### 数学代写|交换代数代写commutative algebra代考|MAST90025

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## 数学代写|交换代数代写commutative algebra代考|Finiteness Conditions and the Snake Lemma

To start with, let us recall the notion of exact sequences of modules over a ring $R$. A sequence of $R$-modules is a chain of morphisms of $R$-modules
$$\ldots \stackrel{f_{n-2}}{\longrightarrow} M_{n-1} \stackrel{f_{n-1}}{\longrightarrow} M_{n} \stackrel{f_{n}}{\longrightarrow} M_{n+1} \stackrel{f_{n+1}}{\longrightarrow} \ldots$$
where the indices are varying over a finite or an infinite part of $\mathbb{Z}$. We say that the sequence satisfies the complex property at $M_{n}$ (more specifically, at position $n$ ) if we have $f_{n} \circ f_{n-1}=0$ or, in equivalent terms, im $f_{n-1} \subset \operatorname{ker} f_{n}$. Furthermore, the sequence is said to be exact at $M_{n}$ if, in fact, im $f_{n-1}=$ ker $f_{n}$. If the sequence satisfies the complex property at all places $M_{n}$ (of course, except at those where the sequence might terminate), it is called a complex. Likewise, the sequence is called exact, if it is exact at all places. For example, a morphism of $R$-modules $f: M^{\prime} \longrightarrow M$ is injective if and only if the sequence
$$0 \longrightarrow M^{\prime} \stackrel{f}{\longrightarrow} M$$
is exact; here 0 denotes the zero module and $0 \longrightarrow M^{\prime}$ the zero mapping, the only possible $R$-homomorphism from 0 to $M^{\prime}$. On the other hand, $f$ is surjective if and only if the sequence
$$M^{\prime} \stackrel{f}{\longrightarrow} M \longrightarrow 0$$
is exact; $M \longrightarrow 0$ is the zero mapping, the only possible $R$-homomorphism from $M$ to 0 . Exact sequences of type

$$0 \longrightarrow M^{\prime} \stackrel{f}{\longrightarrow} M \stackrel{g}{\longrightarrow} M^{\prime \prime} \longrightarrow 0$$
are referred to as short exact sequences. The exactness of such a sequence means:
(1) $f$ is injective,
(2) $\operatorname{im} f=\operatorname{ker} g$,
(3) $g$ is surjective.
Thus, for a short exact sequence as above, we can view $M^{\prime}$ as a submodule of $M$ via $f$ and we see, using the Fundamental Theorem of Homomorphisms $1.4 / 6$, that $g$ induces an isomorphism $M / M^{\prime} \longrightarrow M^{\prime \prime}$. Conversely, every submodule $N \subset M$ gives rise to the short exact sequence
$$0 \longrightarrow N \longrightarrow M \longrightarrow M / N \longrightarrow 0$$
Another type of short exact sequences can be built from the direct sum of two $R$-modules $M^{\prime}$ and $M^{\prime \prime}$, namely
(*)
$$0 \longrightarrow M^{\prime} \longrightarrow M^{\prime} \oplus M^{\prime \prime} \longrightarrow M^{\prime \prime} \longrightarrow 0$$

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

As we have seen in $1.5 / 8$, a ring is called Noetherian if all its ideals are finitely generated or, equivalently by $1.5 / 9$, if its ideals satisfy the ascending chain condition. The aim of the present chapter is to show that the Noetherian hypothesis, as simple as it might look, nevertheless has deep impacts on the structure of ideals and their inclusions, culminating in the theory of Krull dimension, to be dealt with in Section 2.4.

To discuss some standard examples of Noetherian and non-Noetherian rings, recall from Hilbert’s Basis Theorem $1.5 / 14$ that all polynomial rings of type $R\left[X_{1}, \ldots, X_{n}\right]$ in finitely many variables $X_{1}, \ldots, X_{n}$ over a Noetherian ring $R$ are Noetherian. The result extends to algebras of finite type over a Noetherian ring $R$, i.e. $R$-algebras of type $R\left[X_{1}, \ldots, X_{n}\right] / \mathfrak{a}$ where $\mathfrak{a}$ is an ideal in $R\left[X_{1}, \ldots, X_{n}\right]$. In particular, algebras of finite type over a field $K$ or over the ring of integers $\mathbb{Z}$ are Noetherian. One also knows that all rings of integral algebraic numbers in finite extensions of $\mathbb{Q}$ are Noetherian (use Atiyah-MacDonald $[2], 5.17)$, whereas the integral closure of $\mathbb{Z}$ in any infinite algebraic extension of $\mathbb{Q}$ is not; see Section $3.1$ for the notion of integral dependence and in particular $3.1 / 8$ for the one of integral closure. Also note that any polynomial ring $R[\mathfrak{X}]$ in an infinite family of variables $\mathfrak{X}$ over a non-zero ring $R$ will not be Noetherian. Other interesting examples of non-Noetherian rings belong to the class of (general) valuation rings, as introduced in $9.5 / 13$.

