### 数学代写|交换代数代写commutative algebra代考|MTH4312

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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|Special results for Noetherian rings

Next are some excerpts of the second classical period of Noetherian rings. Although they will have no direct impact in parts of the book, the proofs are masterpieces worth recording here.

Theorem 2.5.6 (I. S. Cohen, 1950). If every prime ideal of the ring $A$ is finitely generated then $A$ is Noetherian.

Proof. If there is some nonfinitely generated ideal then the family of such ideals is nonempty and is easily seen to be inductive. Let $P$ be a maximal element of this family.
One claims that $P$ is a prime ideal. Indeed, suppose there are $a, b \in A \backslash P$ such that $a b \in P$. In particular, $P$ is properly contained in the ideal $(P, a)$, hence by the maximality of $P$, the latter ideal is finitely generated. Clearly, one may choose a set of generators of $(P, a)$ of the form $x_{1}+b_{1} a, \ldots, x_{n}+b_{n} a$, with $x_{i} \in P, b_{i} \in A$. On the other hand, the quotient $P: a$ is also finitely generated since $b \in(P: a) \backslash P$. However, at is easy to see, $P=\left(x_{1}, \ldots, x_{n}, a(P: a)\right)$, so $P$ is finitely generated, which gives a contradiction.

By the main hypothesis of the statement, $P$ is finitely generated and this repeated contradiction shows that there could not be any nonfinitely generated ideal to start with.

The second result was proved simultaneously, but independently, by P. Eakin ([48]) and M. Nagata ([113]). About 33 years later, Nagata gave a new proof ([114]). Quite recently, a more encompassing result has been given by P. Jothilingam ([89]) The assertion of the theorem involves the notion of a (finitely generated) module, as well as the concept of integral extension, for which one refers to Chapter 3 and to Section 2.2, respectively. The proof below is the first argument given by Nagata, which still looks the clearest, if not the shortest.

Theorem 2.5.7 (Eakin-Nagata). Let A be a subring of a Noetherian ring R. If R is finitely generated as A-module, then $A$ is Noetherian.

Proof. The argument is divided in several reduction steps. As a natural start, one wishes to induct on the number of generators of $R$ as an $A$-module. The problem is that by writing $R=A b_{1}+\cdots+A b_{m}$ as a finitely generated $A$-module, for certain $b_{i} \in R$, an intermediate submodule such as $A b_{1}+\cdots+A b_{m-1}$ is Noetherian (as a submodule of a Noetherian ought to be), but has no structure of a ring. To fix it, one takes the $A$-subalgebra $A\left[b_{1}, \ldots, b_{m-1}\right]$. Clearly, the latter is still finitely generated as an $A$-module. With this, one is reduced to the following.

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

Coming back to the last two conditions in Lemma 2.5.1, there is a sort of dual statement.
Lemma 2.5.10. The following conditions for a ring $R$ are equivalent:
(i) (Descending chain condition) Every chain of ideals $I_{1} \supset I_{2} \supset \cdots$ is stationary, i. e., there exists an index $i$ such that $I_{i}=I_{i+1}=\cdots$.
(iii) (Minimum condition) Every nonempty family of ideals of $R$ has a minimal element (i. e., an ideal belonging to the family not containing properly any other ideal in the family).
The verification of this equivalence is left to the reader.
Definition 2.5.11. A ring $R$ is called Artinian or an Artin ring (after Emil Artin) if it satisfies the above equivalent conditions.

The basic theory of Artin rings is a lot more involved than the Noetherian counterpart. Here are some of its “strange” basic properties.
Proposition 2.5.12. Let $R$ be an Artin ring. Then:
(1) If $R$ is a domain, then it must be a field.
(2) Every prime ideal of $R$ is maximal.
(3) If $R$ is a local ring, then every nonunit is nilpotent.
(4) The annihilator of a proper (i. e., nonzero) minimal ideal of $R$ is a prime ideal.
(5) Any ideal of $R$ is a (finite) product of prime ideals.
(6) The set of maximal (resp., prime) ideals of an Artin ring is finite.
Proof. (1) Let $0 \neq a \in R$. Then the descending chain $(a) \supset\left(a^{2}\right) \supset \cdots$ stabilizes. Say, $\left(a^{n}\right) \subset\left(a^{n+1}\right)$. Then $a^{n}=a^{n+1} b$, for some $b \in B$. Cancelling $a^{n}$ yields $a b=1$, hence $a$ is a unit.
(2) For any ideal $I \subset R$, the ring $R / I$ is again Artinian as is easily verified. In particular, if $P \subset R$ is a prime ideal then $R / P$ is a field by (1), hence $P$ is a maximal ideal.
(3) Recall that $R$ has a unique maximal ideal $\mathrm{m}$. Let $a \in R$ be a nonunit, $l$. $e, a \in \mathrm{m}$. Consider again a stable value $\left(a^{n}\right)$ of the descending chain $(a) \supset\left(a^{2}\right) \supset \cdots \cdot$ As in the proof of (1), one gets $a^{n}(a b-1)=0$, for some $b \in R$. But $a b-1 \notin \mathfrak{m}$ since $a \in \mathfrak{m}$, i. e., $a b-1$ is a unit. It follows that $a^{n}=0$.
(4) Let $I \subset R$ be a proper minimal ideal and let $a:=0: I$ denote its annihilator. Let $c, d \in R$ such that $c d \in a$. If neither $c \in a$ nor $d \in a$ then both $c I$ and $d I$ are nonzero deals contained in $I$. By minimality, one must have $c I=I=d I$. Multiplying through by $d$, yields $c d I=d I=I \neq 0$. Therefore, $c d \notin \mathfrak{a}-$ a contradiction.
(5) If $I \subset R$ is an ideal, then $R / I$ is again Artinian, as one easily sees. Therefore, one can assume that $I={0}$.

