### 数学代考|代数几何代写algebraic geometry代考|Primary Decomposition

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## 数学代考|代数几何代写algebraic geometry代考|Primary Decomposition

An ideal $q$ in a ring $R$ is called primary if $q \neq R$, and whenever $x y \in q$, we have either $x \in q$ or $y^{n} \in q$ for some $n \in \mathbb{N}$. This definition may seem unnatural at first, because of its asymmetry. In particular, it is not the same thing as a power of a prime ideal (see Exercises 27 and 28). It is, however, obviously true that for a primary ideal $q$, the radical $p=\sqrt{q}$ is prime. We often call $q$ a $p$-primary ideal.

It turns out that the concept of a primary ideal behaves better than many similar notions. Perhaps it could be motivated by noting that an ideal $q$ is primary if and only if in the ring $R / q$, every zero divisor $x$ (which, recall, means a non-zero element $x$ for which there is a nonzero element $y$ with $x y=0$ ) is nilpotent (i.e. satisfies $x^{n}=0$ for some $n \in \mathbb{N}$ ).
Note that this implies the following
5.1.1 Lemma If $q$ is an ideal in a ring $R$ such that $m=\sqrt{q}$ is a maximal ideal, then $q$ is $m$-primary.

Proof Every element of the image $\bar{m}$ of $m$ in $R / q$ is, by assumption, nilpotent. Therefore $\bar{m}=\operatorname{Nil}(R / q)=J a c(R / q)$. Therefore, the ring $R / q$ is local, and every element not in $\bar{m}$ is a unit, and hence cannot be a zero divisor. Thus, every zero divisor in $R / q$ is nilpotent, as we needed to prove.

To further demonstrate the utility of primary ideals, consider the concept of decomposition of ideals: A decomposition of an ideal $I \neq R$ in a ring $R$ is an expression of the form
$$I=J_{1} \cap \cdots \cap J_{n}$$
where $J_{1}, \ldots, J_{n} \neq R$ are ideals. An ideal $I \neq R$ is called indecomposable if it cannot be expressed as $I=J \cap K$ for ideals $J, K \supsetneq I$. Recall that since a Noetherian ring satisfies the ascending chain condition $(\mathrm{ACC}$ ) with respect to ideals, there cannot be an infinite sequence of ideals
$$I_{1} \subsetneq I_{2} \subsetneq I_{3} \ldots \ldots$$

## 数学代考|代数几何代写algebraic geometry代考|Artinian Rings

A commutative ring $R$ is called Artinian if its ideals satisfy the descending chain condition (or DCC), i.e. if every sequence of ideals
$$I_{1} \supseteq I_{2} \supseteq \ldots$$
in $R$ is eventually constant. We will see that this is actually a very restrictive condition (more so than the ACC for ideals).
5.2.1 Lemma An integral domain $R$ which is Artinian is a field.
Proof Let $0 \neq x \in R$. Then by the DCC, $\left(x^{n}\right)=\left(x^{n+1}\right)$ for some $n \in \mathbb{N}$. Therefore, $x^{n}$ is a multiple of $x^{n+1}$, and since $R$ is an integral domain, $x$ is a unit.

Since a quotient of an Artinian ring is obviously Artinian, every prime ideal in an Artinian ring $R$ is maximal, and hence $\operatorname{dim}(R)=0$. Also, obviously, $R$ satisfies (4.3.3).
5.2.2 Lemma The nilradical of an Artinian ring $R$ is nilpotent, i.e. there exists $a k \mathbb{N}$ such that $\operatorname{Nil}(R)^{k}=0$.

Proof By the DCC, there is some $k \in \mathbb{N}$ such that $a=\operatorname{Nil}(R)^{k}=\operatorname{Nil}(R)^{k+1}$. We will show that $a=0$. Assume this is false. Note that then $a \cdot a \neq 0$, i.e. there exists an element $x \in a$ with $x \cdot a \neq 0$. By the DCC, we may further assume that if this is also true with $x$ replaced by $x y$ for some $y \in R$, then $(x)=(x y)$. But now if $x \cdot a \neq 0$, then $x \cdot a \cdot a=x \cdot a \neq 0$, so indeed, there exists a $y \in a$ such that $x y \cdot a \neq 0$. Thus, $(x)=(x y)$, and inductively, $(x)=\left(x y^{n}\right)$ for every $n \in \mathbb{N}$. However, $y$ is by assumption nilpotent, and hence $x=0$, which is a contradiction.
5.2.3 Proposition An Artinian ring $R$ is a product of finitely many local Artinian rings.
(Note that since a product of finitely many Artinian rings is obviously Artinian, this is an if and only if condition.)

