Question about proof of 9.5 from Atiyah's Introduction to Commutative Algebra? I don't understand the proof of 9.5, but I know what's separable extension. Could someone give me the complete proof? Or recommend some materials that are helpful to understand the proof for me to read? (I didnt get the logic about the whole proof.)
Theorem 9.5. The ring of integers in an algebraic number field K is a Dedekind domain.
Proof. K is a separable extension of Q (because the characteristic is zero), hence by (5.17) there is a basis $v_1$ .. . , $v_n$ of K over Q such that $A\subseteq \sum Z_{v_j}$. Hence A is finitely generated as a Z-module and therefore Noetherian. Also A is integrally closed by (5.5). To complete the proof we must show that every nonzero prime ideal $\mathfrak p $ of A is maximal, and this follows from (5.8) and (5.9):
(5.9) shows that $\mathfrak p \cap Z \neq 0$, hence $\mathfrak p \cap Z$  is a maximal ideal of Z and therefore$\mathfrak p $ is maximal in A by (5.8).
The following parts are the parts that make me confused. Could someone explain more about them?
Part1. hence by (5.17) there is a basis $v_1$ .. . , $v_n$ of K over Q such that $A\subseteq \sum Z_{v_j}$
Part2. To complete the proof we must show that every nonzero prime ideal $\mathfrak p $ of A is maximal, and this follows from (5.8) and (5.9):
(5.9) shows that $\mathfrak p \cap Z \neq 0$,
 A: Comment: "Sorry. 1. Could you explain more about what is O_k? 2. Why we can assure that the intersection of I and Z is a non-zero prime ideal?"
Answer: If $\mathbb{Q} \subseteq K$ is a finite extension of fields, let $A:=\mathcal{O}_K \subseteq K$ be the integral closure of $R:=\mathbb{Z}$ in $K$.  It follows $A$ is a finitely generated $R$-module (this is essentially 5.17 in AM). Let $I \subseteq A$ be a non-zero prime ideal
and let $0 \neq y \in I$ be an element. Since $y$ is integral over $R$ there is a monic polynomial $f(t):=t^n+a_1t^{n-1}+ \cdots + a_n$ with $a_i \in R$ and $a_n \neq 0$ and $f(y)=0$. It follows
$$ a_n=-(y^n+a_1y^{n-1}+ \cdots + a_2y) \in I \cap R$$
since $I$ is an ideal and $a_{n-i}y^i \in I$ for all $i$. It follows $0 \neq a_n \in I \cap R$ hence $J:=I\cap R \neq 0$. Let $J=(p)$ with $p$ a non zero prime number.
We get a map of rings
$$ k:=R/(p) \rightarrow A/I $$
and since $A$ is a finitely generated $R$-module it follows $A/I$ is a finite dimensional vector space over $k$. Since $I$ is a prime ideal it follows $L:=A/I$ is an integral domain. Let $0 \neq x \in L$ be any element. Since $L$ has finite dimension over $k$ we get a finite dimensional $k$-vector space
$$T:=k\{1,x,x^2,..,x^n\}$$
and a surjective map $g:k[t] \rightarrow T$ defined by $f(t)=x$. The kernel $(p(t))$ is generated by a non -zero irreducible polynomial $p(t)$ (since $k[t]$ is a PID) and hence $(p(t))$ must be a maximal ideal. It follows $T \cong k[t]/(p(t))$ must be a field and hence $x$ has a multiplicative inverse in $L$. Hence $L$ is a field, a finite extension of $k$.
Note: There is in fact a classification of the field $L$: If $\mathbb{F}_p:=\mathbb{Z}/(p)\mathbb{Z}$, there is a unique integer $r \geq 1$ and a unique field $\mathbb{F}_{p^r}$ with $L \cong \mathbb{F}_{p^r}$. Hence the fields $L$ constructed using prime ideals in $A$ are classified.
A: Let $A$ be the set of algebraic integers.
Take set of all bases with algebraic integers $B=\{(b_1,...,b_n):\{b_i\} \text{ is a basis of } $K$ \text{ and } b_i \in A \}$.
Now we have $\Delta(b_1,...,b_n) := \det(Tr(b_i b_j)) \in \mathbb{Z} $ for $(b_1,...,b_n) \in B$. This is because $Tr(z) \in \mathbb{Z}$ for $z \in A$.
