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## Algebraic Number Theory

reference book: Algebraic Number Theory

### Introduction

#### primary concept

free abelian group

G is an abelian group with a basis, then any element in G could be uniquely expressed as an integer combination of finitely basis elements, like a 2-dim lattice in $$\mathbb{Z}$$.

module

module is an algebraic structure which is extension of vector space, scalar multiplication of module could be a ring instead of a field in vector space.

ideal is a kind of module, more details could be seen here.

field of fractions

Frac(R) means a field is defined based on a integral domain R like Q based on Z, which uses an equivalence relation on $$R\times R\backslash \{0\}$$ by letting $$(n, d) \sim (m, b)$$ whenever $$nb =md$$.

fractional ideal

A fractional ideal $$I$$ of R is an R-submodule, and it’s a subgroup of Frac(R). There exists a non-zero $$r \in R$$ let $$rI \subset R$$.

proper ideal

Any ideal of a ring which is strictly smaller than the whole ring.

principal ideal

An ideal which is generated by a single element, like $$(2) \subset \mathbb{Z}$$.

exact sequence

A sequence of group homomorphisms between groups like:

$$G_0 \stackrel{f_1}\rightarrow G_1 \stackrel{f_2}\rightarrow \dots \stackrel{f_n}\rightarrow G_n$$, $$im(f_i) = ker(f_{i+1})$$

$$0$$ or $$1$$ in sequence always mean trivial group $$\{0\}, \{1\}$$.

for sequences whose form like $$0 \rightarrow A \stackrel{f}\rightarrow B \stackrel{g}\rightarrow C \rightarrow 0$$, which is called ‘short exact sequences’. $$f$$ is monomorphism (injective or one-to-one) and $$g$$ is an epimorphism

#### symbol

$$K^{\times}$$: the multiplicative group of units in K.

$$O_K$$ : the ring of integers in field K, which contains all algebraic integers in K.

#### p8

The definition of norm map is in Field norm, $$N_{L/K} (\alpha) = \prod_{\sigma\in Gal(L/K)} \sigma(\alpha)$$

this is multiplicative, the details of norm could be found in p31

#### p10

There’s two errors here

1. $$(1+\sqrt{-5})^2 = -4 + 2\sqrt{-5}$$, but $$(2, 1 + \sqrt{-5})^2$$ still be equal to $$(2)$$.

2. $$(2, 1+\sqrt{-5})(3, 1-\sqrt{-5})$$ should be $$(6, 3+3\sqrt{-5}, 2-2\sqrt{-5})$$

1. as book says, $$(2, 1+\sqrt{-5})^2 = (4, -4 + 2\sqrt{-5}, 2+2\sqrt{-5}) = (2)$$

2. $$(3, 1+\sqrt{-5})(3, 1-\sqrt{-5}) = (9, 3+3\sqrt{-5}, 3-3\sqrt{-5}, 6) \subset (3)$$

$$3 = 9-6 \in (9, 3+3\sqrt{-5}, 3-3\sqrt{-5}, 6)$$

so $$(3, 1+\sqrt{-5})(3, 1-\sqrt{-5}) = (3)$$

3. $$(2, 1+\sqrt{-5})(3, 1+\sqrt{-5}) = (6, 3+3\sqrt{-5}, 2+2\sqrt{-5}, -4 + 2\sqrt{-5})$$

$$6 = (1+\sqrt{-5})(1-\sqrt{-5})$$, $$-4 + 2\sqrt{-5}= (1+\sqrt{-5})^2$$

$$(2, 1+\sqrt{-5})(3, 1+\sqrt{-5}) = (1+\sqrt{-5})$$

so $$(2, 1+\sqrt{-5})(3, 1+\sqrt{-5}) \subset (1+\sqrt{-5})$$

and $1+ =3+3 - (2+2) (2, 1+)(3, 1+)$

so $$(2, 1+\sqrt{-5})(3, 1+\sqrt{-5}) = (1+\sqrt{-5})$$

4. $$(2, 1+\sqrt{-5})(3, 1-\sqrt{-5}) = (6, 3+3\sqrt{-5}, 2-2\sqrt{-5})$$

$$1-\sqrt{-5} = 6 - (3+3\sqrt{-5} + 2-2\sqrt{-5}) \in (6, 3+3\sqrt{-5}, 2-2\sqrt{-5})$$

$$(6, 3+3\sqrt{-5}, 2-2\sqrt{-5}) \subset (1+\sqrt{-5}, 1-\sqrt{-5})$$

#### p11

doubtful point:

units of $$\mathbb{Z}[\sqrt2]$$ are not only $$1+\sqrt 2$$, why $$\mathbb{Z}[\sqrt 2]^{\times} = \{\pm(1+\sqrt 2)^m | m \in \mathbb{Z}\}$$?

maybe in $$\mathbb{Z}[\sqrt 2]^{\times}$$, we could still use addition?

