### 数学代写|交换代数代写commutative algebra代考|MATH2301

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

Given any ring homomorphism $\varphi: R \longrightarrow R^{\prime}$, it is quite convenient for our purposes to view $R^{\prime}$ as an $R$-algebra with respect to $\varphi$; see $1.4 / 3$. In particular, $R^{\prime}$ carries then a structure of an $R$-module, where products of type $r \cdot r^{\prime}$ for $r \in R, r^{\prime} \in R^{\prime}$ are defined by $r \cdot r^{\prime}=\varphi(r) \cdot r^{\prime}$, using the multiplication on $R^{\prime}$.
Definition 1. Let $\varphi: R \longrightarrow R^{\prime}$ be a homomorphism of rings. An element $x \in R^{\prime}$ is called integral over $R$ with respect to $\varphi$, or is said to depend integrally on $R$ with respect to $\varphi$, if it satisfies a so-called integral equation over $R$, i.e. if there are elements $a_{1}, \ldots, a_{n} \in R$ such that
$$x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=0 .$$
The ring $R^{\prime}$ is called integral over $R$, or $\varphi$ is called integral, if each $x \in R^{\prime}$ is integral over $R$.

Furthermore, $\varphi$ is called finite if it equips $R^{\prime}$ with the structure of a finite $R$ – module.

If $R=K$ and $R^{\prime}=K^{\prime}$ are fields, then $\varphi$ is injective and we may view $K$ as a subfield of $K^{\prime}$. Recall that the extension of fields $K \subset K^{\prime}$ is called algebraic if $K^{\prime}$ is integral over $K$.

Remark 2. Let $\varphi: R \hookrightarrow R^{\prime}$ be a monomorphism of integral domains such that $R^{\prime}$ is integral over $R$. Then $R$ is a field if and only if $R^{\prime}$ is a field.

Proof. Assume first that $R$ is a field and let $x \neq 0$ be an element in $R^{\prime}$. Then $x$ satisfies an integral equation over $R$,
$$x^{n}+a_{1} x^{n-1}+\ldots+a_{n}=0, \quad a_{1}, \ldots, a_{n} \in R .$$
Dividing out a suitable power of $x$, we may assume $a_{n} \neq 0$ and, hence, that $a_{n}$ is invertible in $R$. In the field of fractions of $R^{\prime}$ we can multiply the equation by $x^{-1}$. This yields
$$x^{-1}=-a_{n}^{-1}\left(x^{n-1}+a_{1} x^{n-2}+\ldots+a_{n-1}\right) \in R^{\prime}$$
hence $R^{\prime}$ is a field.

## 数学代写|交换代数代写commutative algebra代考|Noether Normalization and Hilbert’s Nullstellensatz

In this section we want to illustrate the concept of integral dependence by discussing polynomial rings $K[X]=K\left[X_{1}, \ldots, X_{n}\right]$ over a field $K$, for a finite set of variables $X=\left(X_{1}, \ldots, X_{n}\right)$. As we know already from $1.5 / 14, K[X]$ is Noetherian. Furthermore, it follows from the Lemma of Gauß (see [3], 2.7/1) that $K[X]$ is factorial and, hence, normal by $3.1 / 10$.

Let $A$ be a $K$-algebra and call a set of elements $x_{1}, \ldots, x_{n} \in$ A algebraically independent over $K$ if the $K$-homomorphism $K\left[X_{1}, \ldots, X_{n}\right] \rightarrow A$ substituting $x_{i}$ for the variable $X_{i}$ is injective, and algebraically dependent otherwise. Furthermore, as before, let us write $K\left[x_{1}, \ldots, x_{n}\right] \subset A$ for the image of such a substitution homomorphism. Recall that $A$ is called a $K$-algebra of finite type if there is a surjective $K$-homomorphism $K\left[X_{1}, \ldots, X_{n}\right] \cdot A$ or, in other words, if there exist elements $x_{1}, \ldots, x_{n} \in A$ such that $A=K\left[x_{1}, \ldots, x_{n}\right]$.
Theorem 1 (Noether’s Normalization Lemma). Let $A$ be a $K$-algebra of finite type. If $A \neq 0$, there exists a finite injective $K$-homomorphism
$$K\left[Y_{1}, \ldots, Y_{d}\right] \longleftrightarrow A$$
for a certain set of variables $Y_{1}, \ldots, Y_{d}$.
The proof of the Normalization Temma needs a terhnical recursion step, which is of interest by itself and which we will prove first.

Lemma 2. Let $K\left[x_{1}, \ldots, x_{n}\right]$ be a $K$-algebra of finite type and consider an element $y \in K\left[x_{1}, \ldots, x_{n}\right]$ given by some expression
$()$ $$y=\sum_{\left(\nu_{1}, \ldots, \nu_{n}\right) \in I} a_{\nu_{1}, \ldots, \nu_{n}} x_{1}^{\nu_{1}} \ldots x_{n}^{\nu_{n}}$$ with coefficients $a_{\nu_{1}, \ldots, \nu_{n}} \in K^{}$, where the summation extends over a finite nonempty index set $I \subset \mathbb{N}^{n}$; in particular, $n \geq 1$. Then there exist elements $y_{1}, \ldots, y_{n-1} \in K\left[x_{1}, \ldots, x_{n}\right]$ such that the canonical monomorphism
$$K\left[y_{1}, \ldots, y_{n-1}, y\right] \hookrightarrow K\left[x_{1}, \ldots, x_{n}\right]$$
is finite.

