### 数学代写|代数数论代写Algebraic number theory代考|Discriminant and Ramification

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## 数学代写|代数数论代写Algebraic number theory代考|Discriminant and Ramification

Finally we arrive at the main result of this chapter, a criterion for ramification, proved by Dedekind in $1882 .$

Theorem 4.37. Suppose $K / k$ is an extension of degree $n$ of number fields. A prime $\mathfrak{p}$ of $k$ ramifies in $K$ if and only if $\mathfrak{p} \mid \mathfrak{o}{K / k}$. Proof. Let $$\mathfrak{p O}=\mathfrak{P}{1}^{c_{1}} \ldots \mathfrak{P}{g}^{c{g}}$$
be the factorization of $\mathfrak{p}$ into powers of distinct primes in $\mathcal{O}$. Since the ring
$$\mathcal{O} / \mathrm{pO}=\mathcal{O} / \mathfrak{P}{1}^{e{1}} \ldots \mathfrak{P}{g}^{e{g}} \cong \mathcal{O} / \mathfrak{P}{1}^{e{1}} \times \ldots \times \mathcal{O} / \mathfrak{P}{g}^{e{g}}$$

$\mathfrak{p}$ is ramified $\Leftrightarrow$ some $e_{j}>1 \Leftrightarrow \mathcal{O} / \mathfrak{P}{j}^{\epsilon{j}}$ is not reduced $\Leftrightarrow \mathcal{O} / \mathfrak{p O}$ is not reduced $\Leftrightarrow \mathfrak{d}{(\mathcal{O} / p \mathcal{O}) /(0 / p)}=(0)$. Thus we need to show that $\mathfrak{d}{(\mathcal{O} / p \mathcal{O}) /(\mathfrak{o} / \mathfrak{p})}=(0) \Leftrightarrow$ $p \mid \mathfrak{o}_{K / k}$.

Let $S=\mathfrak{o} \backslash \mathfrak{p}, A$ the localization of $\mathfrak{o}$ at $\mathfrak{p}, B=S^{-1} \mathcal{O}, \mathfrak{P}=S^{-1} \mathfrak{p}$, the maximal ideal of $A$. Since $\mathcal{O}$ is a finitely generated (but not necessarily free) o-module, $B$ is a finitely generated $A$-module, generated by the same elements. Now since $A$ is a principal ideal domain, $B$ has a basis over $A$, easily seen to consist of $n=[K: k]$ elements $\alpha_{1}, \ldots, \alpha_{n}$. Since $S$ does not intersect any of the prime ideals of $\mathcal{O}$ lying above $\mathfrak{p}$, we have the following diagram:
$$\begin{array}{ccc} \mathcal{O} / \mathfrak{p O} & \cong & B / \mathfrak{P} B \ \mid & & \mid \ 0 / \mathfrak{p} & \cong & A / \mathfrak{P} \end{array}$$
For $\beta$ in $B$, we denote by $\bar{\beta}$ its residue class in $B / \mathfrak{P B}$. The dimension of $\mathcal{O} / \mathfrak{p O}$ over $o / \mathfrak{p}$ is $n$ and so is the dimension of $B / \mathfrak{P B}$ over $A / \mathfrak{P}$. Since $\bar{\alpha}{1}, \ldots, \bar{\alpha}{n}$ generate $B / \mathfrak{P B}$ over $A / B$, by comparing dimensions, they must form a basis of $B / \mathfrak{P} B$ over $A / \mathfrak{P}$. Thus by Theorem $4.26$ and the diagram above, $\mathfrak{d}{(\mathcal{O} / \mathrm{pO}) /(\mathrm{o} / \mathrm{p})}=(0)$ if and only if $\Delta\left(\alpha{1}, \ldots, \alpha_{n}\right)=0$. Thus we show that $\mathfrak{p} \mid \mathfrak{d}{K / k}$ if and only if $\Delta\left(\alpha{1}, \ldots, \alpha_{n}\right) \in \mathfrak{P}$.

First, let $\Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right) \in \mathfrak{P}$. If $\left{\beta_{1}, \ldots, \beta_{n}\right}$ is a basis of $K$ over $k$ consisting of elements in $\mathcal{O}$, then
$$\beta_{i}=\sum_{j=1}^{n} a_{i j} \alpha_{j} \quad\left(a_{i j} \in A\right)$$
which shows that $\Delta\left(\beta_{1}, \ldots, \beta_{n}\right)=\operatorname{det}\left(\operatorname{Tr}\left(\alpha_{i} \alpha_{j}\right)\right) \cdot\left(\operatorname{det}\left(a_{i j}\right)\right)^{2} \in \mathcal{O} \cap \mathfrak{P}=\mathfrak{p}$. Hence, $\mathfrak{o}{K / k} \subseteq \mathfrak{p}$, i.e. $\mathfrak{p} \mid \mathfrak{d}{K / k}$. Conversely, suppose $\mathfrak{p} \mid \mathfrak{v}{K / k}$. If $\alpha{1}, \ldots, \alpha_{n}$ is a basis of $B$ over $A$, write each $\alpha_{j}=\beta_{j} / s$ with $\beta_{j}$ in $\mathcal{O}$ and $s$ in $S$. Then
$$\Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right)=\operatorname{det}\left(\operatorname{tr}\left(\alpha_{i} \alpha_{j}\right)\right)=\frac{1}{s^{2 n}} \operatorname{det}\left(\operatorname{tr}\left(\beta_{i} \beta_{j}\right)\right)$$
is in $A \mathfrak{o}_{K / k} \subseteq A \mathfrak{p} \subseteq \mathfrak{P}$.