#### exercises

1. $$d$$, a square-free number, $$K = \mathbb{Q}[\sqrt d]$$, we talk about $$O_K$$ here.

​ Let $$\alpha = a+b \sqrt d, b \neq 0, a, b\in \mathbb{Q}$$, then if $$\alpha \in O_K$$,

it should satisfy $$2a \in \mathbb{Z}, a^2-b^2d\in \mathbb{Z}$$

​ If $$a, b \in \mathbb{Z}$$, it’s obviously that the condition are met whatever d is.

​ if $$a \notin \mathbb{Z}$$, it should be a half-integer, set $$a = \frac{1}{2} + a'$$, $$\frac{1-4b^2d}{4}$$ should be an integer, which is equivalent to $$4b^2d \equiv 1 \mod 4$$. if this work , the coefficient $$4$$ should be cancelled, and $$4b^2d$$ should be an integer, so b is a half-integer too, set $$b = b' + \frac{1}{2}$$, then it turns to $$d \equiv 1\mod 4$$.

​ After all, if $$d \equiv 1 \mod 4$$, $$a, b$$ could be all half-integer or all integer, else $$a, b$$ should be all integer.

2. In $$\mathbb{Z}[\sqrt{-5}]$$, proof $$(6) = (2, 1+\sqrt{-5})^2(3, 1+\sqrt{-5})(3, 1-\sqrt{-5})$$ is a factorization of $$(6)$$ into a product of prime ideals.

1. $$(2, 1+\sqrt{-5})$$ is prime means:

$$\forall a, b\in \mathbb{Z},$$ if $$(2a+(1+\sqrt{-5})b)\ |\ (cd),\ c, d \in \mathbb{Z}[\sqrt{-5}]$$

then $$c \in (2, 1+\sqrt{-5})$$ or $$d \in (2, 1+\sqrt{-5})$$.

Let $$c = c_0 + c_1\sqrt{-5}, d = d_0 + d_1\sqrt{-5}$$, then we have

$$2(2a^2 + 3b^2+2ab)\ |\ (c_0^2+5c_1^2)(d_0^2+5d_1^2)$$

$$2$$ is prime, which means $$(c_0^2 + 5c_1^2) \in (2)$$ or $$(d_0^2 + 5d_1^2) \in (2)$$

so $$(2, 1+\sqrt{-5})$$ is prime.

2. $$(3, 1+\sqrt{-5})$$ is prime means:

$$\forall a, b\in \mathbb{Z},$$ if $$(3a+(1+\sqrt{-5})b)\ |\ (cd),\ c, d \in \mathbb{Z}[\sqrt{-5}]$$

then $$c \in (3, 1+\sqrt{-5})$$ or $$d \in (3, 1+\sqrt{-5})$$.

Let $$c = c_0 + c_1\sqrt{-5}, d = d_0 + d_1\sqrt{-5}$$, then we have

$$3(3a^2 + 2b^2+2ab)\ | \ (c_0^2+5c_1^2)(d_0^2+5d_1^2)$$

$$3$$ is prime, which means $$(c_0^2 + 5c_1^2) \in (3)$$ or $$(d_0^2 + 5d_1^2) \in (3))$$

so $$(3, 1+\sqrt{-5})$$ is prime.

3. In a similar way, we could also prove that $$(3, 1-\sqrt{-5})$$ is prime.

### Commutative Algebra

some details in this chapter would be found here.

#### primary concept

quotient ring

use an equivalence relation $$\sim$$ on a ring $$R$$, $$I$$ is a two-sided ideal in R:

$a b$ if and only if $$(a-b) \in I$$

$$a \mod I$$ is the equivalence class of element $$a \in R$$, $$[a] = a+I:=\{a+r|r\in I\}$$

we call $$R/I$$ a quotient ring which is $$\{a \mod I | a\in R\}$$

zero-element in $$R/I$$ is $$I$$, multiplicateive identity is $$1+I$$

Euclidean domain

A Euclidean domain is an integral domain which can be endowed with at least one Euclidean function.

A Euclidean function on an integral domain R is a function $$f: R\backslash\{0\} \rightarrow \{a\ | \ a \geq 0\}$$ which satisfies:

$$\forall a, b \in R, b \neq 0$$, there exist $$q, r \in R,\quad a = b q + r$$

and $$r= 0$$ or $$f(r) < f(b)$$

algebra

Let $$K$$ be a field, and let $$A$$ be a vector space over K, if there’s a bilinear map $$A \times A \rightarrow A$$ on A, then we call $$A$$ as $$K$$-algebra, and $$K$$ is the base field of $$A$$.

