### 数学代考|代数几何代写algebraic geometry代考| Rational Maps, Smooth Maps and Dimension

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (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

## 数学代考|代数几何代写algebraic geometry代考|Definition of a Rational Map

For two algebraic varieties $X, Y$, a rational map from $X$ to $Y$ is morphism of varieties
$$f: U \rightarrow Y$$
where $U$ is a non-empty Zariski open subset of $X$. The rational map $f$ is considered equal to a rational map
$$g: V \rightarrow Y$$
if
$$f(x)=g(x) \text { for all } x \in U \cap V$$
Recall that (by irreducibility), a non-empty Zariski open set in a variety $X$ is dense which means that its complement does not contain any non-empty open set. This implies that for non-empty Zariski open sets $U, V \subseteq X, U \cap V$ is non-empty and Zariski open.

A rational map $f: X \rightarrow Y$ is called dominant if for $W \subseteq Y$ non-empty Zariski open, the inverse image $f^{-1}(W)$ is non-empty (note that it is by definition open).

## 数学代考|代数几何代写algebraic geometry代考|The Category of Varieties and Dominant Rational Maps

One can consider the category whose objects are varieties, and morphisms are dominant rational maps. An isomorphism in this category is called a birational equivalence. A variety is called rational if it is birationally equivalent to an affine (equivalently, projective) space. Note also that any rational map of varieties that has an inverse as a rational map is necessarily dominant. This means that if we consider the larger category of varieties and all rational maps (not necessarily dominant), it has the same isomorphisms as the category of varieties and rational dominant maps.
$$X \mapsto K(X)$$
where $X$ is a variety and $K(X)$ is its field of rational functions gives rise to an equivalence of categories between the category of varieties and rational dominant maps, and the opposite of the category of fields containing $\mathbb{C}$ which are finitely generated (as fields) over $\mathbb{C}$, and homomorphisms of $\mathbb{C}$-algebras (or, equivalently, homomorphisms of fields which fix $\mathbb{C}$ ). Such fields are sometimes known as function fields over $\mathbb{C}$. Functoriality follows from functoriality of the field of fractions with respect to injective homomorphisms of integral domains.

To go the other way, select, for a function field $K$ over $\mathbb{C}$, elements $x_{1}, \ldots x_{n}$ which generate $K$ as a field containing $\mathbb{C}$. This defines a map
$$h: \mathbb{C}\left[x_{1}, \ldots, x_{n}\right] \rightarrow K$$
Then send $K$ to $Z(I)$ where $I$ is the kernel of $h$, i.e. the ideal of all polynomials $p$ such that $p(h)=0$. This is an affine variety since the ideal $I$ is prime (because the quotient by $I$, which is the image of $h$, is an integral domain). By definition, homomorphisms of fields give rise to rational maps, and the two constructions are inverse to each other.
In particular, a variety $X$ is rational if and only if
$$K(X) \cong \mathbb{C}\left(x_{1}, \ldots, x_{n}\right)$$
as $\mathbb{C}$-agebras for some $n$. (Note: the right hand side means the field of rational functions on $A^{n}$, i.e. the field of fractions of the ring of polynomials $\mathbb{C}\left[x_{1}, \ldots, x_{n}\right]$.) We also say that the field $K(X)$ is rational.

## 数学代考|代数几何代写algebraic geometry代考|Standard Smooth Homomorphisms of Commutative Rings

A homomorphism of commutative rings
$$f: A \rightarrow B$$
is called standard smooth of dimension $k \geq 0$ if $f$ can be expressed as
$$A \rightarrow A\left[x_{1}, \ldots, x_{n}\right] /\left(f_{1}, \ldots, f_{m}\right) \cong B$$
where $n=m+k$, the first homomorphism sends $a \in A$ to $a$, and $f_{i}$ are polynomials such that the ideal in
$$A\left[x_{1}, \ldots, x_{n}\right] /\left(f_{1}, \ldots, f_{m}\right)$$
generated by the determinants of the $m \times m$ submatrices of the Jacobi matrix
$$\left(\begin{array}{ccc} \frac{\partial f_{1}}{\partial x_{1}} & \cdots & \frac{\partial f_{1}}{\partial x_{n}} \ \cdots & \cdots & \cdots \ \frac{\partial f_{m 1}}{\partial x_{1}} & \cdots & \frac{\partial f_{m}}{\partial x_{n}} \end{array}\right)$$
is $A\left[x_{1}, \ldots, x_{n}\right] /\left(f_{1}, \ldots, f_{m}\right)$ (or, equivalently, contains 1$)$. This is equivalent to saying that the ideal in $A\left[x_{1}, \ldots, x_{n}\right]$ generated by the determinants and $f_{1}, \ldots, f_{m}$ contains 1 . If $A$ is a field, this can be tested using Gröbner basis algorithm, which we will learn in the next Section.

## 数学代考|代数几何代写algebraic geometry代考|Definition of a Rational Map

F:在→是

G:在→是

F(X)=G(X) 对全部 X∈在∩在

X↦ķ(X)

H:C[X1,…,Xn]→ķ

ķ(X)≅C(X1,…,Xn)

## 数学代考|代数几何代写algebraic geometry代考|Standard Smooth Homomorphisms of Commutative Rings

F:一种→乙

(∂F1∂X1⋯∂F1∂Xn ⋯⋯⋯ ∂F米1∂X1⋯∂F米∂Xn)

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