## 数学代写|交换代数代写commutative algebra代考|The Classical Nullstellensatz

The Nullstellensatz is a theorem which concerns the systems of polynomial equations over a discrete field. Very informally, its meaning can be described as follows: a geometric statement necessarily possesses an algebraic certificate. Or even: a proof in commutative algebra can (almost) always be summarized by simple algebraic identities if it is sufficiently general.

If we have discrete fields $\mathbf{K} \subseteq \mathbf{L}$, and if $(f)=\left(f_1, \ldots, f_s\right)$ is a system of polynomials in $\mathbf{K}\left[X_1, \ldots, X_n\right]=\mathbf{K}[X]$, we say that $\left(\xi_1, \ldots, \xi_n\right)=(\xi)$ is a zero of $(f)$ in $\mathbf{L}^n$, or a zero of $(f)$ with coordinates in $\mathbf{L}$, if the equations $f_i(\xi)=0$ are satisfied. Let $\mathfrak{f}=\left\langle f_1, \ldots, \bar{f}s\right\rangle{\mathbf{K}[X]}$. Then, all the polynomials $g \in f$ are annihilated in such a $(\xi)$. We therefore equally refer to $(\xi)$ as a zero of the ideal $f$ in $\mathbf{L}^n$ or as having coordinates in $\mathbf{L}$.
We begin with an almost obvious fact.
9.1 Fact Let $\mathbf{k}$ be a commutative ring and $h \in \mathbf{k}[X]$ a monic polynomial of degree $\geqslant 1$.

• If some multiple of $h$ is in $\mathbf{k}$, this multiple is null.
• Let $f$ and $g \in \mathbf{k}[X]$ of respective formal degrees $p$ and $q$. If $h$ divides $f$ and $g$, then $\operatorname{Res}_X(f, p, g, q)=0$.
We now present a generalization of the basic elimination Lemma 7.5.
9.2 Lemma (Elimination of a variable between several polynomials) Let $f, g_1, \ldots$, $g_r \in \mathbf{k}[X](r \geqslant 1)$, with $f$ monic of degree $d$.

Let $\mathfrak{f}=\left\langle f, g_1, \ldots, g_r\right\rangle$ and $\mathfrak{a}=\mathfrak{f} \cap \mathbf{k}$ (this is the elimination ideal of the variable $X$ in f). Also let
$$\begin{gathered} g(T, X)=g_1+T g_2+\cdots+T^{r-1} g_r \in \mathbf{k}[T, X], \ R(T)=R\left(f, g_1, \ldots, g_r\right)(T)=\operatorname{Res}X(f, g(T, X)) \in \mathbf{k}[T], \ \mathfrak{b}=\Re\left(f, g_1, \ldots, g_r\right) \stackrel{\text { def }}{=} \mathrm{c}{\mathbf{k}, T}\left(R\left(f, g_1, \ldots, g_r\right)(T)\right) \subseteq \mathbf{k} . \end{gathered}$$

1. The ideal $\mathfrak{b}$ is generated by $d(r-1)+1$ elements and we have the inclusions
$$\mathfrak{b} \subseteq \mathfrak{a} \subseteq \sqrt{\mathfrak{b}}=\sqrt{\mathfrak{a}} .$$
More precisely, let $e_i=1+(d-i)(r-1), i \in \llbracket 1 . . d \rrbracket$, then for arbitrary elements $a_1, \ldots, a_d \in \mathfrak{a}$, we have
$$a_1^{e_1} a_2^{e_2} \cdots a_d^{e_d} \in \Re\left(f, g_1, \ldots, g_r\right)$$

## 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|经典的《零》

Nullstellensatz是一个定理，它与离散场上的多项式方程组有关。非正式地说，它的含义可以这样描述:一个几何命题必须具有代数证明。甚至:交换代数中的一个证明，如果它足够普遍，(几乎)总是可以用简单的代数恒等式来概括

• 如果$\mathbf{k}$中有$h$的某个倍数，则该倍数为空。
• 分别让$f$和$g \in \mathbf{k}[X]$的正式学位$p$和$q$。如果$h$除以$f$和$g$，则$\operatorname{Res}_X(f, p, g, q)=0$ .
我们现在提出基本消去引理7.5.
9.2引理(消去几个多项式之间的变量)令$f, g_1, \ldots$, $g_r \in \mathbf{k}[X](r \geqslant 1)$，用$f$的次$d$的monic。