#### symbol

${\rm Hom}$: For two modules $$M, N$$ over a ring $$R$$, $${\rm Hom}_R(M, N)$$ denotes the set of all module homomorphisms from $$M$$ to $$N$$.

#### p17

In the first one of Nakayama’s Lemma, $$\mathfrak{a}$$ should be an ideal contained in all maximal ideals of $$A$$, this make $$1-a_1$$ not be in $$\mathfrak{m}$$, more details could be seen here.

#### p18

$$\mathfrak{a}^e \in S^{-1}A$$ means $$\{a/s \ | \ a \in \mathfrak{a}, s\in S\}$$

For $$\mathfrak{a}$$ an a ideal of $$S^{-1}A$$, $$\mathfrak{a} \cap A$$ means all integer in $$\mathfrak{a}$$.

#### p19

'$$\mathfrak{p}$$ a prime ideal in $$S^{-1}A \Rightarrow \mathfrak{p} \cap A$$ is a prime ideal in $$A$$ disjoint from $$S$$’ is not clear, $$\mathfrak{p} = \{\frac{a}{b} \ | \ a\in \mathfrak{p}', b \in S\}$$ has to satisfies $$\mathfrak{p}' \cap S = \varnothing$$

$$A_{\mathfrak{p}}$$ means $$(A-\mathfrak{p})^{-1}A$$, more details could be seen here.

Conversely thinking, if there’s another genertors $$w$$ in $$\mathfrak{p}A_{\mathfrak{p}}$$, then $w (A-)^{-1}(A-)$, it would make $$\mathfrak{p}A_{\mathfrak{p}}$$ and $$A$$ equal. We could use some examples in $$\mathbb{Z}$$ to help us understand.

To understand $$(m) + (n) = \mathbb{Z}$$ if and only if $$m$$ and $$n$$ are relatively prime. We could think back to $$GCD$$ on $$\mathbb{Z}$$, if $$GCD(m, n) = 1$$, then $$\exists\ s, t, \ sm+tn = 1$$.

#### p20

$$\mathfrak{a} \mathfrak{b} = \{\sum_ia_ib_i\ | \ a_i\in \mathfrak{a}, b_i\in \mathfrak{b}\}, c\in \mathfrak{a}_1 \cap \mathfrak{a}_2$$, which means $$a_1c+a_2c \in \mathfrak{a_1} \mathfrak{a_2}$$.

the detailed proof of $$A/(\mathfrak{a}_1 \cap \mathfrak{a}_2) \simeq A/\mathfrak{a}_1 \times A/ \mathfrak{a}_2$$ could be seen here.

#### p21

doubtful point: element in $$\otimes_iM_i$$ should be like $$(m_0, m_1, \dots)$$, but why the isomorphism map be $$(\sum m_i) \otimes (\sum n_j) \rightarrow \sum m_i \otimes n_j$$

#### p22

proof for $M/aM (A/)_A M$ could be seen here. The point is why $$im(\mathfrak{a} \otimes_A M) \cap M = \mathfrak{a}M = \{\sum a_im_i \ | \ a_i \in \mathfrak{a}, m_i \in M\}$$, firstly I think for different bilinear map $$f$$, then $$\mathfrak{a} \otimes_A M$$ should be different. But then I realized that all the rings are isomorphic, so we could use an simple bilinear map $$f(a, b) = a \cdot b$$, then we get $$im(\mathfrak{a} \otimes_A M) \cap M= \{\sum a_im_i \ | \ a_i \in \mathfrak{a}, m_i \in M\}$$.

For counter-example about $$M \rightarrow N$$ is injective but $$M \otimes_A P \rightarrow N \otimes_A P$$ is not.

We know that $$M \otimes_A P \rightarrow N \otimes_A P = (M\rightarrow N)\otimes_AP$$

Let $$A = \mathbb{Z}, M = \mathbb{Z}, N = m\mathbb{Z}, P = \mathbb{Z}/ m\mathbb{Z}$$, then $$M \rightarrow N$$ s injective, and $$M \otimes_A P \rightarrow N \otimes_A P$$ is a zero map.

#### p23

the reason of using $$m \rightarrow 1\otimes m$$ to denote the map from $$M$$ to $$B \otimes_A M$$ is that $$B\otimes_A M \simeq \{bm\ | \ b\in B, m\in M \} \simeq \{1 \otimes m\ | \ m \in M\}$$

Because I’m not familiar with field, so maybe compelete notes on this part later.

