Questions regarding statements in Algebraic Number Theory (2nd ed.) by Mollin I'm reading Algebraic Number Theory (2nd edition) by Richard Mollin, and I have several questions regarding various statements made in the book. Unfortunately, the different statements are not really related, which might suggest to split them up into different questions. On the other hand, I guess that most of the questions will have fairly simple answers and so I'm not sure what value they would have individually. Please let me know if I should split them up into separate questions. Anyway, here are the questions.


*

*Regarding prime decomposition it is stated:

Theorem 1.30 If $\mathfrak D_F$ is the ring of integers of a quadratic field $F = \mathbb Q(\sqrt D)$, and $p \in \mathbb Z$ is a prime, then the following holds
  $$(p) = p\mathfrak D_F = \begin{cases} \mathcal {P_1P_2} & \dots\\ & \\ \mathcal P & \text{if } \dots \text{ or } p = 2, D \equiv 5 \pmod 8, \\ & \text{where $\mathcal P$ is a prime $\mathfrak D_F$-ideal with $N(\mathcal P) = p^2$,} \\ & \\
\mathcal P^2 & \dots
\end{cases}  $$

Here $N(\mathcal P)$ denotes the norm of $\mathcal P$, defined as the smallest natural number in $\mathcal P$ (exercise 1.58, p. 54).
What I don't understand is that we would have $N(\mathcal P) = p^2$ in the second case? For instance if we consider the ideal $(2)$, then this would contain the number 2, no? But the theorem seems to claim that the smallest natural number in $(2)$ is 4. What am I missing? 

*Regarding the definition of totally real and complex fields (exercise 2.1, p. 62):

$\dots$ Note that if $\mathbb Q(\alpha_j) \subset \mathbb R$ for all $F$-conjugates of $F$, then $F$ is called a totally real field and if $\mathbb Q(\alpha_j) \subset \mathbb C - \mathbb R$, then $F$ is called  totally complex. 

The definition of a totally real field seems to be in agreement with what I have read elsewhere, but isn't the definition of a totally complex field wrong? I.e., $\mathbb Q(\alpha_j)$ certainly contains some elements in $\mathbb R$, no? It's just that for a totally complex field, it won't be wholly contained in $\mathbb R$. Thus, I would instead have expected something like $\mathbb Q(\alpha_j) \cap (\mathbb C - \mathbb R) \neq \emptyset $.

*In the proof of the same exercise (2.1, p. 62), he writes:

Suppose that $\theta$ is an embedding of $F$ in $\mathbb C$ with $\theta(\alpha) = \beta$. Since
  $$ 0 = m_{\alpha,\mathbb Q}(\alpha) = \sum_{j=0}^{d-1}q_j\alpha^j \text{ with } q_j \in \mathbb Q $$
  then
  $$  0 = \theta(0) = \theta \left( \sum_{j=0}^{d-1}q_j\alpha^j \right) \overset{*}{=}  \sum_{j=0}^{d-1}q_j\theta(\alpha^j) = \sum_{j=0}^{d-1}q_j\beta^j. $$

In the equality denoted $*$, I assume that he is using the fact that any embedding $\theta$ fixes $\mathbb Q$ pointwise. But doesn't that implicitly require that embeddings---and in particular, ring-homomorphisms---send the multiplicative identity to the identity? But his definition of ring-homomorphisms doesn't require that 1 gets sent to 1! (Definition A.10, p. 327)

*Regarding prime factorization of gcd and lcm of ideals he writes the following (I have shortened and simplified a bit in order to focus on my question):

Theorem 1.19 Suppose $D$ is a Dedekind domain and $I$, $J$ are $D$-ideals with prime factorization
  $$ I = \prod_{j=1}^r \mathcal P_j^{a_j}  \text{ and } J = \prod_{j=1}^r \mathcal P_j^{b_j},$$
  then 
  $$ \gcd(I, J) =  \prod_{j=1}^r \mathcal P_j^{m_j}, $$
  where $m_j = \min(a_j, b_j).$

The definition of $\gcd$ is: $\gcd(I,J) \overset{\mathrm {def}}{=} I + J$,
and the ideals $I$ and $J$ are said to be relatively prime if $\gcd(I,J) = I + J = D$. He then proceed with the proof as follows.

