### 数学代考|代数几何代写algebraic geometry代考|The Definition of Algebraic Varieties

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• Foundations of Data Science 数据科学基础

## 数学代考|代数几何代写algebraic geometry代考|Affine Algebraic Sets

The starting point of algebraic geometry is studying solutions of systems of polynomial equations in several variables over a field. (A field is an algebraic structure with operations of addition and multiplication which satisfy all the formal properties of the real numbers,

i.e. commutativity and associativity of both operations, the existence of $0 \neq 1$ with their usual properties, distributivity, and the existence of an additive inverse-or minus signas well as multiplicative inverses of non-zero elements). Systems of polynomial equations in variables $x_{1}, \ldots, x_{n}$ can always be written in the form
$$\begin{array}{r} p_{1}\left(x_{1}, \ldots, x_{n}\right)=0 \ \ldots \ p_{m}\left(x_{1}, \ldots, x_{n}\right)=0 \end{array}$$
where $p_{1}, \ldots, p_{m}$ are polynomials. Solutions of the Eqs. (1.1.1) are $n$-tuples of elements $\left(x_{1}, \ldots, x_{n}\right)$ of the given field which satisfy the equations. Such $n$-tuples are also called zeros of the polynomials $p_{1}, \ldots, p_{n}$. Sets of zeros of sets of polynomials are called affine algebraic sets.

The set of all polynomials in $n$ variables over a field forms a commutative ring, which means that it has operations of addition and multiplication satisfying all the formal properties of integers, i.e. commutativity and associativity of both operations, the existence of 0 and 1 , distributivity, and the existence of an additive inverse. One defines not necessarily commutative rings by dropping the assumption that multiplication be commutative (we then must require that 1 be a left and right unit and that left and right distributivities hold). In this book, by a ring, we shall mean a commutative ring, unless specified otherwise.

Solutions of (1.1.1), or zeros of the polynomials $p_{1}, \ldots, p_{m}$, are also zeros of all linear combinations
$$a_{1} p_{1}+\cdots+a_{m} p_{m}$$
where $a_{1}, \ldots, a_{m}$ are arbitrary polynomials.
The elements (1.1.2) form the ideal generated by $p_{1}, \ldots, p_{m}$, which is denoted by
$$\left(p_{1}, \ldots, p_{m}\right)$$
An ideal in a commutative ring is a subset which contains 0 , is closed under $+$, and multiples by elements of the ring. By the Hilbert basis theorem, which we prove in Sect. 4 (Theorem 4.2.1), the ring of polynomials in $n$ variables over a field is Noetherian, which means that every ideal is finitely generated (i.e. generated by finitely many elements). Because of this, it is sufficient to consider systems of finitely many polynomial equations (1.1.1).

Note that a commutative ring $R$ is Noetherian if and only if it satisfies the ascending chain condition (ACC) with respect to ideals. To satisfy the ACC with respect to subsets of a certain kind means that there does not exist an infinite chain
$$I_{1} \subsetneq I_{2} \subsetneq \cdots \subsetneq I_{n} \subsetneq \cdots$$

of such sets. Thus, we claim that a ring $R$ is Noetherian if and only if (1.1.3) does not occur in $R$ where $I_{n}$ are ideals. To see this, if $R$ is not Noetherian, it has an ideal $I$ which is not finitely generated, so having picked, by induction, elements $r_{1} \ldots, r_{n} \in I$, they cannot generate $I$, so we can pick $r_{n+1} \in I \backslash\left(r_{1}, \ldots, r_{n}\right)$. Thus, $R$ fails the ACC for ideals. On the other hand, if $R$ fails the ACC for ideals, then we have ideals (1.1.3) in $R$. Assume, for contradiction, that $R$ is Noetherian. Let
$$I=\bigcup_{n} I_{n} .$$
Then the ideal is finitely generated, say, by elements $r_{1}, \ldots, r_{k}$. Thus, there exists an $n$ such that $r_{1}, \ldots, r_{k} \in I_{n}$, which implies $I_{n}=I$, which is a contradiction.

## 数学代考|代数几何代写algebraic geometry代考|Complex Numbers

Zeros of polynomials behave better when we work in the field $\mathbb{C}$ of complex numbers than in the field $\mathbb{R}$ of real numbers. The field $\mathbb{C}$ contains the number $i$ which has the property
$$i^{2}=-1$$
and more generally, a complex number can be uniquely written as $a+b i$ where $a, b$ are real numbers. Addition and multiplication are then determined by the properties of a field. Division is possible because $(a+b i)(a-b i)=a^{2}+b^{2}$, and we can thus make the denominator real.
Thus, the polynomial equation
$$x^{2}+1=0$$
has solutions in $\mathbb{C}$, namely $i$ and $-i$, while it has no solution over the field of real numbers $\mathbb{R}$.

It turns out that more generally, every non-constant polynomial in one variable with coefficients in $\mathbb{C}$ has at least one zero (we also say root). A field which satisfies this property is called algebraically closed. The fact that $\mathbb{C}$ is algebraically closed is known as the fundamental theorem of algebra. In the first three sections of this chapter, we will assume from now on that we are working over the field $\mathbb{C}$. More generally, in much of what we say (excluding connections with analysis), we could work over any algebraically closed field.

## 数学代考|代数几何代写algebraic geometry代考|Nullstellensatz

The fact that a non-constant polynomial over $\mathbb{C}$ always has a root can be generalized to several variables as follows: Let $I$ be an ideal in the ring $\mathbb{C}\left[x_{1}, \ldots, x_{n}\right]$ of polynomials in $n$ variables over $\mathbb{C}$. Let $X=Z(I)$ be the affine algebraic set which is the set of zeros of

the ideal $I$. If $I=\left(f_{1}, \ldots, f_{m}\right)$, we also write
$$Z\left(f_{1}, \ldots, f_{m}\right)=Z(I)$$
Let, on the other hand, $I(X)$ be the ideal of all polynomials which are zero on $X$ (i.e. $p\left(x_{1}, \ldots, x_{n}\right)=0$ for every $\left.\left(x_{1}, \ldots, x_{n}\right) \in X\right)$. Then
$$I(X)=\sqrt{I}$$
where the right hand side of (1.1.4) is called the radical of $I$ and consists of all polynomials $p$ for which $p^{k} \in I$ for some non-negative integer $k$. Equation (1.1.4) is called the Nullstellensatz, and is due to Hilbert. In German, Nullstelle means zero, literally “zero place,” a point at which a polynomial is zero. In English, as we already remarked, such a point is called just a “zero,” which can be confusing.

As many facts in algebraic geometry, a proof of the Nullstellensatz requires certain methods from algebra. The kind of algebra relevant to the foundations of algebraic geometry is known as commutative algebra, to which we will keep returning throughout this book. The Nullstellensatz will be restated and proved in Sect. $4.3$ below.

## 数学代考|代数几何代写algebraic geometry代考|Affine Algebraic Sets

p1(X1,…,Xn)=0 … p米(X1,…,Xn)=0

(1.1.1) 的解，或多项式的零点p1,…,p米, 也是所有线性组合的零

(p1,…,p米)

X2+1=0

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