#### exercises

1. If $$S$$ is saturated, we assumpt that its complement isnot a union of prime ideals, which means there’s at least one prime ideal $$I$$ satisfies:

$$\exists I_0 \subseteq S,\ I_0 \not \subset A\backslash S,\quad \ I\backslash I_0 \subseteq A\backslash S,\ I\backslash I_0\not \subset S$$

then because of the property of being saturated , $$\exists i_0 \in I_0, i_0 \in S$$, $$i_0$$ is prime, so $$\{1, i_0, i_0^2, \dots, i_0^n, \dots \} \subseteq S$$. Let $$U$$ be the set of all the units, because $$1 \in S$$, then $$U \subseteq S$$. For all the prime ideals with one generator $(i_0), (i_1), ,$ the set $$\{1, i_j, i_j^2, \dots, i_j^n, \dots \} = A\backslash U - \cup_{ k\neq j}(i_k)$$.

so in this case, all the elements in $$I\backslash I_0$$ could still be allocated to other prime ideals, the assumption is false.

If complement of $$S$$ is a union of prime ideals, then according to above proof, S should be like $$A- \cup_{ k\notin J}(i_k) = \cup_{j\in J}\{1, i_j, i_j^2, \dots,i_j^n,\dots \} \cup U$$, then $$\forall s \in S$$, the form of it is $$\prod_{j\in J} i_j^{e_j}, e_j \in \mathbb{Z}$$, clearly it’s saturated.

2. According to 1, we know that $$S' = S$$ could satisfy the need.

For $$S^{-1}A = \{\frac{a}{s} \ | \ a\in A, s \in S\}$$, the form of $$\frac{a}{s}$$ is $$\frac{a}{\prod_{j\in J} i_j^{e_j}} , e_j \in \mathbb{Z}$$.

so for prime ideal $$i_w \in A$$, $$(\frac{1}{i_w})$$ is still a prime ideal in $$S^{-1}A$$.

### Rings of Integers

#### primary concept

separable extension

for an algebraic field enxtension $$E \supseteq F$$ is called a separable extension if $$\forall \alpha \in E$$, its minimal polybomial over $$F$$ is a separable polynomial (the polynomial’s roots are distinct, without multiple root).

dual space

For a vector space $$V$$ over a field $$F$$, dual space $$V^{\lor}$$ is the set of all linear maps from $$V$$ to $$F$$.

Kronecker delta

$$\delta_{ij} = 0$$ if $$i \neq j$$ else $$1$$.

stem field

$$K$$ a field, and $$f \in K[x]$$ is an irreducible polynomial. The ring $$L = K[x]/(f)$$ is a field, with a distinguished root $$\overline x = x + (f)$$ of $$f$$. The stem field of $$f$$ is the pair $$(L, \overline x)$$.

#### symbol

$$(k_{ij})$$: sometime means a matrix $$M$$ satisfy $$M_{ij} = k_{ij}$$

#### p25

symmetric polynomial means all the coefficients in the polynomial has equal status.

#### p26

Cause $$h(X)$$ is a product of monomials, so we could take $$g(\alpha_{\sigma(1)}, \dots,\alpha_{\sigma(n))})$$ as $$g_i(\alpha_1, \dots, \alpha_n)$$, then the coefficient polynomials would be symmetric after expanding the product.

The point is: for any integral elements $,$, $$\alpha +\beta, \alpha \beta$$ are still integral elements.

For the first proof, the chain of thought should be:

‘symmetric polynomials are polynomials of symmetric elementary polynomials’ $$\Rightarrow$$

‘plug integral elements in any polynomial in $$A[X]$$ would get other integral elements.’ $$\Rightarrow$$

then we use polynomials $$x+y / xy$$ to get ‘all integral elements form a ring’.

At the end of the second proof, the point is $$\alpha \beta M, (\alpha \beta) M$$ all in $$M$$.

#### p28

another proof of Proposition 2.9 could be seen here. $$a, b$$ are not random elements in $$A$$.