.

$$\begin{gathered} g(T, X)=g_1+T g_2+\cdots+T^{r-1} g_r \in \mathbf{k}[T, X], \ R(T)=R\left(f, g_1, \ldots, g_r\right)(T)=\operatorname{Res}X(f, g(T, X)) \in \mathbf{k}[T], \ \mathfrak{b}=\Re\left(f, g_1, \ldots, g_r\right) \stackrel{\text { def }}{=} \mathrm{c}{\mathbf{k}, T}\left(R\left(f, g_1, \ldots, g_r\right)(T)\right) \subseteq \mathbf{k} . \end{gathered}$$

$$\mathfrak{b} \subseteq \mathfrak{a} \subseteq \sqrt{\mathfrak{b}}=\sqrt{\mathfrak{a}} .$$

$$a_1^{e_1} a_2^{e_2} \cdots a_d^{e_d} \in \Re\left(f, g_1, \ldots, g_r\right)$$

## 数学代写|交换代数代写交换代数代考|数场理想的乘法理论

8.19定义一个环的理想a $\mathbf{A}$被认为是可逆的，如果存在一个理想$\mathfrak{b}$和一个正则元$a$，使得$\mathfrak{a} \mathfrak{b}=\langle a\rangle$ .
8.20事实假设a是一个环的可逆理想$\mathbf{A}$ .

1. 理想a有限生成
2. $\mathfrak{a}$ 由 $k$ 元素和if $\mathfrak{a} \mathfrak{b}=\langle a\rangle$ 那就买普通的吧 $\mathfrak{b}$ 由 $k$ 元素。此外 $\mathfrak{b}=(\langle a\rangle: a)$
3. 我们有规则a $\subseteq \subseteq \mathfrak{a} \Rightarrow \mathrm{c} \subseteq \mathfrak{o}$ 为了所有的理想 $\mathrm{c}$ 和 $\mathrm{D}$.
4. $\mathfrak{c} \subseteq$ 有唯一存在 $\mathfrak{d}$ 如此这般 $\mathfrak{d} \mathfrak{a}=\mathfrak{c}$，即 $\mathfrak{d}=(\mathfrak{c}: \mathfrak{a})$，如果 $\mathfrak{c}$ 是有限生成的，是吗 $\mathfrak{0}$.

D;如果$\mathfrak{a} \mathfrak{c} \subseteq \mathfrak{a} \mathfrak{d}$乘以$\mathfrak{b}$，我们得到$a \mathfrak{c} \subseteq a \mathfrak{d}$，由于$a$是正则的，这意味着$\mathbf{c} \subseteq \mathfrak{0}$

1. 如果$\mathfrak{a} \mathfrak{b}=\langle a\rangle$，我们找到两个有限生成的理想$\mathfrak{a}1 \subseteq \mathfrak{a}$和$\mathfrak{b}_1 \subseteq \mathfrak{b}$，这样$a \in \mathfrak{a}_1 \mathfrak{b}_1$和$\mathfrak{a} \mathfrak{b}=\langle a\rangle \subseteq \mathfrak{a}_1 \mathfrak{b}_1 \subseteq \mathfrak{a} \mathfrak{b}_1 \subseteq \mathfrak{a} b{\text {. From the above, we deduce }}$等式$\mathfrak{a}_1 \mathfrak{b}_1=\mathfrak{a} \mathfrak{b}_1=\mathfrak{a} \mathfrak{b}$。从第3项$\mathfrak{b}=\mathfrak{b}_1$。类似地，$\mathfrak{a}=\mathfrak{a}_1$ .
2. 如果$\mathfrak{a}=\left\langle a_1, \ldots, a_k\right\rangle$，我们发现$b_1, \ldots, b_k \in \mathfrak{b}$使$\sum_i a_i b_i=a$ .
通过第1项与$\mathfrak{a}_1=\mathfrak{a}$和$\mathfrak{b}_1=\left\langle b_1, \ldots, b_k\right\rangle$的推理，我们得到等式

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