## 数学代写|交换代数代写commutative algebra代考|Noetherian and Artinian rings

（i）（有限基础）每个理想R是有限生成的。
(ii) (升链条件) 每条理想链我1⊂我2⊂⋯是平稳的，即存在一个索引一世这样我一世=我一世+1=⋯.
(iii)（最大条件）每个非空理想族R有一个最大的元素（即，一个属于家庭的理想，没有适当地包含在家庭中的任何其他理想中）。

(二)⇒(iii) 给定一个非空的家庭F的理想R. 挑一些我属于F. 如果我是最大元素F， 完毕。否则，选择我2在F适当地包含我1:=我. 以这种方式进行，可以找到一系列适当的包含物我1⊂我2⊂⋯. 通过假设，这个序列稳定，比方说，在索引一世≥1. 然后我一世是最大元素F.
㈢⇒（我让我⊂R成为一个理想。考虑家庭F的有限生成的理想R包含在我. 清楚地，F是非空的，因为，和. g.，零理想（由空集生成）属于它。根据假设，F有一个最大元素，比如说，Ĵ⊂我. 宣称：Ĵ=我. 为了让b∈我是任意元素。然后是放大的理想Ĵ,b)仍然属于F. 但Ĵ是最大的，因此(Ĵ,b)=Ĵ， IE，b∈Ĵ.

## 数学代写|交换代数代写commutative algebra代考|Special results for Noetherian rings

P. Eakin ([48]) 和 M. Nagata ([113]) 同时但独立地证明了第二个结果。大约 33 年后，Nagata 给出了一个新的证明（[114]）。最近，P. Jothilingam ([89]) 给出了一个更全面的结果。该定理的断言涉及（有限生成的）模块的概念，以及积分扩展的概念，参见第3 和第 2.2 节，分别。下面的证明是 Nagata 给出的第一个论点，即使不是最短的，它看起来仍然是最清晰的。

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

（i）（降链条件）每条理想链我1⊃我2⊃⋯是平稳的，即存在一个索引一世这样我一世=我一世+1=⋯.
(iii) （最小条件）每一个非空的理想族R具有最小元素（即，属于该家庭的理想，不包含该家庭中的任何其他理想）。

Artin 环的基本理论比 Noetherian 的对应物要复杂得多。以下是它的一些“奇怪”的基本属性。

(1) 如果R是域，那么它一定是域。
(2) 的每个素理想R是最大的。
(3) 如果R是一个局部环，那么每个非单位都是幂零的。
(4) 一个适当的（即非零的）最小理想的湮灭子R是一个主要理想。
(5) 任何理想R是素理想的（有限）乘积。
(6) Artin 环的最大（或素数）理想集是有限的。

(2) 对于任何理想我⊂R， 戒指R/我再次是 Artinian，这很容易验证。特别是，如果磷⊂R是一个素理想R/磷是 (1) 的一个域，因此磷是一个极大理想。
(3) 回想一下R有一个独特的最大理想米. 让一个∈R成为一个非单位，l. 和,一个∈米. 再考虑一个稳定的值(一个n)降链的(一个)⊃(一个2)⊃⋯⋅如 (1) 的证明，得一个n(一个b−1)=0， 对于一些b∈R. 但一个b−1∉米自从一个∈米， IE，一个b−1是一个单位。它遵循一个n=0.
(4) 让我⊂R是一个适当的最小理想，让一个:=0:我表示它的歼灭者。让C,d∈R这样Cd∈一个. 如果两者都没有C∈一个也不d∈一个然后两者C我和d我是包含在非零交易我. 通过最低限度，一个人必须有C我=我=d我. 乘以d, 产量Cd我=d我=我≠0. 所以，Cd∉一个−一个矛盾。
(5) 如果我⊂R是一个理想，那么R/我正如人们很容易看到的那样，它又是 Artinian。因此，可以假设我=0.

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