For proposition 2.11, firstly I’m confused about ‘some $$a_i \in A$$’, if $$is integral over A, then all $$a_i$$ should be in $$A$$. The rest part of proof are clearly, we know all integral elements form a ring, so if all the roots are integral over $$A$$, and so are all the coefficients. The proof in book has proved that if an element $$\alpha$$ of $$L$$ is integral over $$A$$, then all the coefficients of its minimal polynomial over K are in $$A$$. And conversely, I know ‘some’ works that if there’s no coefficient of $$\alpha$$ minimal polynomial over $$K$$ is in $$A$$, then $$\alpha$$ would not be integral over $$A$$. Also, I’m confused about proposition 2.13, any A-algebra should be a vector space, then it should be a A-module, why need to prove that. #### p31 There’s an example could help us to understand the determinant of a basis of module works here. The proof of ‘index of $$N$$ is $$det(a_{ij})$$’ could be seen here. We could associate it with the relation with determinant of lattice and sublattice (In $$\mathbb{Z}^n$$, determinant of sublattice is multiple of determinant of lattice.). trace and norm Consider $$L/K$$ as an extension of fields of degree $$n$$. The trace of $$x \in L$$ is the trace of the $$K$$-linear endomorphism $$m_x: L\rightarrow L, m_x(y) = xy$$, the norm of $$x$$ is $$det(m_x)$$. Computational procedure (for element  x L, basis is $$\{e_0,\dots, e_{n-1}\}$$): 1. calculate vector: $$\vec m_{xy} = x (e_0y_0, \dots, e_{n-1}y_{n-1})$$ 2. calculate matrix: $$m_x : m_x \times \vec m_{xy} = (y_0, \dots, y_{n-1})$$ 3. calculate trace and norm: $$Tr(x) = Tr(m_x), Nm(x) = det(m_x)$$ For proposition 2.19, it build a field $$K[\beta]$$ in the middle to complete the proof. $$L = K[\beta]$$ in the proof is not the previous $$L$$. For $$L/K$$ with basis $$\{1, \beta, \beta^2, \dots, \beta^{n-1}\}$$, we could get trace and norm as follow$$ f(x) = x^n - 1x^{n-1} - - {n-1}x - _n \

= _1, f(_i) = 0, i = 1, 2, , n-1\

m_x = $\begin{bmatrix} 0 &\cdots & 0 & \alpha_n\\ 1 & \cdots & 0 & \alpha_{n-1}\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \cdots & 1 & \alpha_1 \end{bmatrix}$

$$use Vieta’s formulas, we get $$Tr(\beta) = \alpha_{1} = \sum \beta_i, Nm(\beta) =(-1)^n\alpha_n = \prod \beta_i$$, and 2.19 extend it to some $$L$$ with degree bigger than $$n$$. #### p32 In corollary 2.20, $$\sigma_i$$ are all homomorphisms, $$\sigma_i \beta$$ means $$\sigma_i(\beta)$$ In this page, I’m confused the difference between module and extension, because in $$Tr_{B/A}$$, somtimes $$B$$ is an extension over $$A$$, sometimes $$B$$ is $$A$$-module. The point is that we could think that if $$B$$ is a $$A$$-module which is a 2-dims vector space, then we couldn’t define the polynomial for $$B/A$$, how could we calculate $$Tr_{B/A}$$. Then we could see here, the second answer gives a construction for vector space like module, then we know the $$Tr$$ is just an operator, if meet its property, it could exist, in general case, it’s called the Hattori-Stallings trace. #### p33 How to construct the isomorphism from $$V^{\lor}$$ to $$V$$ could be seen here. If got any more question about discriminant, you may get the answer here. #### p35 For corollary 2.30, I think we could know the basis $$\{\beta_1, \dots, \beta_m\}$$ of $$L$$ could all in $$O_L$$, so the subring of $$L$$ which is finitely generated as a $$\mathbb{Z}$$-module is also generated by basis in $$O_L$$, clearly $$O_L$$ is the largest subring. #### p36 In Remark 2.33, $$Tr(\beta \beta_i) \in A , i=1, \dots, m$$ means that a ‘dual’ basis $$\{\beta_1', \dots, \beta_m'\}$$ is in $$C^*$$, then we know the conclusion from 2.29. #### p39 If $$D(1, \alpha, \dots, \alpha^{m-1})$$ doesn’t contain any square factor, then if should be $$disc(O_K/\mathbb{Z})$$, and $$|O_K| = |\mathbb{Z}[\alpha]|$$, then we knwo $$O_K = \mathbb{Z}[\alpha]$$. sign function here just means the plus or minus of integer. According to 2.34, we know $$D(1, \beta, \dots, \beta^{m-1}) = \prod_{1\leq i\leq j\leq m}(\beta_i-\beta_j)^2$$, then we know 2.40(a). detailed proof of Stickllberger’s Theorem could be seen here. proof for a^2 , 1, a  if $$\exists a\in \mathbb{Z}, a^2 \equiv 2 \mod 4$$, then $$a^2 = 4k+2 = 2(2k+1)$$, the right side only has one 2 factor, so there’s a contradiction. if $$\exists a \in \mathbb{Z}, a\equiv 3\mod 4$$, then $$a^2-1 = 4k+2 \Rightarrow (a+1)(a-1)=2(2k+1)$$, the left side has two 2 factor, but the right side only has one. #### p41 I use sagemath to do the same work as PARI do in the book. #### exercise 1. To find an example, we want $$\alpha = (a+b\sqrt 5)(c+d\sqrt 5) = (e+f\sqrt 5)(g + h\sqrt 5)$$ with $$a \neq e, a \neq g$$, so the variables should satisfy $$ac + 5bd = eg + 5hf, ad + bc = eh+ fg$$. While, it’s too bother, let’s just use $$-4 = -2 \cdot 2 = (1+\sqrt 5)\cdot (1-\sqrt 5)$$. 2. If $$f(X)$$ is reducible in $$K[X]$$, which means $$f(X) = f_0(X)f_1(X)...fn(X)$$ with $$f_i(X) \in K[X]$$. We know that $$f(X) \in A[X]$$, so all the roots of $$f(X)$$ are integral over $$A$$. For $$f_0(X)$$, the roots of $$f_0$$ are all integral over A, so all the coefficients should be integral over $$A$$, then they are all in $$A$$. 3. Let $$L = K[\beta]$$, $$\beta$$ is one root of $$f(X) \in K[X]$$. From 2.34, we know that $$disc(L/K) = (-1)^{\frac{m(m-1)}{2}}\prod_i(\prod_{j\neq i} (\beta_i-\beta_j))$$, $$\beta_1=\beta, \beta_2, \dots, \beta_n$$ are the roots of $$f(X)$$, because $$L/K$$ is not separable, then there exists $$\beta_s =\beta_t$$, $$disc(L/K) = 0$$. 4. $$\mathfrak{a} \neq (2)$$ is clearly, $$\mathfrak{a}^2 = (4, 2+2\sqrt{-3}, -2+2\sqrt{-3})$$, $$-2+2\sqrt{-3} = 2+2\sqrt{-3} - 4$$, then $$\mathfrak{a}^2 = (4, 2+2\sqrt{-3}) = (2)\mathfrak{a}$$. In $$\mathbb{Z}[\sqrt{-3}]$$, there exists $$10 = 2 \cdot 5 = (1+\sqrt{-3})(1-\sqrt{-3})$$, so it couldn’t factor uniquely. 5. For $$\alpha \in A[\beta] \cap A[\beta ^{-1}]$$, there will be two polynomials $$f(X) \in A[\beta][X]$$, $$g(X)\in A[\beta^{-1}][X]$$, with f() = g() = , and the coefficients of two polynomials are all in $$A$$, we let $$deg(f) = m, deg(g) = n$$ We set $$M = A + A\beta + \dots + A\beta^{m+n}$$, then we wonder if $$\alpha \beta^k$$ is in $$M$$. 1. $$0 \leq k < n$$: $$\alpha \beta^k = \beta^kf(\beta) \in M$$ 2. $$n\leq k \leq m+n$$: $$\alpha \beta^k = \beta^{k-n} \alpha \beta^{n} \in M$$ then we know that $$\alpha \beta^k \in M$$, so $$\alpha M \in M$$, then$$ is integral over $$A$$, more details could be seen here.