To approach the subject of Krull dimension for Noetherian rings, the technique of primary decomposition, developed in Section 2.1, is used as a key tool. We will show in $2.1 / 6$ that such a primary decomposition exists for all ideals a of a Noetherian ring $R$. It is of type
$$\mathfrak{a}=\mathfrak{q}{1} \cap \ldots \cap \mathfrak{q}{r}$$
where $\mathfrak{q}{1}, \ldots, \mathfrak{q}{r}$ are so-called primary ideals in $R$. Primary ideals generalize the notion of prime powers in principal ideal domains, whereas the concept of primary decomposition generalizes the one of prime factorization.

Looking at a primary decomposition $(*)$, the nilradicals $\mathfrak{p}{i}=\operatorname{rad}\left(\mathfrak{q}{i}\right)$ are of particular significance; they are prime in $R$ and we say that $\mathfrak{q}{i}$ is $\mathfrak{p}{i}$-primary. As any finite intersection of $\mathfrak{p}$-primary ideals, for any prime ideal $\mathfrak{p} \subset R$, is $\mathfrak{p}$-primary again (see $2.1 / 4$ ), we may assume that all $\mathfrak{p}{1}, \ldots, \mathfrak{p}{\tau}$ belonging to the primary decomposition () are different. In addition, we can require that the decomposition $()$ is minimal in the sense that it cannot be shortened any further. In such a situation we will show in $2.1 / 8$ that the set of prime ideals $\mathfrak{p}{1}, \ldots, \mathfrak{p}{r}$ is uniquely determined by $\mathfrak{a}$; it is denoted by Ass $(\mathfrak{a})$, referring to the members of this set as the prime ideals associated to $a$. There is a uniqueness assertion for some of the primary ideals $\mathfrak{q}_{i}$ as well (see $2.1 / 15$ ), although not all of them will be unique in general.

## 数学代写|交换代数代写commutative algebra代考|Primary Decomposition of Ideals

Let $R$ be a principal ideal domain. Then $R$ is factorial and any non-zero element $a \in R$ admits a factorization $a=\varepsilon p_{1}^{n_{1}} \ldots p_{r}^{n_{r}}$ with a unit $\varepsilon \in R^{*}$, pairwise non-

equivalent prime elements $p_{i} \in R$, and exponents $n_{i}>0$, where these quantities are essentially unique. Passing to ideals, it follows that every ideal $\mathfrak{a} \subset R$ admits a decomposition
$$\mathfrak{a}=\mathfrak{p}{1}^{n{1}} \cap \ldots \cap \mathfrak{p}{r}^{n{r}}$$
with pairwise different prime ideals $\mathfrak{p}{i}$ that are unique up to order, and exponents $n{i}>0$ that are unique as well. The purpose of the present section is to study similar decompositions for more general rings $R$, where the role of the above prime powers $\mathfrak{p}{i}^{n{i}}$ is taken over by the so-called primary ideals. In the following we start with a general ring $R$ (commutative and with a unit element, as always). Only later, when we want to show the existence of primary decompositions, $R$ will be assumed to be Noetherian. For a generalization of primary decompositions to the context of modules see Serre [24], I.B.

Definition 1. A proper ideal $\mathfrak{q} \subset R$ is called a primary ideal if $a b \in \mathfrak{q}$ for any elements $a, b \in R$ implies $a \in \mathfrak{q}$ or, if the latter is not the case, that there is an exponent $n \in \mathbb{N}$ such that $b^{n} \in \mathfrak{q}$.

Clearly, any prime ideal is primary. Likewise, for a prime element $p$ of a factorial ring, all powers $(p)^{n}, n>0$, are primary. But for general rings we will see that the higher powers of prime ideals may fail to be primary. Also note that an ideal $\mathfrak{q} \subset R$ is primary if and only if the zero ideal in $R / \mathfrak{q}$ is primary. The latter amounts to the fact that all zero-divisors in $R / \mathfrak{q}$ are nilpotent. More generally, if $\pi: R \longrightarrow R / \mathfrak{a}$ is the canonical projection from $R$ onto its quotient by any ideal $\mathfrak{a} \subset R$, then an ideal $\mathfrak{q} \subset R / \mathfrak{a}$ is primary if and only if its preimage $\pi^{-1}(\mathfrak{q})$ is primary in $R$.

## 数学代写|交换代数代写commutative algebra代考|Finiteness Conditions and the Snake Lemma

…⟶Fn−2米n−1⟶Fn−1米n⟶Fn米n+1⟶Fn+1…

0⟶米′⟶F米

0⟶米′⟶F米⟶G米′′⟶0

（1）F是单射的，
(2)在里面⁡F=克尔⁡G,
(3) G是主观的。

0⟶ñ⟶米⟶米/ñ⟶0

(*)

0⟶米′⟶米′⊕米′′⟶米′′⟶0

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