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

In this section, one confronts the role of the associated primes with the theory of primary decomposition. The latter was a brilliant achievement of Emmy Noether and constitutes a great simplification of prime and primary ideal theory in Noetherian rings.
2.6.1 The nature of the components
The basic insight of E. Noether consisted in starting out with a stronger notion than that of a primary ideal.

Definition 2.6.1. Let $R$ be a ring. An ideal $I \subset R$ is irreducible if it is not the proper intersection of two ideals, that is, whenever there are ideals $I_{1}, I_{2} \subset R$ such that $I=$ $I_{1} \cap I_{2}$, then either $I_{1}=I$ or $I_{2}=I$.

The terminology is inspired from the classical case of an irreducible polynomial. Noether showed the following.
Lemma 2.6.2. Let $R$ denote a Noetherian ring. Then:
(1) (Satz II) Any ideal is the intersection of a finite set of irreducible ideals.
(2) (Satz VI) An irreducible ideal is primary.
Proof. (1) $\Lambda s s u m e ~ t h e ~ a s s e r t i o n ~ i s ~ f a l s e, ~ s o ~ t h e ~ f a m i l y ~ o f ~ i d e a l s ~ o f ~$ sertion fails has a maximal element $I \subset R$. Since $I$ is not irreducible, one must have an intersection $I=I_{1} \cap I_{2}$ where both factors contain $I$ properly. By the maximality of $I$, both $I_{1}$ and $I_{2}$ must be finite intersections of irreducible ideals, hence so is $I$-a contradiction.
(2) Let $I \subset R$ be irreducible, but not primary. By definition, there dre elements $a, b \in R$ such that $a b \in I$, but neither $a \in I$ nor $b^{l} \in I$ for every integer $l \geq 1$. Then the ideals $(I, a)$ and $\left(I, b^{l}\right)$ (for all $l \geq 1$ ) contain $I$ properly.

## 数学代写|交换代数代写commutative algebra代考|The Lasker–Noether fundamental theorem

Theorem 2.6.3 (Primary decomposition). Let $I \subset R$ be an ideal of a Noetherian ring $R$. For any reduced primary decomposition $I=\bigcap_{t=1}^{m} \mathcal{P}_{t}$, one has:

(a) $\left{\sqrt{\mathcal{P}{1}}, \ldots, \sqrt{\mathcal{P}{m}}\right}=\operatorname{Ass}(R / I)$.
(b) For any other reduced primary decomposition $I=\bigcap_{i=1}^{m} \mathcal{Q}{i}$, one has $$\left{\mathcal{P}{i} \mid \sqrt{\mathcal{P}{i}} \in \operatorname{Min}(R / I)\right}=\left{\mathcal{Q}{i} \mid \sqrt{\mathcal{Q}{i}} \in \operatorname{Min}(R / I)\right}$$ Proof. Let $I=\bigcap{I} \mathcal{P}{i}$ stand for a reduced primary decomposition and set $P{i}=\mathcal{P}{i}$. (a) Let $P \in \operatorname{Ass}(R / I)$ be an associated prime of $R / I$. Say, $P=I:(x)$, for some $x \in R \backslash I$ (cf. Definition 2.5.17). Then $P=\bigcap{i}\left(\mathcal{P}{i}:(x)\right.$ ). Note that $\mathcal{P}{i}:(x)$ is again $P_{i}$-primary if $x \notin \mathcal{P}{i}$, else it is $(1)=R$. Thus, passing to radicals, $P$ is the intersection of a (necessarily, nonempty) finite subset of the set $\left{P{i}\right}_{i}$. Therefore, $P$ must coincide with one of these prime ideals.