1. All $$\alpha_i \alpha_j, i\neq j$$ would have factor $$(1+\sqrt{10})(1-\sqrt{10}) = -9$$ or $$(1+\sqrt{7})(1-\sqrt{7}) = -6$$, they are all divisible by 3 in $$O_K$$. According to 2.12, we know that if $$\frac{\alpha_i^n}{3}$$ is integral over $$\mathbb{Z}$$, then its norm and trace should all in $$\mathbb{Z}$$, the norm clearly lie in $$\mathbb{Z}$$, but the trace is not. We know $$Tr(\alpha_i^n) = \sum \alpha_j^n = 4^n$$, because for $$f(X) \in \mathbb{Z}[X], f(\alpha_j) = 0$$, then the minimal polynomial $$f_n(X)$$ of $$\alpha_i^n$$ would satisfy $$f_n(\alpha_j^n) = 0$$.

2. If $$\overline g$$ is divisible by $$f$$ in $$F_3[X]$$, then $$\overline g(\alpha) = 0$$ over $$F_3$$, and $$g(X) = \overline g(X) + 3g'(X)$$, so $$g(\alpha) \equiv \overline g(\alpha) \mod 3$$.

if $$g(\alpha)$$ is divisible by 3 in $$\mathbb{Z}[\alpha]$$, then $$\overline g(\alpha) = 0$$ in $$F_3$$, because $$f$$ is minimal polynomial of $$\alpha$$, then $$\overline g$$ is divisible by $$f$$. The point is that whether $$f$$ is still minimal polynomial in $$F_3[X]$$. If there’s another $$f' \in F_3[X], f'(\alpha) = 0$$ with $$deg(f') < deg(f)$$, then $$f'(\alpha) + 3T = 0$$ over $$\mathbb{Z}$$, $$f''(X) = f'(X) + 3T$$ become minimal polynomial, so there’s a contradicition.

