### 数学代写|交换代数代写commutative algebra代考|MTH 7059

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## 数学代写|交换代数代写commutative algebra代考|The Artin-Rees Lemma

We will prove the Artin-Rees Lemma in order to derive Krull’s Intersection Theorem from it. The latter in turn is a basic ingredient needed for characterizing the Krull dimension of Noetherian rings in Section 2.4.

Lemma 1 (Artin-Rees Lemma). Let $R$ be a Noetherian ring, $\mathfrak{a} \subset R$ an ideal, $M$ a finitely generated $R$-module, and $M^{\prime} \subset M$ a submodule. Then there exists an integer $k \in \mathbb{N}$ such that
$$\left(\mathfrak{a}^{i} M\right) \cap M^{\prime}=\mathfrak{a}^{i-k}\left(\mathfrak{a}^{k} M \cap M^{\prime}\right)$$
for all exponents $i \geq k$.
Postponing the proof for a while, let us give some explanations concerning this lemma. The descending sequence of ideals $\mathfrak{a}^{1} \supset \mathfrak{a}^{2} \supset \ldots$ defines a topology on $R$, the so-called a-adic topology; see $6.1 / 3$ for the definition of a topology. Indeed, a subset $E \subset R$ is called open if for every element $x \in E$ there exists an exponent $i \in \mathbb{N}$ such that $x+\mathfrak{a}^{i} \subset E$. Thus, the powers $\mathfrak{a}^{i}$ for $i \in \mathbb{N}$ are the basic open neighborhoods of the zero element in $R$. In a similar way, one defines the $\mathfrak{a}$-adic topology on any $R$-module $M$ by taking the submodules $\mathfrak{a}^{i} M$ for $i \in \mathbb{N}$ as basic open neighborhoods of $0 \in M$. Now if $M^{\prime}$ is a submodule of $M$, we may restrict the a-adic topology on $M$ to a topology on $M^{\prime}$ by taking the intersections $\mathfrak{a}^{i} M \cap M^{\prime}$ as basic open neighborhoods of $0 \in M^{\prime}$. Thus, a subset $E \subset M^{\prime}$ is open if and only if for every $x \in E$ there exists an exponent $i \in \mathbb{N}$ such that $x+\left(\mathfrak{a}^{i} M \cap M^{\prime}\right) \subset E$

However, on $M^{\prime}$ the a-adic topology exists as well and we may try to compare both topologies. Clearly, since $\mathfrak{a}^{i} M^{\prime} \subset \mathfrak{a}^{i} M \cap M^{\prime}$, any subset $E \subset M^{\prime}$ that is open with respect to the restriction of the a-adic topology on $M$ to $M^{\prime}$ will be open with respect to the a-adic topology on $M^{\prime}$. Moreover, in the situation of the Artin-Rees Lemma, both topologies coincide, as follows from the inclusions
$$\left(\mathfrak{a}^{i} M\right) \cap M^{\prime}=\mathfrak{a}^{i-k}\left(\mathfrak{a}^{k} M \cap M^{\prime}\right) \subset \mathfrak{a}^{i-k} M^{\prime}$$
for $i \geq k$.

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

In order to define the dimension of a ring $R$, we use strictly ascending chains $\mathfrak{p}{0} \subsetneq \mathfrak{p}{1} \subsetneq \ldots \subsetneq \mathfrak{p}{n}$ of prime ideals in $R$, where the integer $n$ is referred to as the length of the chain. Remark 1. Let $R$ be a ring and $\mathfrak{p} \subset R$ a prime ideal. Then: (i) The chains of prime ideals in $R$ starting with $\mathfrak{p}$ correspond bijectively to the chains of prime ideals in $R / \mathfrak{p}$ starting with the zero ideal. (ii) The chains of prime ideals in $R$ ending with $\mathfrak{p}$ correspond bijectively to the chains of prime ideals in the localization $R{\mathfrak{p}}$ ending with $\mathfrak{p} R_{\mathfrak{p}}$.
Proof. Assertion (i) is trivial, whereas (ii) follows from $1.2 / 5$.
$$\mathfrak{p}{0} \subsetneq \mathfrak{p}{1} \subsetneq \ldots \subseteq \mathfrak{p}{n},$$ where the $\mathfrak{p}{i}$ are prime ideals in $R$, is denoted by $\operatorname{dim} R$ and called the Krull dimension or simply the dimension of $R$.