Conversely, let $P$ denote the radical of a primary component. In order to show that $P \in \operatorname{Ass}(R / I)$, it suffices to show that $P_{P} \in \operatorname{Ass}\left(R_{P} / I_{P}\right)$. Changing notation, one can now assume that $R$ is local, with unique maximal ideal $\mathrm{m}$, and $I \subset \mathrm{m}$ admits a reduced primary decomposition with an m-primary component $\mathcal{M}$. One wishes to show that $m \in \operatorname{Ass}(R / I)$

Now, the radical of any other primary component is a prime ideal contained in $\mathrm{m}$. Let $x \in \mathfrak{M} \backslash \mathcal{M}$ be an element contained in every other primary component, a choice granted by the reduced nature of the primary decomposition. Then $I:(x)=\mathcal{M}:(x)$, hence $I:(x)$ is m-primary. By Proposition 2.5.18(i), there is an element $y \in \mathrm{m}$ such that $I$ : $(y)$ is an associated prime of $R / I$. But then $I:(y)$ contains a power of $m$, so necessarily $I:(y)=m$. Therefore, $\mathfrak{m} \in \operatorname{Ass}(R / I)$, as was to be shown.
(b) Given $P \in \operatorname{Min}(R / I)$, localizing at $P$, clearly $I_{p}=\mathcal{P}{p}$, where $\mathcal{P}$ denotes the corresponding primary component of the given primary decomposition. If $\mathcal{Q}$ is the $P$-primary component of another reduced primary decomposition, one must have the equality $\mathcal{P}{P}=\mathcal{Q}_{P}$, locally of two $P$-primary ideals. It follows that they are also equal over $R$ (cf. Proposition 2.1.4).

Remark 2.6.4. It had been realized by Noether, if not earlier by Lasker, that the nonminimal primary components in a reduced primary decomposition of an ideal are not uniquely determined by the ideal. Even worse, to any embedded associated prime of $R / I$ there usually correspond infinitely many distinct primary components. Noether gave the following simple example on a footnote of her paper: $I=\left(X^{2}, X Y\right) \subset k[X, Y]$ ( $k$ an infinite field). Then $I=(X) \cap\left(X^{2}, Y+a X\right)$ is a reduced primary decomposition for any $a \in k$.

## 数学代写|交换代数代写commutative algebra代考|Basics on the underlying graded structures

A more comprehensive treatment of graded structures will be considered in Chapter $7 .$ Here, one focus on the following special setup: $R:=k\left[x_{0}, \ldots, x_{n}\right]$ stands for a polyno-mial ring over a field $k$. One endows $R$ with a structure of graded ring, by which one means the decomposition
$$R=\bigoplus_{t \geq 0} R_{t}, \quad R_{t}=k x_{0}^{t}+k x_{0}^{t-1} x_{1}+\cdots+k x_{n}^{t} \subset R .$$
The $k$-vector space $R_{t}$, spanned by the homogeneous polynomials of degree $t$, is called the $t$ th graded part of $R$. An ideal $I \subset R$ is homogeneous if it can be generated by homogeneous polynomials or, equivalently, if $I=\bigoplus_{t \geq 0} I_{t}$, where $I_{t}:=I \cap R_{t}$.

Often a homogeneous polynomial of degree $t$ will be called a $t$-form. Given a homogeneous ideal $I$, an important related degree is the initial degree of $I$, defined to be the least $t \geq 0$ such that $I_{t} \neq 0$.

Perhaps the first feature of homogeneous ideals is that the property of being prime or primary can be verified solely by using homogeneous test elements.

Lemma 2.7.1. Let $I \subset R$ denote a homogeneous ideal. Then $I$ is prime (resp., primary) if given homogeneous elements $f, g \in R$ such that $f g \in I$ then either $f \in I$ or else $g \in I$ (resp., $g^{\ell} \in I$, for some $\ell \geq 1$ ).

