# 数学代写|交换代数代写commutative algebra代考|MATH2301

## 数学代写|交换代数代写commutative algebra代考|Rings of Integers of a Number Field

If $\mathbf{K}$ is a number field its ring of integers is the integral closure of $\mathbb{Z}$ in $\mathbf{K}$.
8.17 Proposition and definition (Discriminant of a number field) I et $\mathbf{K}$ he a mumber field and $\mathbf{Z}$ its ring of integers.

1. An element $y$ of $\mathbf{K}$ is in $\mathbf{Z}$ if and only if $\operatorname{Min}_{\mathbb{Q}, y}(X) \in \mathbb{Z}[X]$.
2. We have $\mathbf{K}=\left(\mathbb{N}^*\right)^{-1} \mathbf{Z}$.
3. Assume that $\mathbf{K}=\mathbb{Q}[x]$ with $x \in \mathbf{Z}$. Let $f(X)=\operatorname{Min}_{\mathbb{Q}, x}(X)$ be in $\mathbb{Z}[X]$ and $\Delta^2$ be the greatest square factor of $\operatorname{disc}_X f$.
Then, $\mathbb{Z}[x] \subseteq \mathbf{Z} \subseteq \frac{1}{\Delta} \mathbb{Z}[x]$.
4. The ring $\mathbf{Z}$ is a free $\mathbb{Z}$-module of rank $[\mathbf{K}: \mathbf{Q}]$.
5. The integer $\operatorname{Disc}_{\mathbf{Z} / \mathbb{Z}}$ is well-defined. We call it the discriminant of the number field $\mathbf{K}$.
D 1. Results from Lemma $8.10$ (Kronecker’s theorem).
6. Let $y \in \mathbf{K}$ and $g(X) \in \mathbb{Z}[X]$ be a nonzero polynomial that annihilates $y$. If $a$ is the leading coefficient of $g, a y$ is integral over $\mathbb{Z}$.
7. Let $\mathbf{A}=\mathbb{Z}[x]$ and $n=[\mathbf{K}: \mathbf{Q}]$. Let $z \in \mathbf{Z}$, which we as $h(x) / \delta$ with $\delta \in \mathbb{N}^*$, $\langle\delta\rangle+\mathrm{c}(h)=\langle 1\rangle$ and $\operatorname{deg} h<n$. We have $\mathbf{A}+\mathbb{Z} z \subseteq \frac{1}{\delta} \mathbf{A}$ and it thus suffices to prove that $\delta^2$ divides $\operatorname{disc}X(f)$. The ring $\mathbf{A}$ is a free $\mathbb{Z}$-module of rank $n$, with the basis $\mathcal{B}_0=\left(1, x, \ldots, x^{n-1}\right)$. Proposition $5.10$ gives $$\operatorname{Disc}{\mathbf{A} / \mathbb{Z}}=\operatorname{disc}{\mathbf{A} / \mathbb{Z}}\left(\mathcal{B}_0\right)=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}0\right)=\operatorname{disc}_X f$$ The $\mathbb{Z}$-module $M=\mathbf{A}+\mathbb{Z} z$ is also free, of rank $n$ with a basis $\mathcal{B}_1$, and we obtain the equalities $$\operatorname{disc}_X f=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}0\right)=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}_1\right) \times d^2,$$
where $d$ is the determinant of the matrix of $\mathcal{B}_0$ over $\mathcal{B}_1$ (Proposition II-5.33 2). Finally, $d=\pm \delta$ by the following Lemma $8.18$, as required.

## 数学代写|交换代数代写commutative algebra代考|The Multiplicative Theory of the Ideals of a Number Field

8.19 Definition An ideal a of a ring $\mathbf{A}$ is said to be invertible if there exist an ideal $\mathfrak{b}$ and a regular element $a$ such that $\mathfrak{a} \mathfrak{b}=\langle a\rangle$.
8.20 Fact Let a be an invertible ideal of a ring $\mathbf{A}$.

1. The ideal a is finitely generated.
2. If $\mathfrak{a}$ is generated by $k$ elements and if $\mathfrak{a} \mathfrak{b}=\langle a\rangle$ with a regular, then $\mathfrak{b}$ is generated by $k$ elements. Furthermore $\mathfrak{b}=(\langle a\rangle: a)$.
3. We have the rule a $\subseteq \subseteq \mathfrak{a} \Rightarrow \mathrm{c} \subseteq \mathfrak{o}$ for all ideals $\mathrm{c}$ and $\mathrm{D}$.
4. If $\mathfrak{c} \subseteq$ a there exists a unique $\mathfrak{d}$ such that $\mathfrak{d} \mathfrak{a}=\mathfrak{c}$, namely $\mathfrak{d}=(\mathfrak{c}: \mathfrak{a})$, and if $\mathfrak{c}$ is finitely generated, so is $\mathfrak{0}$.

D 3. If $\mathfrak{a} \mathfrak{c} \subseteq \mathfrak{a} \mathfrak{d}$ by multiplying by $\mathfrak{b}$ we obtain $a \mathfrak{c} \subseteq a \mathfrak{d}$, and since $a$ is regular, this implies $\mathbf{c} \subseteq \mathfrak{0}$.