For example, fields are of dimension 0, whereas a principal ideal domain is of dimension 1, provided it is not a field. In particular, we have $\operatorname{dim} \mathbb{Z}=1$, as well as $\operatorname{dim} K[X]=1$ for the polynomial ring over a field $K$. Also we know that $\operatorname{dim} K\left\lfloor X_{1}, \ldots, X_{n}\right\rfloor \geq n$, since the polynomial ring $K\left\lfloor X_{1}, \ldots, X_{n}\right\rfloor$ contains the chain of prime ideals $0 \subsetneq\left(X_{1}\right) \subsetneq\left(X_{1}, X_{2}\right) \subsetneq \ldots \subsetneq\left(X_{1}, \ldots, X_{n}\right)$. In fact, we will show $\operatorname{dim} K\left[X_{1}, \ldots, X_{n}\right]=n$ in Corollary 16 below. Likewise, the polynomial ring $K\left[X_{1}, X_{2}, \ldots\right]$ in an infinite sequence of variables is of infinite dimension, whereas the zero ring 0 is a ring having dimension $-\infty$ since, by convention, the supremum over an empty subset of $\mathbb{N}$ is $-\infty$. Any non-zero ring contains at least one prime ideal and therefore is of dimension $\geq 0$.

## 数学代写|交换代数代写commutative algebra代考|Background and Overview

Recall that an extension of fields $K \subset L$ is called algebraic if each element $x \in L$ satisfies an algebraic equation over $K$, i.e. an equation of type
$$x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=0$$
for suitable coefficients $a_{i} \in K$. Replacing $K \longrightarrow L$ by an arbitrary (not necessarily injective) ring homomorphism $\varphi: R \longrightarrow R^{\prime}$, equations of the just mentioned type are still meaningful; they are referred to as integral equations. Furthermore, $R^{\prime}$ is said to be integral over $R$ (via $\varphi$ ) if every element $x \in R^{\prime}$ satisfies an integral equation over $R$.

The fact that any finite extension of fields is algebraic, is fundamental in field theory. In $3.1 / 5$ we will generalize this result to ring extensions and show that any ring homomorphism $\varphi: R \longrightarrow R^{\prime}$ which is finite in the sense that it equips $R^{\prime}$ with the structure of a finite $R$-module, is integral. We obtain this assertion from a quite general characterization of integral dependence in terms of a module setting; see Lemma 3.1/4. The proof is based on Cramer’s rule and is much more laborious than in the field case.

To give a simple example illustrating a basic application of Lemma 3.1/4, consider the polynomial ring $R[X]$ in one variable $X$ over a non-zero ring $R$ and fix a monic polynomial
$$f=X^{n}+a_{1} X^{n-1}+\ldots+a_{n} \in R[X]$$
with coefficients $a_{i} \in R$. For a second variable $Y$ look at the canonical morphism
$$\varphi: R[Y] \longrightarrow R[X], \quad Y \longmapsto f .$$
We claim that $\varphi$ is finite and, hence, integral. Of course, the equation
$$X^{n}+a_{1} X^{n-1}+\ldots+\left(a_{n}-\varphi(Y)\right)=0$$
shows that $X$ is integral over $R[Y]$. From this we conclude by induction that the $R[Y]$-submodule generated by $X^{0}, \ldots, X^{n-1}$ in $R[X]$ contains all powers of $X$ and, hence, coincides with $R[X]$. In other words, $\varphi$ is finite. Alternatively,we could have derived this fact directly from $3.1 / 4$ (ii). Furthermore, using $3.1 / 4$ (iii) or $3.1 / 5$, it follows that $\varphi$ is integral. The latter is a non-trivial fact which cannot be derived by a direct ad hoc computation.

## 数学代写|交换代数代写commutative algebra代考|The Artin-Rees Lemma

(一个一世米)∩米′=一个一世−ķ(一个ķ米∩米′)

(一个一世米)∩米′=一个一世−ķ(一个ķ米∩米′)⊂一个一世−ķ米′

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

p0⊊p1⊊…⊆pn,在哪里p一世是最理想的R, 表示为暗淡⁡R并称为克鲁尔维度或简称为R.

## 数学代写|交换代数代写commutative algebra代考|Background and Overview

Xn+一个1Xn−1+…+一个n=0

F=Xn+一个1Xn−1+…+一个n∈R[X]

Xn+一个1Xn−1+…+(一个n−披(是))=0

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