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I am currently trying to explicitly compute some of the notions introduced in Neukrich's algebraic number theory in the case $\mathbb{Q}(\sqrt{a})$ as an algebraic number field ($a$ square free).

To this end, it is clear that the ring of integers is in this case is given by $\mathbb{Z} + \alpha \mathbb{Z}$ where $\alpha$ is $\sqrt{a}$ if $a = 2,3$ mod $4$ and $1/2 + 1/2\sqrt{a}$ if $a = 1$ mod $4$.

Now, I want to calculate the conductor. In the first case, this should just be $\mathbb{Z}+\sqrt{a}\mathbb{Z}$.

In the second case, $(2)$ should be contained in the conductor. Is there anything more that can be said about the conductor in this case (e.g. an explicit formula)?

How does this all lead to Proposition 8.5 in Neukirch (i.e. how can we apply Proposition 8.3 in Neukirch/why is (p) relatively prime to the conductor if $(p,2a) = 1$)? Isn't this, i.e. $(p,2a) = 1$ a stronger condition?

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In case (2), where $a \equiv 1 \bmod 4$, the conductor is 2$\mathcal{O}$ where $\mathcal{O}$ is the ring of integers of $\mathbb{Q}(\sqrt{a})$, that is $\mathcal{O} = \{ (j+k\sqrt{a})/2 \mid j \equiv k \bmod 2\}$. To see this, note that an element of the conductor is in $\mathbb{Z}[\sqrt{a}],$ so is of the form $j+k\sqrt{a}$. This is in the conductor iff

$\quad (j+k\sqrt{a})\cdot \frac{m + n\sqrt{a}}{2} \in \mathbb{Z}[\sqrt{a}] \quad \forall m,n \mbox{ with } m \equiv n \bmod 2$

This forces $jm + kna \equiv jm + kn \equiv0 \bmod 2$ (since $a \equiv 1 \bmod 4$). Taking $m=n=1$ this shows that $j \equiv k \bmod 2$. The set of such $j + k\sqrt{a}$ is exactly $2\mathcal{O}$.

To see that $p$ (that is $p\mathcal{O}$) is relatively prime to the conductor $2\mathcal{O}$, note that if $p$ is odd then $1 \in p\mathcal{O} + 2\mathcal{O}$.

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  • $\begingroup$ My original solution was overly complex (maybe that's it didn't get any votes!). I've updated it. $\endgroup$ Commented Oct 11, 2023 at 23:33

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