conductor of a quadratic extension of a number field Let $K$ be an arbitrary number field, $\Delta$ an integer of $K$ which is a square mod $4$, let $L=K(\sqrt \Delta)$, and $\mathfrak{f}$ the conductor of $L/K$. I am wondering about the precise relation of $\mathfrak{f}$ and $\Delta$.
For $K=\mathbb Q$ we know that $\mathfrak{f}$ equals $(\Delta_0)$, where $\Delta_0$ is the absolute disriminant of $L$. So one might wonder whether in general $\mathfrak{f}=\Delta/\mathfrak{g}^2$, where $\mathfrak{g}$ is the largest ideal whose square divides $\Delta$ and such that $\Delta$ is a square modulo $4 \mathfrak{g}^2$.
 A: For any finite Galois extension $L/K$ with group $G$, the conductor $\mathcal F (L/K)$ can be defined in terms of Artin conductors $\mathcal F(\chi)$ attached to all the characters $\chi$ of $G$, see e.g. Cassels-Fröhlich, chap. 6 (by Serre), §4.4. The relation to the discriminant $\Delta (L/K)$ is given by the so called « conductor-discriminant formula » of Artin-Hasse. If $L/K$ is abelian, this reduces to $\Delta (L/K)$ = the product over all characters of $G$ of the $\mathcal F(\chi)$, and in the quadratic case, to discriminant = conductor (op. cit. p.160). 
A: ), except for the last statement I think what you said is correct. Your notation is a bit confusing since  $\Delta$ is usually reserved for the discriminant, and as Nguyen wrote, for quadratic extensions the discriminant and the finite part of the conductor are equal as ideals by the conductor-discriminant formula.
So let me write $L = K(\sqrt{\alpha}\,)$. If $\alpha$ is a square modulo $4$, then the primes above $2$ are not ramified. Since the odd part of the discriminants of quadratic extensions are squarefree, you get the finite part of the conductor by dividing $(\alpha)$ by the largest square of the ideal dividing it, just as you have guessed. In fact if $(\alpha) = {\mathfrak d} \cdot {\mathfrak g}^2$, then the primes in ${\mathfrak d}$ are ramified and must
divide the discriminant by the decomposition law in Kummer extensions (Hecke's book is a good reference).
In general, the conductor also contains the ramified infinite primes, so in this case you have to multiply the finite part by all infinite primes at which $\alpha$ is negative.
A: So it remains to compute the discriminant ideal $\Delta$ = $ \Delta (L/K)$ (I keep this usual notation) of the quadratic extension $L = K (\sqrt d)$. Denote by $R, S$ resp. the rings of integers of $K, L$. Rather to compute local Hilbert symbols as you suggest, it seems easier to use Fröhlich’s description of $\Delta$ in terms of  the « module index » as in Cassels-Fröhlich, chap.1, §3.
For any finite extension $L/K$ of number fields, for two finitely generated  $R$-submodules $M, N$ of $L$ which span the $K$-vector space $L$, I recall quickly the definition of the fractional ideal denoted $[M :N]$. Since $R$ is Dedekind, the localized rings $R_P$ at all finite primes $P$ are discrete v.r., which implies in particular that the localized modules $M_P, N_P$ are $R_P$-free of the same rank, hence linearly isomorphic, and the local index $[M :N]_P$ can be defined as a determinant. The global index $[M :N]$ is then the fractional ideal whose localization at any $P$ is $[M :N]_P$ (op. cit., p.10). The discriminant $\Delta (M)$ is by definition the module index $[D(M) : M]$, where $D(M)$ is the dual module of $M$ w.r.t. to the trace form (the reference to $R$ is omitted), and one has the formula $\Delta (N) = \Delta (M) . [M :N]^2$ (propos.4). If $M = R[a]$ for $a$ in $S$, a particular case of the comparison different/discriminant reads : $\Delta (R[a]) =  Norm(g’(a))R$, where $g(X)$ is the minimal polynomial of $a$, $g’(X)$ its derivative, $Norm$ the norm of $L/K$ (propos.6). Applying the formula of propos.4, one gets  $\Delta (L/K) = Norm(g’(a))R . [R[a] :S]^2$. In the quadratic case here, one obtains $\Delta (L/K) = (4d)R . [R[\sqrt d ] :S]^2$. I haven’t had the courage to check this against your formula involving the ideal $\mathfrak g$ .