## 数学代写|代数数论代写Algebraic number theory代考|Lattices in R

If $\boldsymbol{a} \in \mathbb{R}^{n}$ and $r>0$, we call the subset
$$B_{r}(\boldsymbol{a})=\left{\boldsymbol{x} \in \mathbb{R}^{n} \mid \operatorname{dist}(\boldsymbol{x}, \boldsymbol{a})=|\boldsymbol{x}-\boldsymbol{a}|0, such that X \cap B_{r}(\boldsymbol{a})={\boldsymbol{a}}. Consider \mathbb{R}^{n} as an Abelian group under addition. A lattice in \mathbb{R}^{n} is a discrete subgroup L \neq{0} of \mathbb{R}^{n}. Let d be the dimension of the subspace of \mathbb{R}^{n} spanned by elements of a lattice L \subseteq \mathbb{R}^{n}. Clearly, d \leq n. We call d the rank of the lattice L. A lattice L \subseteq \mathbb{R}^{n} is a full lattice if its rank is n. Remark 5.1. Topologically speaking, L \subseteq \mathbb{R}^{n} is a full lattice if and only if the quotient space \mathbb{R}^{n} / L is compact. EXERCISE Show that L \subseteq \mathbb{R}^{n} is a lattice if and only if it is a \mathbb{Z}-module$$
L=\mathbb{Z} v_{1} \oplus \ldots \oplus \mathbb{Z} v_{d}
$$57 for some vectors \boldsymbol{v}{1}, \ldots, \boldsymbol{v}{d} in L. The expression (5.2) means that each \boldsymbol{v} in L has a unique representation$$
\boldsymbol{v}=a_{1} \boldsymbol{v}{1}+\cdots+a{d} \boldsymbol{v}{d} $$with a{j} \in \mathbb{Z}. Hint: If d=1, choose \boldsymbol{v}{1} \neq \mathbf{0}, a vector in L nearest to \mathbf{0}. This is possible, because L is discrete. Then, clearly every vector v in L has a unique representation v=a \boldsymbol{v}{1} for a in \mathbb{Z}, for if \boldsymbol{v}=(a+r) \boldsymbol{v}{1} with 0{1} \in L, contradicting the choice of \boldsymbol{v}_{1}. For d>1, use induction on d. We now give a characterization for a lattice to be full, which is more suitable for our purpose. Theorem 5.2. A lattice L \subseteq \mathbb{R}^{n} is full if and only if there is a bounded set Y \subseteq \mathbb{R}^{n} such that$$
\mathbb{R}^{n}=\cup_{\boldsymbol{v} \in L}(\boldsymbol{v}+Y)
$$Here, \boldsymbol{v}+Y={v+\boldsymbol{y} \mid \boldsymbol{y} \in Y}. Before proving the proposition, we define a useful term. Definition 5.3. Let$$
L=\mathbb{Z} \boldsymbol{v}{1} \oplus \ldots \oplus \mathbb{Z} \boldsymbol{v}{n}
$$be a full lattice in \mathbb{R}^{n}. The set$$
P=\left{c_{1} \boldsymbol{v}{1}+\cdots+c{n} \boldsymbol{v}{n} \mid 0 \leq c{j}<1\right}
$$is called a fundamental parallelepiped of L. It depends on the \mathbb{Z}-basis \left{v_{1}, \ldots, \boldsymbol{v}{n}\right} of L. Clearly P is bounded and$$ \mathbb{R}^{n}=\cup{\boldsymbol{v} \in L}(\boldsymbol{v}+P),