Proof. One proves the case of a prime ideal, leaving the case of a primary ideal to the reader as being similarly handled. Here, one argues with the initial degree of a polynomial. Let $f, g \in R$ such that $f \in I$ nor $g \in I$. Write $f=f_{u}+\cdots, g=g_{v}+\cdots$, with $f_{u} \neq 0$, $g_{v} \neq 0$. Let $f_{u+u_{0}}$ and $g_{v+v_{0}}$ denote the respective first homogeneous constituents not belonging to $I$. By assumption on homogeneous test elements, one has $f_{u+u_{0}} g_{v+v_{0}} \notin I$. By homogeneity of $I$, it follows that
$$\left(f-\left(f_{u}+\cdots+f_{u+u_{0}-1}\right)\right)\left(g-\left(g_{v}+\cdots+g_{v+v_{0}-1}\right)\right) \notin I .$$
But since by construction, $f-\left(f_{u}+\cdots+f_{u+u_{0}-1}\right)$ and $g-\left(g_{v}+\cdots+g_{v+v_{0}-1}\right)$ belong to $I$, necessarily $f g \notin I$, as was to be shown.
One next collects the main operationwise properties of homogeneous ideals.

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

2.6.1 组成部分的本质
E. Noether 的基本见解在于从一个比基本理想更强大的概念开始。

(1) (Satz II) 任何理想都是一组有限的不可约理想的交集。
(2) (Satz VI) 一个不可约的理想是首要的。

(2) 让我⊂R是不可约的，但不是主要的。根据定义，有 dre 元素一个,b∈R这样一个b∈我, 但两者都不一个∈我也不bl∈我对于每个整数l≥1. 然后是理想(我,一个)和(我,bl)（对所有人l≥1） 包含我适当地。

## 数学代写|交换代数代写commutative algebra代考|The Lasker–Noether fundamental theorem

（一个）\left{\sqrt{\mathcal{P}{1}}, \ldots, \sqrt{\mathcal{P}{m}}\right}=\operatorname{Ass}(R / I)\left{\sqrt{\mathcal{P}{1}}, \ldots, \sqrt{\mathcal{P}{m}}\right}=\operatorname{Ass}(R / I).
(b) 对于任何其他简化的初级分解我=⋂一世=1米问一世, 一个有

\left{\mathcal{P}{i} \mid \sqrt{\mathcal{P}{i}} \in \operatorname{Min}(R / I)\right}=\left{\mathcal{Q}{ i} \mid \sqrt{\mathcal{Q}{i}} \in \operatorname{Min}(R / I)\right}\left{\mathcal{P}{i} \mid \sqrt{\mathcal{P}{i}} \in \operatorname{Min}(R / I)\right}=\left{\mathcal{Q}{ i} \mid \sqrt{\mathcal{Q}{i}} \in \operatorname{Min}(R / I)\right}证明。让我=⋂我磷一世代表简化的初级分解和集合磷一世=磷一世. (a) 让磷∈屁股⁡(R/我)是的关联素数R/我. 说，磷=我:(X)， 对于一些X∈R∖我（参见定义 2.5.17）。然后磷=⋂一世(磷一世:(X)）。注意磷一世:(X)又是磷一世- 主要的如果X∉磷一世, 否则是(1)=R. 因此，传递给自由基，磷是集合的（必然是非空的）有限子集的交集\left{P{i}\right}_{i}\left{P{i}\right}_{i}. 所以，磷必须符合这些主要理想之一。

(b) 给定磷∈敏⁡(R/我), 定位于磷， 清楚地我p=磷p， 在哪里磷表示给定初级分解的相应初级分量。如果问是个磷-另一个简化的初级分解的初级组件，必须具有相等性磷磷=问磷, 局部的两个磷——基本理想。因此它们也相等于R（参见提案 2.1.4）。

## 数学代写|交换代数代写commutative algebra代考|Basics on the underlying graded structures

R=⨁吨≥0R吨,R吨=ķX0吨+ķX0吨−1X1+⋯+ķXn吨⊂R.

(F−(F在+⋯+F在+在0−1))(G−(G在+⋯+G在+在0−1))∉我.

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