Proof of Theorem $20$ in Chapter $3$, Marcus' Number Fields Theorem $20$ (Chapter $3$) in Marcus' Number Fields says:

Theorem $20$. Let $K,L$ be number fields with $K\subset L$, and let $R,S$ be the rings of integers in $K,L$ respectively (so $R = \mathbb A \cap K, S = \mathbb A \cap L.)$ Then, every prime $Q$ of $S$ lies over a unique prime $P$ of $R$; every prime $P$ of $R$ lies under at least one prime $Q$ of $S$.

The proof uses Theorem $19$, which I shall state for convenience:

Theorem $19$. Let $P$ be a prime (non-zero prime ideal) of $R$, $Q$ a prime of $S$. Then, the following conditions are equivalent:

*

*$Q \mid PS$

*$Q \supset PS$

*$Q \supset P$

*$Q \cap R = P$

*$Q \cap K = P$
When the above conditions hold, we say that $Q$ lies over $P$, or $P$ lies under $Q$.


I shall reproduce the proof and ask questions.

Proof of Theorem $20$: The first part is clearly equivalent to showing that $Q ∩ R$ is a prime in $R$. This follows easily from the definition of prime ideal and the observation that $1 \notin Q$. Fill in the details, using a norm argument to show that $Q ∩ R$ is nonzero.


*

*How does one show $Q \cap R \ne \{0\}$? We want to find some $x\in (Q \cap R) \setminus \{0\}$ such that $N^K_{\Bbb Q}(x) \ne 0$. We could even consider $N^L_K(x)$ instead, which should be in $K$ since $x \in Q \subset S = \mathbb A \cap L$.


For the second part, the primes lying over $P$ are the prime divisors of $PS$; thus, we must show that $PS \ne S$, so that it has at least one prime divisor.


*

*The primes lying over $P$ (by condition $1$ of Theorem $19$) divide $PS$. Since $S$ is a Dedekind domain, every non-zero ideal factors uniquely into prime ideals. This factorization of $PS$ is non-trivial (i.e., $\ne S$) if and only if $PS\ne S$. Just want to confirm the correctness of my reasoning.


Equivalently, we must show $1 \notin PS$. (We know $1 \notin P$, but why can't $1 = α_1β_1 +· · ·+α_rβ_r, α_i ∈ P, β_i ∈ S$?) To show $1 \notin PS$, we invoke Lemma $2$ for Theorem $15$: There exists $γ ∈ K − R$ such that $γP ⊂ R$. Then $γPS ⊂ RS = S$. If $1 ∈ PS$, then $γ ∈ S$. But then $γ$ is an algebraic integer, contradicting $γ ∈ K − R$.

Thanks!
 A: Your reasoning in your second bullet point is correct. As for the question in your first bullet point, suppose $\alpha \in Q$ with $\alpha$ nonzero. We have $[L : K] = n$, $[K(\alpha) : K] = d$, where $d$ is the degree of the minimal polynomial for $\alpha$ and $d \mid n$. Then $N^{K(\alpha)}_K(\alpha) = \sigma_1(\alpha) \cdots \sigma_d(\alpha)$, where $\sigma_1 \dots, \sigma_d$ are the $d$ embeddings of $K(\alpha)$ into $\mathbb{C}$ that fix $K$ pointwise. Since this product is (up to sign) the constant term of the minimal polynomial for $\alpha$ over $K$ and $\alpha$ is an algebraic integer, this product is in $R$. Each $\sigma_i$ extends to $n/d$ embeddings of $L$ into $\mathbb{C}$, so $N^L_K(\alpha) = \left (N^{K(\alpha)}_K \right)^{n/d}$, hence is in $R$. On the other hand, $N^L_K(\alpha) = \alpha \sigma_2(\alpha) \cdots \sigma_n(\alpha)$, hence $\sigma_2(\alpha) \cdots \sigma_n(\alpha) = N^L_K(\alpha)/\alpha \in L$. Since every $\sigma_i(\alpha)$ is an algebraic integer and the algebraic integers form a ring $\mathbb{A}$, $\sigma_2(\alpha) \cdots \sigma_n(\alpha) \in \mathbb{A}$ as well, hence in $S$. It follows that $N^L_K(\alpha) \in Q$, so $Q \cap R \neq \{0\}$.
