### 数学代写|交换代数代写commutative algebra代考|MATH33O3

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

Let us discuss some examples demonstrating how rings of the just mentioned types occur in a natural way. We start with a problem from Number Theory.
Proposition. For a prime number $p>2$ the following conditions are equivalent:
(i) Therc cxist intcgers $a, b \subset \mathbb{Z}$ satisfying $p=a^{2}+b^{2}$.
(ii) $p \equiv 1 \bmod 4$.
In order to explain the proof, assume first that (i) is given and consider an integer $a \in \mathbb{Z}$. Then, $\bmod 4$, the square $a^{2}$ is congruent to 0 (if $a$ is even) or to 1 (if $a$ is odd). Therefore an odd number of type $a^{2}+b^{2}$ with integers $a, b \in \mathbb{Z}$ is always congruent to $1 \bmod 4$ so that (ii) follows.

To derive (i) from condition (ii), we use a trick. Namely, we enlarge the ring of integers $\mathbb{Z}$ by passing to the ring of integral Gauß numbers $\mathbb{Z}[i]$. Thus, for a prime number $p>2$ satisfying $p \equiv 1 \bmod 4$, we have to solve the equation $p=x^{2}+y^{2}$ in $\mathbb{Z}$ or, using the factorization $x^{2}+y^{2}=(x+i y)(x-i y)$, the equation
$$p=(x+i y)(x-i y)$$
in $\mathbb{Z}[i]$. To do so, we use the following auxiliary results:
Proposition (Gauß). $\mathbb{Z}[i]$ is an Euclidean ring with respect to the degree function
$$\delta: \mathbb{Z}[i]-{0} \longrightarrow \mathbb{N}, \quad z \longmapsto \delta(z):=\left|z^{2}\right|$$
In particular, $\mathbb{Z}[i]$ is factorial.
Proposition (Wilson). Every prime number $p$ satisfies $(p-1) ! \equiv-1 \bmod p$.
The result of Gauß can easily be checked by relying on the fact that every complex number $c \in \mathbb{C}$ can be approximated by a complex number $z \in \mathbb{Z}[i]$ satisfying $|c-z| \leq \frac{1}{2} \sqrt{2}$; see [3], Section 2.4, for a more detailed argumentation. To derive the result of Wilson, look at the finite field $\mathbb{F}{p}=\mathbb{Z} / p \mathbb{Z}$, whose elements are denoted by $0,1, \ldots, p-1$ for simplicity. Every element $\alpha \in \mathbb{F}{p}$ satisfies $\alpha^{p}=\alpha$, as can easily be verified by induction using the simplified binomial formula $(\alpha+1)^{p}=\alpha^{p}+1^{p}=\alpha^{p}+1$. Therefore the elements of $\mathbb{F}{p}$ are precisely the zeros of the polynomial $X^{p}-X \in \mathbb{F}{p}[X]$, and all these zêros are simple. In particular, we have
$$X^{p-1}-1=(X-1)(X-2) \ldots(X-(p-1))$$
in $\mathbb{F}_{p}[X]$. Comparing coefficients, this yields

\begin{aligned} 1 \cdot 2 \cdot \ldots \cdot(p-1) &=(-1)^{p-1} \cdot(-1) \cdot(-2) \cdot \ldots \cdot(-(p-1)) \ &=(-1)^{p-1} \cdot(-1)=-1 \end{aligned}
at least for $p$ odd, but clearly also for $p=2$ since then $-1=1$. This establishes the result of Wilson.

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

The present chapter is devoted to discussing some basic notions and results on rings and their modules. Except for a few preliminary considerations, all rings will be meant to be commutative and to admit a unit element 1 . Like a field, a ring comes equipped with two laws of composition, namely addition “+” and multiplication “”, which behave in the same way as is known from the case of fields. The only difference is that non-zero elements of a ring $R$ do not need to admit multiplicative inverses in $R$, a default that has far-reaching consequences. A prominent example of such a ring is $\mathbb{Z}$, the ring of integers. But we can easily construct more intricate types of rings. Let $k$ be a field and write $R$ for the cartesian product of $k$ with itself, i.e. $R=k \times k$. Defining addition and multiplication on $R$ componentwise by
\begin{aligned} \left(\alpha_{1}, \alpha_{2}\right)+\left(\beta_{1}, \beta_{2}\right) &=\left(\alpha_{1}+\beta_{1}, \alpha_{2}+\beta_{2}\right), \ \left(\alpha_{1}, \alpha_{2}\right) \cdot\left(\beta_{1}, \beta_{2}\right) &=\left(\alpha_{1} \cdot \beta_{1}, \alpha_{2} \cdot \beta_{2}\right) \end{aligned}
we see that $R$ becomes a ring. The equation $(1,0) \cdot(0,1)=(0,0)$ implies that $R$ contains non-trivial zero divisors, whereas $(1,0)^{n}=(1,0)$ for $n>0$ shows that $R$ contains idempotent elements that are different from the unit element $(1,1)$. However, in this case there are no non-trivial nilpotent elements, i.e. elements $\left(\alpha_{1}, \alpha_{2}\right)$ different from $(0,0)$ such that $\left(\alpha_{1}, \alpha_{2}\right)^{n}=(0,0)$ for some exponent $n$. On the other hand, non-trivial nilpotent elements will occur if we take
$$\left(\alpha_{1}, \alpha_{2}\right) \cdot\left(\beta_{1}, \beta_{2}\right)=\left(\alpha_{1} \cdot \beta_{1}, \alpha_{1} \cdot \beta_{2}+\alpha_{2} \cdot \beta_{1}\right)$$
as multiplication on $R$.
For rings $R$ of general type the notion of ideals is fundamental. An ideal in $R$ is just an additive subgroup $\mathfrak{a} \subset R$ that is stable under multiplication by elements of $R$. Historically ideals were motivated by the aim to extend unique factorization results from the ring of integers $\mathbb{Z}$ to more general rings of algebraic numbers. However, as this did not work out well in the conventional setting, Kummer invented his concept of “ideal numbers”, which was then generalized by Dedekind, who introduced the notion of ideals as known today. A further natural step is to pass from ideals to modules over rings, therehy arriving at a simultaneous generalization of ideals in rings and of vector spaces over fields.