We now have that $\det(Tr(b_i b_j)) = \det(\sum_k \sigma_k(b_i b_j)) = \det((\sum_k \sigma_k(b_i) \sigma_k(b_j))) = \det(\sigma_k(b_j))^2$.
Since $(b_1,...,b_n)$ is a basis, $\det(Tr(b_i b_j)) = \det(\sigma_k(b_j))^2 \neq 0$. (Refer to proposition 7.4.4 of https://faculty.math.illinois.edu/~r-ash/Algebra/Chapter7.pdf)
Now choose  $(v_1,...,v_n) = \min_{(b_1,...,b_n) \in B} \Delta(b_1,....,b_n)$.
Note that since $\Delta(b_1,....,b_n) \in \mathbb{Z}$ and $\Delta(b_1,....,b_n) \neq 0$, we have that $\min_{(b_1,...,b_n) \in B} \Delta(b_1,....,b_n) \neq 0$.
If $A \subsetneq \sum_i \mathbb{Z} v_i$ then there exists an $a \in A$ such that $a = \sum_i q_i v_i$ with wlog $q_1 \notin \mathbb{Z}$ and $q_i \in \mathbb{Q}$, $\forall i \in [n]$. We can then write $q_1 = m+r$ where $m \in \mathbb{Z}$ and $0<r<1$ and $r \in \mathbb{Q}$. Now $(a-mv_1,v_2,...,v_2)$ is a basis made of algebraic integers with $\Delta(a-mv_1,v_2,...,v_n) = \Delta(rv_1+q_2 v_2+...+q_n v_n,v_2,...,v_n) = det([\sigma_i(rv_1+q_2 v_2+...+q_n v_n),\sigma_i(v_2),...,\sigma_i(v_n) : 1 \leq i \leq n])^2 = r^2 \times det([\sigma_i(v_1),\sigma_i(v_2),...,\sigma_i(v_n) : 1 \leq i \leq n])^2  = r^2 \times \Delta(v_1,...,v_n).$
This is a contradiction since $0<r<1$ as we have a new basis made of algebraic integers with a smaller discriminant. Hence $A \subseteq \sum_i \mathbb{Z} v_i$.
Note that we need to show that $B$ is non-empty. This is because we can take a basis $\{w_1,...,w_n\}$ for $K$ over $\mathbb{Q}$ and scale $w_i$ by $m_i \neq 0 \in \mathbb{Z}$ so that $\{m_1 w_1,...,m_n w_n\} \subseteq A$. For example, Let $\beta_r(x_1,...,x_n)$ be $r^{th}$ elementary symmetric function. Let $\ell_{ij} \times \beta_r(\sigma_1(w_i),\sigma_2(w_i),...,\sigma_n(w_i)) \in \mathbb{Z}$ then the choice  $m_i = \prod_{j=1}^n \ell_{ij}$ works since this makes the minimal polynomial of $m_i w_i$ have coefficients from integers.
Infact we have proved that there is a common basis made of algebraic integers for $A$ over $\mathbb{Z}$ and $K$ over $\mathbb{Q}$. This is called integral basis.
The above proof is taken from https://mathcourses.nfshost.com/archived-courses/mat-521-2016-spring/lectures/20-integral-bases.pdf
A: Part 1: In Proposition 5.17, take $A=\mathbb Z$, $K=\mathbb Q$, $L=K$. Then $B$ (as in the Proposition) is the integral closure of $\mathbb Z$ in $K$, which is the ring of integers.
Part 2:
For the claim that $\mathfrak p^c\neq 0$, let's assume otherwise. Since $0$ is also prime, we would have that $0^c=\mathfrak p^c=0$, and hence $0=\mathfrak p$ by Corollary 5.9
In Corollary 5.8, take $A=\mathbb Z$, $B=$ the ring of integers. Since $\mathfrak p$ is prime, we know that $\mathfrak p^c=\mathfrak p\cap\mathbb Z\neq 0$, so we can write $\mathfrak p^c=(p)$ for some prime number $p$. Since $(p)$ is maximal, it follows from Corollary 5.8 that $\mathfrak p$ is maximal.