## 数学代写|交换代数代写commutative algebra代考|The Cohen-Seidenberg Theorems

In the present section we fix an integral ring homomorphism $\varphi: R \longrightarrow R^{\prime}$ and discuss the relationship between prime ideals in $R$ and $R^{\prime}$. Given a (prime) ideal $\mathfrak{P} \subset R^{\prime}$, we write $\mathfrak{P} \cap R$ for the restricted (prime) ideal $\varphi^{-1}(\mathfrak{P}) \subset R$.
Proposition 1. Let $\varphi: R \longrightarrow R^{\prime}$ be an integral ring homomorphism.
(i) A prime ideal $\mathfrak{P} \subset R^{\prime}$ is maximal if and only if its restriction $\mathfrak{p}=\mathfrak{P} \cap R$ is maximal in $R$.
(ii) Let $\mathfrak{P}{1}, \mathfrak{P}{2} \subset R^{\prime}$ be prime ideals satisfying $\mathfrak{P}{1} \cap R=\mathfrak{P}{2} \cap R$. Then $\mathfrak{P}{1} \subset \mathfrak{P}{2}$ implies $\mathfrak{P}{1}=\mathfrak{P}{2}$.

Proof. In the situation of (i), the map $\varphi$ induces an integral monomorphism of integral domains $R / \mathfrak{p} \longrightarrow R^{\prime} / \mathfrak{P}$, and we see from $3.1 / 2$ that $R / \mathfrak{p}$ is a field if and only if $R^{\prime} / \mathfrak{P}$ is a field. Hence, $\mathfrak{p}$ is maximal in $R$ if and only if $\mathfrak{P}$ is maximal in $R^{\prime}$.

Now let $\mathfrak{P}{1}, \mathfrak{P}{2} \subset R^{\prime}$ be two prime ideals as required in (ii), namely such that the induced ideals $\mathfrak{p}=\mathfrak{P}{1} \cap R$ and $\mathfrak{P}{2} \cap R$ coincide in $R$. Let $S=R-\mathfrak{p}$ and consider the ring homomorphism $R_{S} \longrightarrow R_{\varphi(S)}^{\prime}$ induced from $\varphi$, which is integral by $3.1 / 3$ (ii) since $\varphi$ is integral. Then, by $1.2 / 7$, the ideal $S^{-1} \mathfrak{p}$ generated by $\mathfrak{p}$ in $R_{S}$ is maximal. Likewise we can consider the ideals $S^{-1} \mathfrak{P}{1}$ and $S^{-1} \mathfrak{P}{2}$ generated by $\mathfrak{P}{1}$ and $\mathfrak{Y}{2}$ in $R_{\varphi(S)}^{\prime}$. These are prime by $1.2 / 5$ since $\varphi(S)$ is disjoint from $\mathfrak{F}{1}$ and $\mathfrak{P}{2}$. Then $\mathfrak{q}{1}=S^{-1} \mathfrak{P}{1} \cap R_{S}$ and $\mathfrak{q}{2}=S^{-1} \mathfrak{P}{2} \cap R_{S}$ are prime ideals in $R_{S}$ that contain $S^{-1} \mathfrak{p}$. Because $S^{-1} \mathfrak{p}$ is maximal in $R_{S}$, we get $S^{-1} \mathfrak{p}=\mathfrak{q}{1}=\mathfrak{q}{2}$. Thus, $S^{-1} \mathfrak{F}{1}$ and $S^{-1} \mathfrak{F}{2}$ are two ideals in $R_{\varphi(S)}^{\prime}$ whose restrictions to $R_{S}$ are maximal. But then, by (i), we know that $S^{-1} \mathfrak{P}{1}$ and $S^{-1} \mathfrak{F}{2}$ are maximal in $R_{\varphi(S)}^{\prime}$ and it follows $S^{-1} \mathfrak{P}{1}=S^{-1} \mathfrak{F}{2}$ from $\mathfrak{P}{1} \subset \mathfrak{F}{2}$. Since $\mathfrak{F}{1}=S^{-1} \mathfrak{F}{1} \cap R^{\prime}$ and $\mathfrak{F}{2}=S^{-1} \mathfrak{F}{2} \cap R^{\prime}$ by $1.2 / 5$, we conclude $\mathfrak{F}{1}=\mathfrak{F}{2}$, as desired.

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

Xn+一个1Xn−1+…+一个n=0.

Xn+一个1Xn−1+…+一个n=0,一个1,…,一个n∈R.

X−1=−一个n−1(Xn−1+一个1Xn−2+…+一个n−1)∈R′

## 数学代写|交换代数代写commutative algebra代考|Noether Normalization and Hilbert’s Nullstellensatz

ķ[是1,…,是d]⟷一个

Normalization Temma 的证明需要一个 terhnical recursion 步骤，这本身就很有趣，我们将首先证明它。

()

ķ[是1,…,是n−1,是]ķ[X1,…,Xn]

## 数学代写|交换代数代写commutative algebra代考|The Cohen-Seidenberg Theorems

(i) 一个主要理想磷⊂R′当且仅当它的限制是最大的p=磷∩R是最大的R.
(ii) 让磷1,磷2⊂R′成为满足的首要理想磷1∩R=磷2∩R. 然后磷1⊂磷2暗示磷1=磷2.

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