Proof. Since $\gcd(I,J) = I + J$, then
  $$ \gcd(I, J) =  \prod_{j=1}^r \mathcal P_j^{a_j} + \prod_{j=1}^r \mathcal P_j^{b_j} = \prod_{j=1}^r \mathcal P_j^{m_j} \left( \prod_{j=1}^r \mathcal P_j^{a_j - m_j} + \prod_{j=1}^r \mathcal P_j^{b_j - m_j}  \right). $$
  However, for each $j$, one of $a_j - m_j$ or $b_j - m_j$ is zero, so the right-hand sum is $D$ since the two summands are relatively prime.

Isn't the last argument circular? I.e., he states that because they are relatively prime, their sum is $D$. But in order to show that they are relatively prime, he would have to show that their sum is $D$!

*In exercise 2.40, p. 82, he asks:

Let $R$ be a Dedekind domain, and let $I$ be an $R$-ideal with 
  $$ I =  \prod_{j=1}^r \mathcal P_j^{a_j},$$ 
  for distinct prime $R$-ideals $\mathcal P_j$. Prove that
  $$ \left | R / I \right|  = \prod_{j=1}^r \left| R / \mathcal P_j \right|^{a_j} $$

By the Chinese Remainder Theorem, I guess that implicitly he want's us to show that $\left| R / \mathcal P_j \right|$ is finite (I'm not seeing how to interpret the last equality if this is not the case). However, isn't this false? I.e., while I know that $R /I$ is finite for rings of integers, I don't think this holds for general Dedekind domains. In fact, the answers to this question seem to demonstrate exactly that. But what does the exercise ask you to prove then? 
 A: 1 .  That definition of norm seems wrong.  In general, if $I$ is a nonzero ideal of $\mathcal O_K$ for $K$ a finite extension of $\mathbb Q$, then the norm $N(I)$ is defined to be the number of elements in quotient ring $\mathcal O_K/I$, which is finite.  In particular, if $K/\mathbb Q$ is quadratic, and $p$ is a prime number which remains prime in $\mathcal O_K$, i.e. $p\mathcal O_K = \mathcal P$, then one can show that $\mathcal O_K/p\mathcal O_K$ has $p^2$ elements.  
2 .  You're right.
3 .  If $\theta: F \rightarrow F'$ is a homomorphism of fields (not necessarily sending the identity to the identity), then there are two possibilities.  (1) $\theta(x) = 0$ for all $x \in F$, or (2) $\theta$ is an embedding, i.e. is injective, and will necessarily send $1_F$ to $1_{F'}$.  The proof is easy.
4 .  It is circular, but you can fix the proof by arguing directly that if $I$ and $J$ are ideals with $\nu_P(I)$ or $\nu_P(J) = 0$ for each prime $P$ (where $\nu_P(I)$ is the power of $P$ occurring in the factorization of $I$), then $I + J = D$.
This is easiest done by localizing.  For each prime $P$, $(I+J)_P = I_P + J_P$ is an ideal of the localized ring $D_P$ which has unique maximal ideal $P_P$.  If $I = P_1^{e_1} \cdots P_r^{e_r}$, then $I_{P_1} = P_{1P_1}^{e_1}$, so the hypothesis on $I$ and $J$ tells you that for each $P$, either $I_P$ or $J_P$ is equal to $D_P$, hence $I_P + J_P = D_P$.  
Then a basic result from module theory tells you that since $(I+J)_P = D_P$ for every maximal ideal $P$ of $D$, you can conclude that $I+J = D$.  
5 .  The statement
$$|R/I| = \prod\limits_{j=1}^r |R/\mathcal P_j|^{a_j}$$
still makes sense if all the cardinalities involed are infinite.  If $A$ and $B$ are sets, then $|A|  |B|$ means the cardinality of the set $A \times B$.  If one of $A$ or $B$ is infinite, then this is just the cardinality of the larger set.  The CRT tells you that
$$|R/I| = \prod\limits_{j=1}^r |R/\mathcal P_j^{a_j}|$$
so the problem becomes to show that there is a bijection between $R/\mathcal P^n$ and $(R/\mathcal P)^n$.
