$D = \mathbb{Z} + \mathbb{Z} \sqrt{d}$ where $d \equiv 1 \bmod 4$, show that the principal ideal $2D$ is ramified. $D = \mathbb{Z} +  \mathbb{Z} \sqrt{d}$  where $d \equiv 1 \bmod 4$. Show that there is a unique prime ideal of $D$ that contains the principal ideal $2D$.
I would like help showing $2D$ is ramified, that is the first step I believe to solving this problem. I would also appreciate any links to a proof that $p$ is ramified in the ring of integers if $d \equiv 2, 3 \bmod 4$
For reference, my definition of ramified is $\langle p \rangle = \mathfrak P^2$ for some prime ideal $\mathfrak P$ of the ring of integers.
 A: You actually have to look at it modulo 8. First, though, we agree that if $d \equiv 1 \bmod 4$, then $D$ includes numbers of the form $$\frac{a}{2} + \frac{b \sqrt d}{2},$$ with $a, b \in \mathbb Z$, right? Then that number is an algebraic integer with polynomial $$x^2 - ax + \left(\frac{a^2}{4} + \frac{db^2}{4}\right).$$
If $d \equiv 1 \bmod 8$, then $$\frac{1}{4} + \frac{d}{4}$$ is even. So if $$\frac{1}{2} + \frac{\sqrt d}{2} \in \mathfrak P,$$ so is $$\frac{1}{2} - \frac{\sqrt d}{2}.$$ Which means ramification.
But if instead $d \equiv 5 \bmod 8$, then $$\frac{1}{4} + \frac{d}{4}$$ is odd. Which means that $$1 + \sqrt d = 2\left(\frac{1}{2} + \frac{\sqrt d}{2}\right),$$ and so we can't rule out that those two multiplicands could be prime.
A couple of examples will make this a heck of a lot clearer. Note that $$N\left(\frac{1}{2} + \frac{\sqrt{41}}{2}\right) = -10,$$ and therefore that number must be contained in an ideal of norm 2. Indeed $$\left(\frac{1}{2} + \frac{\sqrt{41}}{2}\right) \in \left\langle \frac{7}{2} + \frac{\sqrt{41}}{2} \right\rangle.$$
Now consider $$N\left(\frac{1}{2} + \frac{\sqrt{-35}}{2}\right) = 9.$$ This domain is not a unique factorization domain, and yet we see that $$\frac{a}{2} + \frac{b \sqrt{-35}}{2},$$ with $a$ and $b$ both odd, must always have odd norm. Since norm 2 is impossible, it's a short walk to proving that $\langle 2 \rangle$ is a prime, not ramified, ideal.
Hope this helps.

The case of $\langle 2 \rangle$ when $d$ is even is much too easy that I'd be some kind of jerk to give a hint instead of the outright answer: $\langle 2 \rangle = \langle 2, \sqrt d \rangle^2$. As for $d \equiv 3 \bmod 4$, look at the norm of $1 + \sqrt d$, and whether that number is in the same ideal as $1 - \sqrt d$.