3. We know that $$f_i(\alpha) \equiv \overline f_i(\alpha) \mod 3$$, then $$\overline f_i \overline f_j(\alpha) \equiv \alpha_i\alpha_j \mod 3$$, from 1 we know $$\alpha_i\alpha_j \equiv 0\mod 3$$, so $$\overline f | \overline f_i \overline f_j, i\neq j$$, and $$\overline f$$ not divide $$\overline f_i^n$$, because if it happens, then $$\overline f_i^n(\alpha) = 0$$, which means $$\alpha_i^n \equiv 0 \mod 3$$, which is a contradiction.

4. $$\overline f$$ has at least 4 irreducible factors on $$F_3[X]$$, which means $$f$$ has at least 4 irreducible factors on $$\mathbb{Z}[X]$$, and $$f$$ has degree at most 4, so $$f$$ just has these 4 irreducible factors, $$f$$ will be $$f(X) = (X-t_0)(X-t_1)(X-t_2)(X-t_3)$$, we know $$f(\alpha) = 0$$, so $$t_i = \alpha$$, then $$X-t_i$$ is the minipoly of $$\alpha$$, here’s a contradiction.

6. $$S^{-1}A = \{\frac{a}{s}\ |\ a\in A, s\in S \}$$, $$\forall f(X) \in S^{-1}A[X]$$, $$f(X)$$ could be written as $$\frac{\sum_{i=0}^na_i'X^i}{\prod_{i=0}^ns_i}$$, if $$f(x_0) = 0$$, then there exists $$f'(X)$$ with $$f'(x_0) = 0$$ on $$A$$, so $$S^{-1}B$$ is still integral closure of $$S^{-1}A$$.

7. $$A_\mathfrak{p} = \{\frac{a}{s}\ | \ a\in A, s\in A\backslash \mathfrak{p}\}$$, and $$\mathfrak{p}A_\mathfrak{p} = (A \backslash \mathfrak{p}) ^{-1} \mathfrak{p} = \{\frac{p}{s} \ | \ p\in \mathfrak{p}, s \in A\backslash \mathfrak{p}\}$$.

$$A_\mathfrak{p}/ \mathfrak{p}A_\mathfrak{p} = \{\frac{a}{s} \ | \ a\in A/\mathfrak{p}A, s\in A/\mathfrak{p} \}$$，then $$\forall p \in A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$$ can be written in the form $$p = \frac{a}{b}, a, b\in A$$.

### Dedekind Domains & Factorization

We could know that the first main task of this book is to summarize what a kind of ring could have the property which is factorable uniquely.

#### p49

In alternative proof, there’s an error that $$a \in \mathfrak{a}, b\in \mathfrak{b}$$, not $$A$$ and $$B$$.

In lemma 3.10, the map is $$a + \mathfrak{p}^m \rightarrow a + \mathfrak{q}^m$$, firstly I thought it should be a map $$A/\mathfrak{p}^m \rightarrow A/\mathfrak{q}^m$$, but then I realized that $$\mathfrak{q}^m$$ is not ideal of $$A$$, so this quotient ring couldn’t exist. Then I take a look at the equivalence, we want to prove the map is one-to-one, which equals $$\forall a_0, a_1\in A$$, if $$a_0 + \mathfrak{q}^m = a_1+\mathfrak{q}^m$$, then $$a_0+\mathfrak{p}^m=a_1 + \mathfrak{p}^m$$. If $$a_0 = a_1$$, it’s obvious, if not, then $$a_0 \neq a_1, a_0+\mathfrak{q}^m = a_1 + \mathfrak{q}^m$$, hence $$a_0, a_1 \in A\cap \mathfrak{q}^m$$, so in the proof we have to show that $$\mathfrak{q}^m \cap A = \mathfrak{p}^m$$.

#### p50

Firstly I’m confused about $$s^{-1}$$ in $$A_\mathfrak{p}/\mathfrak{q}^m$$, the element in $$A_\mathfrak{p}/\mathfrak{q}^m$$ should be in the form of $$a + \mathfrak{q}^m$$, $$s$$ is just an element in $$A\backslash \mathfrak{p}$$, so $$s^{-1}$$ is also a set $$\{b \ | \ bs + q = 1, q \in \mathfrak{p}^m\}$$. $$ba = \frac{a}{s}(1-q)$$, then $$\frac{a}{s}(1-q) + \mathfrak{q}^m = \frac{a}{s} + \mathfrak{q}^m$$, then I know that “invertible” means $$b + \mathfrak{q}^m$$ is the inverse of $$s + \mathfrak{q}^m$$.

In the proof, firstly we need to know the isomorphism from the ideals of $$A/\mathfrak{b}$$ to the ideals of $$A$$ containing $$\mathfrak{b}$$, details could be seen here.

