### 数学代写|交换代数代写commutative algebra代考|MATH3303

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|交换代数代写commutative algebra代考|Commutative rings and ideals

The most fundamental object of this book is a commutative ring having a multiplicative identity element. Throughout the text, one refers to it simply as a ring.

A ring homomorphism (or simply, a homomorphism) is a map $\varphi: R \rightarrow S$ between rings, which besides being compatible with the two operations, is also required to map the multiplicative identity element of $R$ to the one of $S$. If no confusion arises, one usually denotes the multiplicative identity of any ring by 1 , even if there is more than one ring involved in the discussion. A ring homomorphism $R \rightarrow S$ that admits an inverse ring homomorphism $S \rightarrow R$ is called an isomorphism. As is easily seen, any bijective homomorphism is an isomorphism.

The kernel of $\varphi$ is the set $\operatorname{ker} \varphi:={a \in R \mid \varphi(a)=0}$. It is easy to see that $\operatorname{ker} \varphi$ is an ideal of $R$ and induces an injective homomorphism $R / \operatorname{ker} \varphi \hookrightarrow S$. Because of Proposition 1.1.2 below, one often moves over to the subring $R / \operatorname{ker} \varphi$ for the sake of an argument.

Given an arbitrary homomorphism $\varphi: R \rightarrow S$, one can move back and forth between ideals of $S$ and of $R$ : given an ideal $J \subset S$, the inverse image $\varphi^{-1}(J) \subset R$ is an ideal of $R$, while given an ideal $I \subset R$ one obtain the ideal of $S$ generated by the set $\varphi(I)$. The first such move is called a contraction-a terminology that rigorously makes better sense when $R \subset S$; in the second move, the ideal generated by $\varphi(I)$ is called the extended ideal of $I$.

A subgroup of the additive group of a ring $R$ is called a subring provided it is closed under the product operation of $R$ and contains the multiplicative identity of $R$.

An element $a \in R$ is said to be a zero-divisor if there exists $b \in R, b \neq 0$, such that $a b=0$; otherwise, $a$ is called a nonzero divisor. In this book, a nonzero divisor will often be referred to as a regular element. A sort of extreme case of a zero-divisor is a nilpotent element $a$, such that $a^{n}=0$ for some $n \geq 1$.

One assumes a certain familiarity with these notions and their elementary manipulation.

A terminology that will appear very soon is that of an $R$-algebra to designate a ring $S$ with a homomorphism $R \rightarrow S$.

## 数学代写|交换代数代写commutative algebra代考|Sum of ideals

The set theoretic union of two ideals $I, J \subset R$ is not an ideal, unless one of them is contained in the other. So, one takes the ideal generated by $I \cup J$-this is called the ideal sum of the two ideals and is denoted by $I+J$ or $(I, J)$. The second notation was largely favored in parts of the classical literature and is the one to be employed in this book. On the other hand, the first notation and the terminology are largely justified by the fact that a typical element of $I+J$ has the form $a+a^{\prime}$, with $a \in I$ and $a^{\prime} \in J$, thus sharing the goodies of the notion of summing two subgroups of an additively written Abelian group or two subspaces of a vector space. In particular, an arbitrary expression $a+a^{\prime}$ uniquely determines its summands if and only if $I \cap J={0}$. In the case of Abelian groups or vector spaces, this condition implies direct sum $I \oplus J$. However, the burden carried by the ring multiplication and by the ideal theoretic main property cause the null intersection to be a somewhat rare phenomenon since it requires lots of zero-divisors in the ring.

In contrast to the case of ideal intersection, the ideal sum is easily obtained in terms of generators, namely, if $I=(S)$ and $J=\left(S^{\prime}\right)$ then $(I, J)=\left(S \cup S^{\prime}\right)$. Note that, since $S \cap S^{\prime} \subset S \cup S^{\prime}$, there is quite a bit of superfluous generators in the union. The ideal sum notion applies ipsis literis to an arbitrary family of ideals and appears quite often in the argument of a general proof and is a useful construction as such.

## 数学代写|交换代数代写commutative algebra代考|Product of ideals

Given ideals $I, J \subset R$, the set ${a b \mid a \in I, b \in J}$ of products is not an ideal either (unless at least one of them is principal). The ideal generated by this set is called the ideal product and is denoted by IJ. Here, the generators question is rather trivial for if $I=(S)$ and $J=\left(S^{\prime}\right)$ then the ideal product $I J$ is generated by the set $\left{s s^{\prime} \mid s \in S, s^{\prime} \in S^{\prime}\right}$.
Note the relation of the product to the intersection: as $I J$ is contained both in $I R$ and in $J R$, it follows that $I J \subset I \cap J$. Thus, a measure of obstruction as to when $I \cap J={0}$ holds is that $I J={0}$, which says that every element of one ideal is zero-divided by every element of the second ideal, a rather severe condition. At the other end of the spectrum, the equality $I J=I \cap J$ seldom takes place, turning out to be rather a difficult condition of “transversality.”

The ideal product extends easily to a finite family of ideals. A special nevertheless exceedingly important case is that of a constant family $\left{I_{i}\right}_{i=1}^{m}, I_{i}=I(1 \leq i \leq m)$. In this case, the ideal product is called the $m$ th power of the ideal $I$ and is naturally denoted by $I^{m}$. Note that if $I=\left(s_{1}, \ldots, s_{n}\right)$ then $I^{m}$ is generated by the “monomials” of “degree” $m$ in $s_{1}, \ldots, s_{n}$. The question as to how many of these monomial-like generators are actually superfluous turns out to be a rather deep question related to the notion of analytic independence of ideal generators-a tall order in modern commutative algebra.Besides, the chain $R=I^{0} \supset I=I^{1} \supset I^{2} \supset \cdots$ plus the multiplication rule $I^{m} I^{n}=I^{m+n}$ give rise to deep considerations in both commutative algebra and algebraic geometry. The two topics are in fact quite intertwined.

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