### 数学代考|代数几何代写algebraic geometry代考|Regular Functions on an Affine Variety

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## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on an Affine Variety

Regular functions on an affine variety (or, more generally, an affine algebraic set) $X \subseteq \mathbb{A}{C}^{n}$ are also polynomials in the sense that they do not have a denominator. More precisely, we have $$\mathbb{C}[X]=\mathbb{C}\left[x{1}, \ldots, x_{n}\right] / I(X)$$
where $I(X)$ is the ideal of all polynomials which are 0 at every point of $X$. Recall that the division symbol in (1.4.3) denotes cosets, i.e. the elements of the ring (1.4.3) are cosets, which are sets of the form
$$p+I(X)={p+q \mid q \in I(X)}$$
Recall that cosets are the algebraic device for setting the elements of $I(X)$ equal to 0 in the ring (1.4.3), which is what we want, since they are constantly zero as functions on $X$.
To see that (1.4.3) is the correct formula for the ring of regular functions on $X$, first note that the right hand side of (1.4.3) maps injectively into the left hand side by the definition

of the ideal $I(X)$. To show that the map is onto, we need to show that if we cover $X$ with Zariski open sets $U_{i}$ in $\mathbb{A}{C}^{n}$ and exhibit rational functions $g{i} / h_{i}$ where $h_{i} \neq 0$ on $U_{i}, i \in S$, which such that $g_{i} / h_{i}=g_{j} / h_{j}$ on $U_{i} \cap U_{j} \cap X$, then there exists a polynomial $\phi$ which restricts to each $g_{i} / h_{i}$ on $U_{i} \cap X$. To this end, first note that we can assume that the set $S$ is finite. This is because by the Nullstellensatz,
$$1 \in I(X)+\sum_{i \in S} I\left(\mathbb{A}{C}^{n} \backslash U{i}\right)$$
and so the indexing set $S$ can be made finite, since only finite sums of elements are allowed. Suppose, then,
$$S={1, \ldots, n}$$
Now also by the Nullstellensatz, there exist polynomials $a_{1}, \ldots, a_{n}, q$ such that $q \in I(X)$ and
$$a_{1} h_{1}+\cdots+a_{n} h_{n}+q=1$$
Then one verifies that the polynomial
$$\phi=a_{1} g_{1}+\cdots+a_{n} g_{n}$$
restricts to $g_{i} / h_{i}$ on each of the Zariski open sets $U_{i} \cap X$ (See Exercise 4.)

## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on the Complement of a Set of Zeros

Let $X$ be an affine variety and let $f \in \mathbb{C}[X]$. Then the complement $X \backslash Z(f)$ is a special kind of quasiaffine variety. (We will see later that, in some sense, “it is still affine,” although not according to our current definition.) We have
$$\mathbb{C}[X \backslash Z(f)]=\mathbb{C}[X]\left[f^{-1}\right]=\mathbb{C}[X]\left[\frac{1}{f}\right]$$
Note that it is alright to take the reciprocal of $f$, because $X \backslash Z(f)$ does not contain any zeros of $f$. This construction is a special case of localization $S^{-1} R=R\left[S^{-1}\right]$ of a ring $R$ with respect to a subset $1 \in S \subseteq R$ closed under multiplication. On the set of “fractions” $r / s$ (i.e., formally, pairs $(r, s)$ ) with $r \in R$ and $s \in S$, define an equivalence relation where $r_{1} / s_{1} \sim r_{2} / s_{2}$ when there exists some $t \in S$ such that $r_{1} s_{2} t=r_{2} s_{1} t$. The set of equivalence classes is $R\left[S^{-1}\right]$. In our case, we let $S$ be the set of all $f^{n}, n \in \mathbb{N}_{0}$. We can alternately describe
$$R\left[f^{-1}\right]=R[t] /(f t-1)$$

The proof of (1.4.5) is actually essentially the same as in Sect. 1.4.3, with the exception that we have
$$a_{1} h_{1}+\cdots+a_{n} h_{n}=f^{N}$$
for some natural number $N$ (which is true by the Nullstellensatz). We then put
$$\phi=a_{1} g_{1} / f^{N}+\cdots+a_{n} g_{n} / f^{N}$$

## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on a Quasiaffine Variety

Let $0 \neq f_{1}, \ldots, f_{m} \in \mathbb{C}[X]$ where $X$ is an affine variety. Then
$$\mathbb{C}\left[X \backslash Z\left(f_{1}, \ldots f_{m}\right)\right]=\mathbb{C}[X]\left[f_{1}^{-1}\right] \cap \cdots \cap \mathbb{C}[X]\left[f_{m}^{-1}\right]$$
Note that the intersection on the right hand side of (1.4.6) is formed in the field of rational functions $K(X)$, which is the field of fractions of the ring $\mathbb{C}[X]$. The field of fractions $Q R$ of an integral domain $R$ is the localization with respect to the set of all non-zero elements, which is closed under multiplication because $R$ is an integral domain. Also, the canonical map $R \rightarrow Q R$ is injective, by cancellation. Conversely, a subring of a field (more generally an integral domain) is obviously an integral domain. Thus, integral domains are precisely those rings which are subrings of fields.
The reason $\mathbb{C}[X]$ is an integral domain is that $X$ is irreducible. Now we have
$$X \backslash Z\left(f_{1}, \ldots, f_{m}\right)=\bigcup_{i=1}^{m} X \backslash Z\left(f_{i}\right)$$
Thus, we can characterize a regular function on $X \backslash Z\left(f_{1}, \ldots, f_{m}\right)$ by a collection of regular functions on $X \backslash Z\left(f_{i}\right)$ which coincide on intersections, but that is equivalent to coinciding in $K(X)$ since $\mathbb{C}[X]$, and hence also $\mathbb{C}[X]\left[f_{i}^{-1}\right]$, are integral domains. Thus, (1.4.6) follows.

## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on an Affine Variety

p+一世(X)=p+q∣q∈一世(X)

1∈一世(X)+∑一世∈小号一世(一种Cn∖在一世)

φ=一种1G1+⋯+一种nGn

## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on the Complement of a Set of Zeros

C[X∖从(F)]=C[X][F−1]=C[X][1F]

R[F−1]=R[吨]/(F吨−1)

(1.4.5) 的证明实际上与 Sect. 1.4.3，除了我们有

φ=一种1G1/Fñ+⋯+一种nGn/Fñ

## 数学代考|代数几何代写algebraic geometry代考|Regular Functions on a Quasiaffine Variety

C[X∖从(F1,…F米)]=C[X][F1−1]∩⋯∩C[X][F米−1]

X∖从(F1,…,F米)=⋃一世=1米X∖从(F一世)

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