## 数学代写|交换代数代写commutative algebra代考|Rings and Ideals

Let us recall the definition of a ring.
Definition 1. A set $R$ together with two laws of composition ” $+”$ (addition) and “. ” (multiplication) is called a ring (with unity) if the following conditions are satisfied:

(i) $R$ is an abelian group with respect to addition; the corresponding zero element is denoted by $0 \in R$.
(ii) The multiplication is associative, i.e.
$$(a \cdot b) \cdot c=a \cdot(b \cdot c) \quad \text { for } \quad a, b, c \in R$$
(iii) There exists a unit element in $R$, which means an element $1 \in R$ such that $1 \cdot a=a=a \cdot 1$ for all $a \in R$.
(iv) The multiplication is distributive over the addition, i.e. for $a, b, c \in R$ we have
$$a \cdot(b+c)=a \cdot b+a \cdot c, \quad(a+b) \cdot c=a \cdot c+b \cdot c .$$
The ring $R$ is called commutative if the multiplication is commutative.
We list some important examples of rings:
(1) fields,
(2) $\mathbb{Z}$, the ring of integers,
(3) $R[X]$, the polynomial ring in a variable $X$ over a commutative ring $R$,
(4) 0 , the zero ring, which consists of just one element $1=0$; it is the only ring with the latter property.

An element $a$ of a ring $R$ is called invertible or a unit if there exists some element $b \in R$ such that $a b=1=b a$. It follows that the set
$$R^{*}={a \in R ; a \text { is a unit in } R}$$
is a group with respect to the multiplication given on $R$.
An element $a$ of a ring $R$ is called a zero divisor if there exists an element $b \in R-{0}$ such that $a b=0$ or $b a=0$. Furthermore, a commutative ring $R \neq 0$ is called an integral domain if it does not contain (non-trivial) zero divisors, i.e. if $a b=0$ with $a, b \in R$ implies $a=0$ or $b=0$. For example, any field is an integral domain, as well as any subring of a field, such as the ring of integers $\mathbb{Z} \subset \mathbb{Q}$. Also one knows that the polynomial ring $R[X]$ over an integral domain $R$ is an integral domain again. However, by definition, the zero ring 0 is not an integral domain.

For a field $K$, its group of units is $K^{}=K-{0}$. Furthermore, we have $\mathbb{Z}^{}={1,-1}$ and $(R[X])^{}=R^{}$ for an integral domain $R$.

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

(i) Therc cxist intcgers一个,b⊂从令人满意的p=一个2+b2.
(二)p≡1反对4.

p=(X+一世是)(X−一世是)

d:从[一世]−0⟶ñ,和⟼d(和):=|和2|

Gauß 的结果可以很容易地通过依赖于每个复数的事实来检查C∈C可以用一个复数来近似和∈从[一世]令人满意的|C−和|≤122; 有关更详细的论证，请参见 [3]，第 2.4 节。要导出 Wilson 的结果，请查看有限域Fp=从/p从，其元素表示为0,1,…,p−1为简单起见。每一个元素一个∈Fp满足一个p=一个, 可以很容易地通过使用简化的二项式公式进行归纳来验证(一个+1)p=一个p+1p=一个p+1. 因此元素Fp正是多项式的零点Xp−X∈Fp[X]，所有这些零点都很简单。特别是，我们有

Xp−1−1=(X−1)(X−2)…(X−(p−1))

1⋅2⋅…⋅(p−1)=(−1)p−1⋅(−1)⋅(−2)⋅…⋅(−(p−1)) =(−1)p−1⋅(−1)=−1

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

(一个1,一个2)+(b1,b2)=(一个1+b1,一个2+b2), (一个1,一个2)⋅(b1,b2)=(一个1⋅b1,一个2⋅b2)

(一个1,一个2)⋅(b1,b2)=(一个1⋅b1,一个1⋅b2+一个2⋅b1)

## 数学代写|交换代数代写commutative algebra代考|Rings and Ideals

（一世）R是关于加法的阿贝尔群；相应的零元素表示为0∈R.
(ii) 乘法是结合的，即

(一个⋅b)⋅C=一个⋅(b⋅C) 为了 一个,b,C∈R
(iii) 存在一个单位元素R, 这意味着一个元素1∈R这样1⋅一个=一个=一个⋅1对所有人一个∈R.
(iv) 乘法对加法是可分配的，即一个,b,C∈R我们有

（1）字段，
（2）从, 整数环,
(3)R[X], 变量中的多项式环X在交换环上R,
(4) 0 ，零环，仅由一个元素组成1=0; 它是唯一具有后一种属性的环。

R∗=一个∈R;一个 是一个单位 R

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