I am reading Cox's "Primes of the Form $x^2 + ny^2$", and am on Chapter 5 (it's a speedrun from number fields to Hilbert's class field). I am attempting exercise 5.1, and I have done some parts, so I am both asking for verification for my solution as well as hints for the rest.
Things we have stated:
Proposition 5.3. For a number field $K$
(i) $\mathcal{O}_K$ is a subring of $\mathbb{C}$ whose field of fractions is $K$.
(ii) $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $[K : \mathbb{Q}]$.
(a) Show that a nonzero ideal $\mathfrak{a}$ of $\mathcal{O}_K$ contains a nonzero integer $m$. (Hint: ...)
My solution: Let $\alpha \neq 0$ be in $\mathfrak{a}$. Of course it is algebraic, so let the monic integer polynomial $f(x) = a_0 + a_1x + \cdots + a_{n - 1}x^{n - 1} + x^n$ be its minimal polynomial. Now, $\langle \alpha \rangle \subset \mathfrak{a}$. In particular, for all integers $i \geq 1$, the elements $\alpha^i$ are in $\mathfrak{a}$. This then means $\sum_{i \geq 1} a_i \alpha^i \in \mathfrak{a}$, and since $f(\alpha) = 0 \in \mathfrak{a}$, we have $m = a_0 \in \mathfrak{a}$.
(b) Show that $\mathcal{O}_K / \mathfrak{a}$ is finite whenever $\mathfrak{a}$ is a nonzero ideal of $\mathcal{O}_K$. Hint: if $m$ is the integer from (a), consider the surjection $\mathcal{O}_K / m\mathcal{O}_K \to \mathcal{O}_K / \mathfrak{a}$. Use part (ii) of Proposition 5.3 to compute the order of $\mathcal{O}_K / m\mathcal{O}_K$.
My Ideas: From above, we know that $\langle m \rangle \subset \langle \alpha \rangle \subset \mathfrak{a}$, so my intuition tells me that this surjection definitely exists. (In my intuition, everything's a module / vector space, so this surjection is just a projection map?) However, I don't know how to explicitly describe it.
To compute $\left|\mathcal{O}_K / m\mathcal{O}_K\right|$, from the proposition above we know that $\mathcal{O}_K \cong \mathbb{Z}^{[K : \mathbb{Q}]}$, so $\mathcal{O}_K / m\mathcal{O}_K$ is just $\left(\mathbb{Z} / m\mathbb{Z}\right)^{[K : \mathbb{Q}]}$ and the order is $m^{[K : \mathbb{Q}]}$. This part makes sense but feels a little hand wavy? Or is it justified as is?
(c) Use (b) to show that every nonzero ideal of $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $[K : \mathbb{Q}]$.
My solution: Fix a nonzero ideal $\mathfrak{a} \subset \mathcal{O}_K$. We know that $\mathcal{O}_K / \mathfrak{a}$ is finite and $\mathcal{O}_K \cong \mathbb{Z}^{[K : \mathbb{Q}]}$, so $\mathfrak{a}$ has to be a product of $[K : \mathbb{Q}]$ infinite subgroups of $\mathbb{Z}$, i.e. $\mathfrak{a} \cong \prod_{i = 1}^{[K : \mathbb{Q}]} m_i\mathbb{Z}$. This is easy to prove by a simple proof by contradiction.
(d) If we have ideals $\mathfrak{a}_1 \subset \mathfrak{a}_2 \subset \cdots$, show that there is an integer $n$ such that $\mathfrak{a}_n = \mathfrak{a}_{n + 1} = \cdots$. Hint: consider the surjections $\mathcal{O}_K / \mathfrak{a}_1 \to \mathcal{O}_K / \mathfrak{a}_2 \to \cdots$, and use (b).
My solution: Again, my intuition tells me the surjections $\mathcal{O}_K / \mathfrak{a}_i \to \mathcal{O}_K / \mathfrak{a}_{i + 1}$ exists, but I don't know how to construct them. Anyways, I claim that if $\mathfrak{a}_i \neq \mathfrak{a}_{i + 1}$, then $\left|\mathcal{O}_K / \mathfrak{a}_{i + 1}\right| < \left|\mathcal{O}_K / \mathfrak{a}_i\right|$. This holds because for $\alpha \in \mathfrak{a}_{i + 1} \setminus \mathfrak{a}_i$ is a nonzero element in the kernel of the surjection. Since the quotients are finite, it must eventually stop and hence there are no infinite ascending chains.
(e) Use (b) to show that a nonzero prime ideal of $\mathcal{O}_K$ is maximal.
My ideas: Let $\mathfrak{a}$ be a prime ideal of $\mathcal{O}_K$, and suppose that $\mathfrak{a} \subset \mathfrak{b}$ (i.e. $\mathfrak{a}$ is not maximal), which gives $\mathcal{O}_K / \mathfrak{b} \subset \mathcal{O}_K / \mathfrak{a}$. Thinking about everything as $\mathbb{Z}$-modules, we can write $\mathcal{O}_K \cong \prod_{i = 1}^{[K : \mathbb{Q}]} \mathbb{Z}$ as ordered coordinates, and similar that $\mathcal{O}_K / \mathfrak{b} \cong \prod_{i = 1}^{[K : \mathbb{Q}]} \mathbb{Z} / m_i \mathbb{Z}$ and $\mathcal{O}_K / \mathfrak{a} \cong \prod_{i = 1}^{[K : \mathbb{Q}]} \mathbb{Z} / n_i \mathbb{Z}$. By the inclusion, we know that $m_i \mid n_i$, and for at least one $j$, $m_j \neq n_j$. However, for such $j$ we have that $n_j = m_j \cdot \left(\frac{n_j}{m_j}\right)$ i.e. $\mathbb{Z} / n_j \mathbb{Z}$ is not an integral domain, and hence the product ring is not an integral domain, which means $\mathfrak{a}$ is not prime.
For you for your help in advance!
Part a is great.
For the surjections in parts b and d, It is a general thing that if $I\subseteq J$ are ideals in $R$ then we can define $R/I\to R/J$ by $r+I\to r+J$ (Why this is well defined? And why is it is a surjection?). The rest of part b is good.
Part c: I don't know how easy is this (it isn't so trivial that subgroups of a free abelian group are also free), but it follows from the theory of finitely generated abelian groups. For all the details regarding this, you can open any book on groups or the first chapter of Algebra by Lang for example.
Part d and e are great.
Another way to prove e (if you know little about rings), is to prove $R/\mathfrak{a}$ is a field. You already know that $R/\mathfrak{a}$ is finite and integral domain so he must be a field (It's a nice exercise proving that every finite integral domain is a field).
By the way, if you are looking for an Algebraic number theory book that does all of this in general I strongly recommend Neukirch's book.