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# 数学代写|交换代数代写Commutative Algebra代考|Dimension theory I

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## 数学代写|交换代数代写Commutative Algebra代考|Noetherian and Artinian rings

The starting principle of this part is as follows.

Lemma 2.5.1. The following conditions for a ring $R$ are equivalent:
(i) (Finite basis) Every ideal of $R$ is finitely generated.
(ii) (Ascending chain condition) Every chain of ideals $I_{1} \subset I_{2} \subset \cdots$ is stationary, i. e., there exists an index $i$ such that $I_{i}=I_{i+1}=\cdots$.
(iii) (Maximum condition) Every nonempty family of ideals of $R$ has a maximal element (i. e., an ideal belonging to the family not contained properly in any other ideal in the family).

Proof. (i) $\Rightarrow$ (ii) Let $I_{1} \subset I_{2} \subset \cdots$ be given. The set union $I:=\bigcup_{i} I_{i}$ is easily seen to be an ideal of $R$. By assumption, $I=\left(a_{1}, \ldots, a_{m}\right)$ for certain $a_{i} \in R$. Forcefully then, there is an index $i$ such that $I_{i}$ contains the set $\left{a_{1}, \ldots, a_{m}\right}$. Therefore, $I \subset I_{i}$, hence clearly $I_{i}=I_{i+1}=\cdots$.
(ii) $\Rightarrow$ (iii) Let there be given a nonempty family $\mathcal{F}$ of ideal s of $R$. Pick some $I$ belonging to $\mathcal{F}$. If $I$ is a maximal element in $\mathcal{F}$, done. Otherwise, choose $I_{2}$ in $\mathcal{F}$ properly containing $I_{1}:=I$. Proceeding this way, one finds a sequence of proper inclusions $I_{1} \subset I_{2} \subset \cdots$. By assumption, this sequence stabilizes, say, at index $i \geq 1$. Then $I_{i}$ is a maximal element in $\mathcal{F}$.
(iii) $\Rightarrow$ (i) Let $I \subset R$ be an ideal. Consider the family $\mathcal{F}$ of finitely generated ideals of $R$ contained in $I$. Clearly, $\mathcal{F}$ is nonempty since, $e . g$., the zero ideal (generated by the empty set) belongs to it. By assumption, $\mathcal{F}$ has a maximal element, say, $J \subset I$. Claim: $J=I$. For let $b \in I$ be an arbitrary element. Then the enlarged ideal $(J, b)$ still belongs to $\mathcal{F}$. But $J$ is maximal, hence $(J, b)=J$, i. e., $b \in J$.

## 数学代写|交换代数代写Commutative Algebra代考|Associated primes

Let $R$ be a ring and let $I \subset R$ be an ideal. While the set of ring homomorphisms $R \rightarrow R / I$ is quite involved, the set of $R$-linear maps $R \rightarrow R / I$ is much simpler: any such map must be given by multiplication by an element of $R / I$. This leads naturally to the following concept.

A prime ideal $P \subset R$ is called an associated prime of $R / I$ if $P$ is the kernel of an $R$-linear map $R \rightarrow R / I$.

Quite often, by abuse, an associated prime $P \subset R$ of $R / I$ is referred to as an associated prime of the ideal $I$. Since $P=0:{R / I}(\bar{x}) P=I:(x)=I:{R}(x)$, for some nonzero element $\bar{x} \in \bar{R}=R / I$, if only to emphasize the role of $I$ in place of $R / I$, one says that $P$ is the $I$-annihilator of some $x \in R \backslash I$.

If $R$ is a Noetherian ring, the family of such ideals is reasonably under control, according to the following

## 数学代写|交换代数代写COMMUTATIVE ALGEBRA代考|Krull’s principal ideal theorem

The next result is one of the cornerstones of Noetherian ring theory, if not of commutative algebra itself. The present account follows pretty much Krull’s original argument. One needs the not less famous preliminary result.

Let $R$ be a ring with Jacobson radical $\mathfrak{N}$ and let $\mathfrak{a} \subset R$ be a finitely generated ideal. If $\mathfrak{a} \subset \mathfrak{N a}$, then $\mathfrak{a}={0}$.

Proof. Assuming $\mathfrak{a} \neq{0}$, let $\left{a_{1}, \ldots, a_{n}\right}$ be a set of generators of $\mathfrak{a}$ with $n \geq 1$. By assumption, one can write $a_{1}=b_{1} a_{1}+\cdots+b_{n} a_{n}$, for suitable $b_{i} \in \mathfrak{N}$. It follows that $\left(1-b_{1}\right) a_{1} \in\left(a_{2}, \ldots, a_{n}\right)$, hence $a_{1} \in\left(a_{2}, \ldots, a_{n}\right)$ as $1-b_{1}$ is invertible. Thus one can always reduce any set of generators, and eventually get $\mathfrak{a}={0}$.

A more general version of this lemma will be given in Section $5.1$ that bears the names of three mathematicians.

## 数学代写|交换代数代写COMMUTATIVE ALGEBRA代考|NOETHERIAN AND ARTINIAN RINGS

(i) (Finite basis) Every ideal of $R$ is finitely generated.
(ii) (Ascending chain condition) Every chain of ideals $I_{1} \subset I_{2} \subset \cdots$ is stationary, i. e., there exists an index $i$ such that $I_{i}=I_{i+1}=\cdots$.
(iii) (Maximum condition) Every nonempty family of ideals of $R$ has a maximal element (i. e., an ideal belonging to the family not contained properly in any other ideal in the family).

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