1. If $\mathfrak{a} \mathfrak{b}=\langle a\rangle$, we find two finitely generated ideals $\mathfrak{a}1 \subseteq \mathfrak{a}$ and $\mathfrak{b}_1 \subseteq \mathfrak{b}$ such that $a \in \mathfrak{a}_1 \mathfrak{b}_1$ and thus $\mathfrak{a} \mathfrak{b}=\langle a\rangle \subseteq \mathfrak{a}_1 \mathfrak{b}_1 \subseteq \mathfrak{a} \mathfrak{b}_1 \subseteq \mathfrak{a} b{\text {. From the above, we deduce }}$ the equalities $\mathfrak{a}_1 \mathfrak{b}_1=\mathfrak{a} \mathfrak{b}_1=\mathfrak{a} \mathfrak{b}$. Whence $\mathfrak{b}=\mathfrak{b}_1$ by item 3. Similarly, $\mathfrak{a}=\mathfrak{a}_1$.
2. If $\mathfrak{a}=\left\langle a_1, \ldots, a_k\right\rangle$, we find $b_1, \ldots, b_k \in \mathfrak{b}$ such that $\sum_i a_i b_i=a$.
By reasoning as in item 1 with $\mathfrak{a}_1=\mathfrak{a}$ and $\mathfrak{b}_1=\left\langle b_1, \ldots, b_k\right\rangle$ we obtain the equality

## 数学代写|交换代数代写交换代数代考|伽罗瓦理论的初等情况

6.8定义和表示法当一个组$G$操作一个集合$E$时，我们将使用以下表示法

• 其中$x \in E, \operatorname{St}_G(x)=\operatorname{St}(x) \stackrel{\text { def }}{=}{\sigma \in G \mid \sigma(x)=x}$表示$x$的稳定器
• $G . x$表示$G$下$x$的轨道，我们将$G . x=\left{x_1, \ldots, x_k\right}$写为:$\left(x_1, \ldots, x_k\right)$是$G \cdot x$的一个不重复的枚举，其中$x_1=x$
• 对于$F \subseteq E, \operatorname{Stp}_G(F)$或$\operatorname{Stp}(F)$指定$F$的点向稳定器
• 如果$H$是$G$的一个子组，
• 用 $|G: H|$ 的指数 $H$ 在 $G$，
• 用 $\operatorname{Fix}_E(H)=\operatorname{Fix}(H)=E^H$ 由固定的元素的子集 $H$， ${x \in E \mid \forall \sigma \in H, \sigma(x)=x}$，
• writing $\sigma \in G / H$ 意思是我们取一个元素 $\sigma \in G$ 在每一个左胸 $H$ 在 $G$.
当 $G$ 有限群是否作用于环上 $\mathbf{B}$，为 $b \in \mathbf{B}$，我们写
$$\operatorname{Tr}G(b)=\sum{\sigma \in G} \sigma(b), \mathrm{N}G(b)=\prod{\sigma \in G} \sigma(b), \text { and } \mathrm{C}G(b)(T)=\prod{\sigma \in G}(T-\sigma(b)) .$$
如果 $G . b=\left{b_1, \ldots, b_k\right}$， ( $b_i$ 的成对不同)，我们写
$$\operatorname{Rv}{G, b}(T)=\prod{i=1}^k\left(T-b_i\right) .$$这个多项式叫做的解 $b$ (相对于 $G$ )。很明显 $\left(\operatorname{Rv}_{G, b}\right)^r=$ $\mathrm{C}_G(b)$ 用 $r=\left|G: \mathrm{St}_G(b)\right|$.

## 数学代写|交换代数代写对易代数代考|消去论

$\mathbf{k}$ -algebra。0只依赖于由$f_i$生成的$\mathbf{k}[X]$的理想$\mathfrak{a}=\left\langle f_1, \ldots, f_s\right\rangle$，我们也称它们为理想$a$的零

• 理想的$\left\langle f_1, \ldots, f_s\right\rangle_{\mathbf{k}_1}\left[X_1, \ldots, X_n\right]$只依赖于理想的$a$:它是$a$生成的$\mathbf{k}_1\left[X_1, \ldots, X_n\right]$的理想$a_1$
• 消除的理想$\mathfrak{b}_1=\mathfrak{a}_1 \backslash \mathbf{k}_1\left[X_1, \ldots, X_r\right]$只依赖于$\mathfrak{b}$:它是$\mathfrak{b}$生成的$\mathbf{k}_1\left[X_1, \ldots, X_r\right]$的理想
基本消除理论面临两个障碍。第一个是从$\mathfrak{a}$计算$\mathfrak{b}$的难度，即从多项式系统$\left(f_1, \ldots, f_s\right)$计算$\mathfrak{b}$的某个有限发电集的难度。这种计算是通过Gröbner基地的理论实现的，我们在本文中不讨论这个理论。此外，与与结果理论相关联的计算不同，这种计算不是均匀的。

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