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# 数学代写|交换代数代写Commutative Algebra代考|MAT4200 Definitions. First Examples

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## 数学代写|交换代数代写Commutative Algebra代考|Definitions. First Examples

Definition (1.1.1). – A ring is a set A endowed with two binary laws, an addition $(a, b) \mapsto a+b$, and a multiplication $(a, b) \mapsto a b$, satisfying the following axioms.

(i) Axioms concerning the addition law and stating that $(\mathrm{A},+)$ is an abelian group:

• For every $a, b$ in A, $a+b=b+a$ (commutativity of addition);
• For every $a, b, c$ in $\mathrm{A},(a+b)+c=a+(b+c)$ (associativity of addition);
• There exists an element $0 \in \mathrm{A}$ such that $a+0=0+a=a$ for every $a$ in A (neutral element for the addition);
• For every $a \in \mathrm{A}$, there exists an element $b \in \mathrm{A}$ such that $a+b=$ $b+a=0$ (existence of an additive inverse);
(ii) Axioms concerning the multiplication law and stating that $(\mathrm{A}, \cdot)$ is a monoid:
• There exists an element $1 \in \mathrm{A}$ such that $1 a=a 1=a$ for every $a \in \mathrm{A}$ (neutral element for the multiplication);
• For every $a, b$, and $c$ in $\mathrm{A},(a b) c=a(b c$ ) (associativity of multiplication);
(iii) Axiom relating the addition and the multiplication:
• For every $a, b$, and $c$ in $\mathrm{A}, a(b+c)=a b+a c$ and $(b+c) a=b a+c a$ (multiplication distributes over addition).
(iv) One says that the ring A is commutative if, moreover:
• For every $a$ and $b$ in A, $a b=b a$ (commutativity).
With their usual addition and multiplication, integers, real numbers, and matrices are fundamental examples of rings. In fact, the ring axioms specify exactly the relevant computation rules to which one is accustomed to. We shall give more examples in a moment, but we first state a few computation rules which follow from the stated axioms.

## 数学代写|交换代数代写Commutative Algebra代考|Let A be a ring

Endowed with the addition law, A is in particular an abelian group. As a consequence, it admits exactly one zero element, which is usually denoted by 0 , or by $0_{\mathrm{A}}$ if it is necessary to specify the ring of which it is the zero element. Moreover, any element $a \in \mathrm{A}$ has exactly one additive inverse, and it is denoted by $-a$. Indeed, if $b$ and $c$ are two additive inverses, then $b=b+0=b+(a+c)=(b+a)+c=c$.

For every $a \in \mathrm{A}$, one has $a \cdot(0+0)=a \cdot 0+a \cdot 0$, hence $a \cdot 0=0$; similarly, $0 \cdot a=0$. Then, for any $a, b \in \mathrm{A}$, one has $a(-b)+a b=a(-b+b)=a \cdot 0=0$, hence $a(-b)=-(a b)$; similarly, $(-a) b=-(a b)$. Consequently, there is no ambiguity in writing $-a b$ for either $(-a) b,-(a b)$ or $a(-b)$.

For any integer $n$, one defines $n a$ by induction, by setting $0 a=0$, and $n a=a+(n-1) a$ if $n \geq 1$ and $n a=-(-n) a$ if $n \leq-1$. Observe that for any $a, b \in \mathrm{A}$ and any integer $n$, one has $a(n b)=n(a b)=(n a) b$. This is proved by induction on $n$ : if $n=0$, then all three terms are 0 ; if $n \geq 1$, then $a(n b)=a(b+(n-1) b)=a b+a((n-1) b)=a b+(n-1)(a b)=n a b$ and similarly for the other equality; finally, if $n \leq-1$, then $a(n b)=a(-((-n) b))=$ $-a((-n) b)=(-n)(-a b)=n a b .$

Similarly, the multiplicative monoid $(\mathrm{A}, \cdot)$ has exactly one neutral element, usually denoted by 1 , or by $1_{\mathrm{A}}$ if it is necessary to specify the ring, and called the unit element of $\mathrm{A}$.

If $a$ belongs to a ring A and $n$ is any positive ${ }^1$ integer $n \geq 0$, one defines $a^n$ by induction by setting $a^0=1$, and, if $n \geq 1, a^n=a \cdot a^{n-1}$. For any integers $m$ and $n$, one has $a^{m+n}=a^m a^n$ and $\left(a^m\right)^n=a^{m n}$, as can be checked by induction.
However, one should take care that $a^n b^n$ and $(a b)^n$ are generally distinct, unless $a b=b a$, in which case one says that $a$ and $b$ commute.

## 数学代写|交换代数代写Commutative Algebra代考|Definitions. First Examples

(i) 关于加法定律的公理并说明 $(\mathrm{A},+)$ 是一个阿贝尔群:

(ii) 关于乘法定律的公理并说明 $(\mathrm{A}, \cdot)$ 是一个半群:

(iii) 加法和乘法的公理:

(iv) 有人说环 $A$ 是可交换的，如果，此外:

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