For $A/ ,$ $$\mathfrak{q}_i^j = (A\backslash \mathfrak{p}_i)^{-1}\mathfrak{p}^j$$ , and for $$\mathfrak{a}/\mathfrak{b}$$, here we take $$\mathfrak{a}/\mathfrak{b}$$ as an element in $$A/\mathfrak{b}$$, then we know that $$\mathfrak{p}_1^{s_1}\mathfrak{p}_2^{s_2}\dots\mathfrak{p}_m^{s_m} / \mathfrak{b}$$ is also an element in $$A/\mathfrak{b}$$, we take this element in the isomorphism, under first isomorphism, it turns to $$\mathfrak{p}_1^{s_1}\mathfrak{p}_2^{s_2}\dots\mathfrak{p}_m^{s_m} / \mathfrak{p}_1^{r_1} \times \dots \times \mathfrak{p}_1^{s_1}\mathfrak{p}_2^{s_2}\dots\mathfrak{p}_m^{s_m} / \mathfrak{p}_m^{r_m}$$, we could use an example $(2 ) / (4)$ over $$\mathbb{Z}$$ to think of $$\mathfrak{p}_1^{s_1}\mathfrak{p}_2^{s_2}\dots\mathfrak{p}_m^{s_m} / \mathfrak{p}_1^{r_1}$$, then we know that $$\mathfrak{p}_1^{s_1}\mathfrak{p}_2^{s_2}\dots\mathfrak{p}_m^{s_m} / \mathfrak{p}_1^{r_1} \cong \mathfrak{p}_1^{s_1} / \mathfrak{p}_1^{r_1}$$, because the zero of both is ${k _1^{r_1-s_1} | k }$. Then under the second isomorphism, $$\mathfrak{p}^{s_i} \rightarrow \mathfrak{q}^{s_i} : A \rightarrow A_{\mathfrak{p}}$$.

In the last, if $$\mathfrak{q}_i^{s_i} = \mathfrak{q}_i^{t_i}$$ with $$s_i \neq t_i$$, we suppose $$s_i > t_i$$ then there’s an ideal chain $$\mathfrak{q}_i^{t_i} \subseteq \mathfrak{q}_i^{t_i-1} \subseteq \dots \subseteq \mathfrak{q}_i^{s_i} \subseteq \dots \subseteq \mathfrak{q}_i$$, any ideal $_i^{j}$ is in this chain, then $$A_{ \mathfrak{p}_i}/ \mathfrak{q}_i$$ is not a field, so $_i$ is not the maximal ideal, which contradicts the premises.

#### p51

Firstly, I’m confused about $$x\equiv 1 \mod \mathfrak{p}_i, i\neq 1$$, how could we get this. Then I think the point is why $$\mathfrak{p}_1$$ and $$(x)$$ generate the same ideals in $$A_{\mathfrak{p}_i}$$. $$\mathfrak{p}_1$$ generates $$\frac{\mathfrak{p}_1}{A-\mathfrak{p}_i}$$, and $$(x)$$ generates $$\frac{(x)}{A-\mathfrak{p}_i} = \frac{(x_1) + \mathfrak{p}_1^2}{A-\mathfrak{p}_i} = \frac{k_1\mathfrak{p}_1 + \mathfrak{p}_1^2}{A-\mathfrak{p}_i} = \frac{\mathfrak{p}_1(k_1 + \mathfrak{p}_1)}{A-\mathfrak{p}_i}$$, we know that $$x_1 \in \mathfrak{p}_1 - \mathfrak{p}_1^2$$ , so $$k_1 \notin \mathfrak{p}_1$$, $$k_1+\mathfrak{p}_1$$ is fixed, then we have a simple isomorphism $$\sigma(\frac{p}{q}) : \frac{p}{q} \rightarrow \frac{p}{(k_1+\mathfrak{p}_1)q}$$ from $$(x)$$ to $$\mathfrak{p}_1$$.

3.14 and 3.15 use the same kind of proof, in 3.15, we could calculate that $$\mathfrak{a}$$ generates $$\frac{\mathfrak{p}_i^{s_i}}{A-\mathfrak{p}_i}$$, and $$\mathfrak{b} + (a)$$ generates $$\frac{\mathfrak{p}_i^{r_i}}{A-\mathfrak{p}_i} + \frac{(x_i)}{A-\mathfrak{p}_i}$$ in $$A_{\mathfrak{p}_i}$$, then I’m wondering whether $$(a) = \mathfrak{a}$$, but then I recognized that $$a \equiv x_i \mod \mathfrak{p}_i^{r_i}$$, so it need the part $$\frac{\mathfrak{p}_i^{r_i}}{A-\mathfrak{p}_i}$$.

#### p52

The point of proof for 3.18 is to use the (c) property of Dedekind domains.