Proof By Lemma 5.2.1, every prime ideal of $R$ is maximal. Thus, Nil( $R$ ) is an intersection of maximal ideals. By the DCC, it is an intersection of finitely many maximal ideals:
$$\operatorname{Nil}(R)=m_{1} \cap \cdots \cap m_{n} .$$

## 数学代考|代数几何代写algebraic geometry代考|Dimension

Let $A$ be a Noetherian local ring with maximal ideal $m$ and an $m$-primary ideal $q$ with $s$ generators. We are interested in studying powers of the ideal $q$, but for inductive purposes, a more general concept must be introduced. Let $M$ be a finitely generated $A$-module. A $q$-stable filtration on $M$ is a sequence $\mathcal{M}$ of submodules
$$M=M_{0} \supseteq M_{1} \supseteq M_{2} \supseteq M_{3} \supseteq \ldots$$
such that
$$q M_{i} \subseteq M_{i+1}$$
for all $i \in \mathbb{N}$, and there exists a $k$ such that equality arises for all $i \geq k$. The key point about $q$-stable filtrations is the following
5.3.1 Lemma (Artin-Rees Lemma) Let $M$ be a finitely generated A-module with a $q$ stable filtration $\mathcal{M}$, and let $N \subseteq M$ be a submodule. Then the submodules $N_{i}=M_{i} \cap N$ form a q-stable filtration on $N($ denoted by $\mathcal{M} \cap N)$.
Proof Consider the ring
$$A^{}=\bigoplus_{i \in \mathbb{N}_{0}} q^{i}$$ (where we set $q^{0}=A$ ). The ring structure is by the product from $q^{i}$ and $q^{j}$ to $q^{i+j}$. Then the ring $A^{}$ is a finitely generated $A$-algebra, and hence is Noetherian by the Hilbert basis

theorem, and
$$M^{}=\bigoplus_{i \in \mathbb{N}{0}} M{i}$$
is a finitely generated module (since the filtration on $M$ is $q$-stable). Now consider the submodules
$$N_{k}^{}=\bigoplus_{i \leq k}\left(N \cap M_{k}\right) \oplus \bigoplus_{j \in \mathbb{N}} q^{j}\left(N \cap M_{k}\right) .$$
We have $N_{k}^{} \subseteq N_{k+1}^{}$, so by the $\mathrm{ACC}$, equality arises for large enough $k$, which is what we were trying to prove.

We are interested in measuring the growth of the $A$-modules $M / M_{k}$. Since the ring $A / q$ is Artinian, the finitely generated $A / q$-modules $M_{i} / M_{i+1}$ have finite length, and hence the $A$-modules $M / M_{k}$ have finite length. We put
$$\chi_{q}^{\mathcal{M}}(k)=\ell\left(M / M_{k}\right) .$$

## 数学代考|代数几何代写algebraic geometry代考|Primary Decomposition

5.1.1 引理 Ifq是环中的理想R这样米=q是一个极大理想，那么q是米-基本的。

## 数学代考|代数几何代写algebraic geometry代考|Artinian Rings

5.2.1 引理 积分域R这是 Artinian 是一个领域。

5.2.2 引理 Artinian 环的零根R是幂零的，即存在一种ķñ这样零⁡(R)ķ=0.

5.2.3 命题 Artinian 环R是有限多个局部 Artinian 环的乘积。
（请注意，由于有限多个 Artinian 环的乘积显然是 Artinian，因此这是一个当且仅当条件。）

## 数学代考|代数几何代写algebraic geometry代考|Dimension

q米一世⊆米一世+1

5.3.1 引理 (Artin-Rees Lemma) Let米是一个有限生成的 A 模q稳定过滤米， 然后让ñ⊆米成为一个子模块。然后是子模块ñ一世=米一世∩ñ形成一个q-稳定的过滤ñ(表示为米∩ñ).

ñķ=⨁一世≤ķ(ñ∩米ķ)⊕⨁j∈ñqj(ñ∩米ķ).

χq米(ķ)=ℓ(米/米ķ).

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