## Algebraic Number Theory

最近事情比较多，代数数论先断更一段时间，搞完毕设再来续。—— 2022 3.1

reference book: Algebraic Number Theory

### Introduction

#### primary concept

free abelian groupG 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 \(\mathbb{Z}\).

modulemodule 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 fractionsFrac(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 idealA 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 idealAny ideal of a ring which is strictly smaller than the whole ring.

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

exact sequenceA 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

#### 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

\(d\), 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.

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.

\((2, 1+\sqrt{-5})\) 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.

\((3, 1+\sqrt{-5})\) 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.

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 ringuse 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 domainA 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 \(f(r) < f(b)\)

algebraLet \(K\) 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

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.

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 extensionfor 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 spaceFor 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 \(A\).

#### p29

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.).

traceandnormConsider \(L/K\) as an extension of fields of degree \(n\). The

traceof \(x \in L\) is the trace of the \(K\)-linear endomorphism \(m_x: L\rightarrow L, m_x(y) = xy\), thenormof \(x\) is \(det(m_x)\).Computational procedure (for element $ x L$, basis is \(\{e_0,\dots, e_{n-1}\}\)):

- calculate vector: \(\vec m_{xy} = x (e_0y_0, \dots, e_{n-1}y_{n-1})\)
- calculate matrix: \(m_x : m_x \times \vec m_{xy} = (y_0, \dots, y_{n-1})\)
- 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

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 \(-4 = -2 \cdot 2 = (1+\sqrt 5)\cdot (1-\sqrt 5)\).

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 \(A\).

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 \(\beta_s =\beta_t\), \(disc(L/K) = 0\).

\(\mathfrak{a} \neq (2)\) 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.

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 \(M\).

- \(0 \leq k < n\): \(\alpha \beta^k = \beta^kf(\beta) \in M\)
- \(n\leq k \leq m+n\): \(\alpha \beta^k = \beta^{k-n} \alpha \beta^{n} \in M\)

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.

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 \(f_n(\alpha_j^n) = 0\).

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.

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.

\(\overline f\) 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.

\(S^{-1}A = \{\frac{a}{s}\ |\ a\in A, s\in S \}\), \(\forall f(X) \in S^{-1}A[X]\), \(f(X)\) 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 \(S^{-1}A\).

\(A_\mathfrak{p} = \{\frac{a}{s}\ | \ a\in A, s\in A\backslash \mathfrak{p}\}\), 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.