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# 数学代写|交换代数代写Commutative Algebra代考|MATH483

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

1.1 Notation. The symbol $\mathbb{Z}$ will always denote the set of integers; in addition, $\mathbb{N}$ (respectively $\mathbb{N}_0$ ) will always denote the set of positive (respectively non-negative) integers. The set of rational (respectively real, complex) numbers will be denoted by the symbol $\mathbb{Q}$ (respectively $\mathbb{R}, \mathbb{C}$ ).

The symbol $\subseteq$ will stand for ‘is a subset of’; the symbol $\subset$ will be reserved to denote strict inclusion. Thus, for sets $A, B$, the expression $A \subset B$ means that $A \subseteq B$ and $A \neq B$.

The symbol ‘ $\square$ ‘ will be used to denote the end, or absence, of a proof. We shall reserve the symbols
$$X, Y, X_1, \ldots, X_n$$
to denote indeterminates.
We shall denote the number of elements in a finite set $\Omega$ by $|\Omega|$.
A comment should perhaps be made about the distinction between a family and a set. We shall often use round parentheses ( ), as in $\left(a_i\right)_{i \in I}$, to denote a family indexed by the set $I$; here $a_i$ should be thought of as situated in the ‘position’ labelled by $i$; and the family $\left(a_i\right){i \in I}$ is considered to be equal to $\left(b_i\right){i \in I}$ if and only if $a_i=b_i$ for all $i \in I$. One can think of a family $\left(a_i\right){i \in I}$, where $a_i$ lies in the set $A$ for all $i \in I$, as a function from $I$ to $A$ : in this interpretation, the image of $i$ under the function is $a_i$. On the other hand, curly braces {} , as in $$\left{d_1, \ldots, d_n\right} \quad \text { or } \quad{d \in D: \text { statement } P(d) \text { is true }} \text {, }$$ will often be used to indicate sets. A set is completely determined by its members, and no concept of ‘position’ is involved when the members of the set are displayed within braces. The distinction between a family and a set parallels that between a function and its image. To illustrate the distinction, let $d_1=d_2=1$ and $d_3=3$. Then the family $\left(d_i\right){i=1}^3$ can be thought of as the ordered triple $(1,1,3)$, whereas the set $\left{d_1, d_2, d_3\right}$ is just the 2 -element set ${1,1,3}={1,3}$.

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

2.1 Definition and Lemma. Let $R$ and $S$ be commutative rings, and let $f: R \rightarrow S$ be a ring homomorphism. Then we define the kernel of $f$, denoted $\operatorname{Ker} f, b y$
$\operatorname{Ker} f:=\left{r \in R: f(r)=0_S\right}$.
Note that
(i) $0_R \in \operatorname{Ker} f$, so that $\operatorname{Ker} f \neq \emptyset$;
(ii) whenever $a, b \in \operatorname{Ker} f$, then $a+b \in \operatorname{Ker} f$ also; and
(iii) whenever $a \in \operatorname{Ker} f$ and $r \in R$, then $r a \in \operatorname{Ker} f$ also.

The above lemma provides motivation for the definition of ideal in a commutative ring, but before we give the definition, we record a fundamental fact about kernels of homomorphisms of commutative rings.
2.2 Lemma. Let $R$ and $S$ be commutative rings, and let $f: R \rightarrow S$ be a ring homomorphism. Then $\operatorname{Ker} f=\left{0_R\right}$ if and only if $f$ is injective.

Proof. $(\Rightarrow)$ Let $r, r^{\prime} \in R$ be such that $f(r)=f\left(r^{\prime}\right)$. Then $r-r^{\prime} \in$ Ker $f=\left{0_R\right}$.
$(\Leftrightarrow)$ Of course, $0_R \in \operatorname{Ker} f$, by 2.1(i). Let $r \in \operatorname{Ker} f$. Then $f(r)=0_S=$ $f\left(0_R\right)$, so that $r=0_R$ since $f$ is injective.
2.3 Definition. Let $R$ be a commutative ring. A subset $I$ of $R$ is said to be an ideal of $R$ precisely when the following conditions are satisfied:
(i) $I \neq \emptyset$;
(ii) whenever $a, b \in I$, then $a+b \in I$ also; and
(iii) whenever $a \in I$ and $r \in R$, then $r a \in I$ also.
It should be clear to the reader that an ideal of a commutative ring $R$ is closed under subtraction. Any reader experienced in non-commutative ring theory should note that we shall not discuss ideals of non-commutative rings in this book, and so we shall have no need of the concepts of left ideal and right ideal.

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

1.1符号。符号$\mathbb{Z}$将始终表示整数集;此外，$\mathbb{N}$(分别为$\mathbb{N}_0$)将始终表示一组正整数(分别为非负整数)。有理数(分别为实数和复数)的集合将用符号$\mathbb{Q}$(分别为$\mathbb{R}, \mathbb{C}$)表示。

$$X, Y, X_1, \ldots, X_n$$

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

2.1定义和引理。设$R$和$S$为交换环，设$f: R \rightarrow S$为环同态。然后定义$f$的核，记为$\operatorname{Ker} f, b y$
$\operatorname{Ker} f:=\left{r \in R: f(r)=0_S\right}$。

(i) $0_R \in \operatorname{Ker} f$，以便$\operatorname{Ker} f \neq \emptyset$;
(ii)只要$a, b \in \operatorname{Ker} f$，那么$a+b \in \operatorname{Ker} f$也;和
(iii)只要$a \in \operatorname{Ker} f$和$r \in R$，那么$r a \in \operatorname{Ker} f$也。

2.2引理。设$R$和$S$为交换环，设$f: R \rightarrow S$为环同态。那么$\operatorname{Ker} f=\left{0_R\right}$当且仅当$f$是单射的。

$(\Leftrightarrow)$当然，$0_R \in \operatorname{Ker} f$，通过2.1(i)。让$r \in \operatorname{Ker} f$。然后是$f(r)=0_S=$$f\left(0_R\right)$，所以$r=0_R$既然$f$是注入的。
2.3定义。设$R$为可交换环。当满足下列条件时，我们称$R$的子集$I$为$R$的理想子集:
(i) $I \neq \emptyset$;
(ii)只要$a, b \in I$，那么$a+b \in I$也;和
(iii)只要$a \in I$和$r \in R$，那么$r a \in I$也。

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