$(2)$ The prime ideal $\mathfrak p$ is ramified in $L$ if and only if either $(i)$ $\ell \nmid v_\mathfrak p(\alpha)$ or $(ii)$ $\ell \mid v_\mathfrak p(\alpha)$ and the congruence $$ x^\ell \equiv \alpha \bmod \mathfrak p^{z(\mathfrak p,\ell)-1+v_\mathfrak p(\alpha)} $$ is not solvable in $K$.
$(3)$ For all $\mathfrak p$ we have $(\ell-1) \mid v_\mathfrak p(\mathfrak d(L/K))$, and $v_\mathfrak p(\mathfrak d(L/K)) = \ell-1$ if and only if $\ell \nmid v_\mathfrak p(\alpha)$ and $\mathfrak p \nmid \ell$.