Units in complex quadratic fields $\mathbb Q(\sqrt D)$, with $D \equiv 1 \pmod 4$

Question

I'm reading Algebraic Number Theory (2nd edition) by Richard Mollin, and I have some questions regarding the statement and proof of Theorem 1.29 (page 47):

Theorem 1.29. If $F = \mathbb Q(\sqrt D)$ is a complex quadratic field, then $$\mathfrak U_F = \mathfrak U_{\mathfrak D_F} = \begin{cases} \langle \frac{1+\sqrt{-3}}{2} \rangle & \text{if } D = -3, \\ \langle \sqrt{-1} \rangle & \text{if } D = -1, \\ \langle -1 \rangle & \text{otherwise}. \end{cases} $$

Here $\mathfrak D_F$ is the ring of integers of $F$; and $\mathfrak U_F$ and $\mathfrak U_{\mathfrak D_F}$ are the corresponding multiplicative groups of invertible numbers in $F$ and $\mathfrak D_F$, respectively.

1. First of all, isn't the first part of this statement simply wrong? I.e., since $F$ is a field, $\mathfrak U_F$ is simply all the non-zero elements of $F$. But $\mathfrak D_F \subset F$ is a ring, and certainly contains non-invertible elements no? Hence $\mathfrak U_F \neq \mathfrak U_{\mathfrak D_F}$?

2. When considering the case $D \equiv 1 \pmod 4$, Mollin writes the following:

   Proof. By Theorem 1.28 we may write $u = a + b\sqrt D\in \mathfrak U_{\mathfrak D_F}$, with $\sigma a, \sigma b \in \mathbb Z$, where $\sigma$ is defined as in the proof of Theorem 1.28.

   $\dots$

   Now we assume that $D \equiv 1 \pmod 4$, so $a^2 - Db^2 = 4$ for $a,b \in \mathbb Z$. $\dots$

   Since $ u \in \mathfrak U_{\mathfrak D_F}$ is invertible iff $N(u) = \pm 1$ (where $N$ is the norm), this would lead to the equation $N(a + b\sqrt D) = (a + b\sqrt D)(a - b\sqrt D) = a^2 - Db^2 = 1$ (since we have a complex quadratic field, the norm is always positive). So where is the 4 coming from in the above statement?

   I know that for $D \equiv 1 \pmod 4$, we have $\mathfrak D_F = \mathbb Z\left[\frac{1 + \sqrt D}{2} \right]$, so I have sneaky feeling that the equation $a^2 - Db^2 = 4$ could be derived in some trivial manner from this fact and using the multiplicative property of the norm. But I am not seeing it.

Answer 1

1. Based on the usage, it seems that Mollin uses $\mathfrak{U}_F$ as a shorthand for $\mathfrak{U}_{\mathfrak{D}_F}$ (I couldn't find anywhere in the book where he defines his notation). Otherwise you are correct that the units in $F$ consists of all nonzero elements, while the units of the ring of integers is much smaller (in fact finite in the case of interest).

2. Note that every element of $\mathbf{Z}\left[\frac{1+ \sqrt{D}}{2}\right]$ can be written as $\frac{a+b\sqrt{D}}{2}$ for some $a,b \in \mathbf{Z}$ with $a \equiv b \pmod 2$. Hence to find units in $\mathfrak{D}_F$, we are actually looking for solutions to $$N\left(\frac{a + b \sqrt{-D}}{2}\right) = 1$$ and satisfying the same parity condition above. Computing the norm, this means finding solutions to $$\frac{a^2 - Db^2}{4} = 1,$$ noting that the parity condition is unnecessary because $a^2 - Db^2$ would be odd (and hence indivisible by 4) if $a$ and $b$ had opposite parities.