$$a disjoint union of translates v+P of P by elements of L. Proof. If L is full, we can take Y to be a fundamental parallelepiped of L. Conversely, suppose a bounded set Y \subseteq \mathbb{R}^{n} exists with the property (5.3) and L is not full. We show that this leads to a contradiction. Let W be the subspace of \mathbb{R}^{n} spanned by the vectors in L. Then d= \operatorname{dim} W<n. Consider \mathbb{R}^{n} as an inner product space with the dot product of vectors. Choose a unit vector \boldsymbol{v}{d+1} (by the Gram-Schmidt Process) which is perpendicular to every vector of W. Let r>0 such that Y \subseteq B{r}(\mathbf{0}). It is easy to see that if \boldsymbol{w}=a \boldsymbol{v}{d+1} is a vector in \mathbb{R}^{n} with a>r, then \boldsymbol{w} \notin \cup{\boldsymbol{v} \in L}(\boldsymbol{v}+Y). This is a contradiction. ## 代数数论代考 ## 数学代写|代数数论代写Algebraic number theory代考|Discriminant and Ramification 最后，我们得出了本章的主要结果，即分枝标准，由 Dedekind 在1882. 定理 4.37。认为ķ/ķ是学位的延伸n的数字字段。一个素数p的ķ延伸到ķ当且仅当p∣○ķ/ķ. 证明。让 p○=磷1C1…磷GCG 是因式分解p成不同素数的幂○. 自从戒指 ○/p○=○/磷1和1…磷G和G≅○/磷1和1×…×○/磷G和G p是分枝的⇔一些和j>1⇔○/磷jεj不减少⇔○/p○不减少⇔d(○/p○)/(0/p)=(0). 因此我们需要证明d(○/p○)/(○/p)=(0)⇔ p∣○ķ/ķ. 让小号=○∖p,一个的本地化○在p,乙=小号−1○,磷=小号−1p, 的最大理想一个. 自从○是一个有限生成的（但不一定是自由的）o-module，乙是一个有限生成的一个-module，由相同的元素生成。现在自从一个是一个主理想域，乙有一个基础一个，很容易看出由n=[ķ:ķ]元素一个1,…,一个n. 自从小号不与任何主要理想相交○躺在上面p，我们有下图： ○/p○≅乙/磷乙 ∣∣ 0/p≅一个/磷 为了b在乙，我们表示为b¯它的残基类在乙/磷乙. 的维度○/p○超过○/p是n的维度也是如此乙/磷乙超过一个/磷. 自从一个¯1,…,一个¯n产生乙/磷乙超过一个/乙，通过比较维度，它们必须形成一个基础乙/磷乙超过一个/磷. 因此由定理4.26和上图，d(○/p○)/(○/p)=(0)当且仅当Δ(一个1,…,一个n)=0. 因此我们证明p∣dķ/ķ当且仅当Δ(一个1,…,一个n)∈磷. 首先，让Δ(一个1,…,一个n)∈磷. 如果\left{\beta_{1}, \ldots, \beta_{n}\right}\left{\beta_{1}, \ldots, \beta_{n}\right}是一个基础ķ超过ķ由元素组成○， 然后 b一世=∑j=1n一个一世j一个j(一个一世j∈一个) 这表明Δ(b1,…,bn)=这⁡(Tr⁡(一个一世一个j))⋅(这⁡(一个一世j))2∈○∩磷=p. 因此，○ķ/ķ⊆p， IEp∣dķ/ķ. 相反，假设p∣在ķ/ķ. 如果一个1,…,一个n是一个基础乙超过一个，写每个一个j=bj/s和bj在○和s在小号. 然后 Δ(一个1,…,一个n)=这⁡(tr⁡(一个一世一个j))=1s2n这⁡(tr⁡(b一世bj)) 在一个○ķ/ķ⊆一个p⊆磷. ## 数学代写|代数数论代写Algebraic number theory代考|Lattices in R 如果一个∈Rn和r>0, 我们称子集$$
B_{r}(\boldsymbol{a})=\left{\boldsymbol{x} \in \mathbb{R}^{n} \mid \operatorname{dist}(\boldsymbol{x }, \boldsymbol{a})=|\boldsymbol{x}-\boldsymbol{a}|0,s在CH吨H一个吨X \cap B_{r}(\boldsymbol{a})={\boldsymbol{a}}.C○ns一世d和r\mathbb{R}^{n}一个s一个n一个b和l一世一个nGr○在p在nd和r一个dd一世吨一世○n.一个l一个吨吨一世C和一世n\mathbb{R}^{n}一世s一个d一世sCr和吨和s在bGr○在pL \ neq {0}○F\mathbb{R}^{n}.大号和吨db和吨H和d一世米和ns一世○n○F吨H和s在bsp一个C和○F\mathbb{R}^{n}sp一个nn和db是和l和米和n吨s○F一个l一个吨吨一世C和L \subseteq \mathbb{R}^{n}.Cl和一个rl是,d \leq n.在和C一个lld吨H和r一个nķ○F吨H和l一个吨吨一世C和大号.一个l一个吨吨一世C和L \subseteq \mathbb{R}^{n}一世s一个F在lll一个吨吨一世C和一世F一世吨sr一个nķ一世sn\$。

57

Rn=∪在∈大号(在+是)

P=\left{c_{1} \boldsymbol{v}{1}+\cdots+c{n} \boldsymbol{v}{n} \mid 0 \leq c{j}<1\right}P=\left{c_{1} \boldsymbol{v}{1}+\cdots+c{n} \boldsymbol{v}{n} \mid 0 \leq c{j}<1\right}

Rn=∪在∈大号(在+磷),

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