### 数学代写|交换代数代写commutative algebra代考|MTH2141

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## 数学代写|交换代数代写commutative algebra代考|Integral closure of ideals

The subject was originally approached by H. Prüfer, later taken up by several authors, including Krull. One important source of problems is the theory of complete ideals in 2-dimensional regular local rings developed by O. Zariski in [169, Appendix 5], while a full update is in the book [150], parts of which were used in its essence in the following brief account.

Definition 2.2.12. Let $R$ be a ring and $I \subset R$ an ideal. An element $a \in R$ is said to integral over $I$ if it satisfies a polynomial $f(x) \in R[x]$ of the form
$$x^{m}+a_{m-1} x^{m-1}+\cdots+a_{1} x+a_{0}$$
for some $a_{i} \in I^{i}$, for all $i$.
The following is an analogue of Proposition 2.2.1.
Proposition 2.2.13. Let $R$ be a ring and $I \subset R$ an ideal. Given $a \in R$, the following are equivalent:
(i) $a \in \tilde{I}$.
(ii) There exists a finitely generated $R$-module $M$ such that $a \in I M:{R} M$ and such that $0:{R} M \subset \sqrt{0}:_{R} a$

Proof. (i) $\Rightarrow$ (ii) Since an equation of integral dependence of $a$ involves finitely many elements of $I$, it is clear that there exists a finitely generated subideal $J \subset I$ such that $a$ is still integral over it. But an equation of integral dependence of $a$ of degree $n$ implies that $a^{n} \in\left(J^{n}, J^{n-1}, \ldots, J a^{n-1}\right)=J(J, a)^{n-1}$. From this follows that $(J, a)^{n} \subset J(J, a)^{n-1}$,

hence
$$a(J, a)^{n-1} \subset(J, a)^{n} \subset J(J, a)^{n-1} \subset I(J, a)^{n-1}$$
Setting $M:=(J, a)^{n-1}$ is a solution of the problem, since if $b \in R$ kills $M$ then in particular it kills $a^{n-1}$, hence $b \in \sqrt{0}:_{R} a$.
(ii) $\Rightarrow$ (i) Apply the same method as in the proof of Proposition 2.2.1, (iii) $\Rightarrow$ (i), to a finite set of generators of $M$, using the hypothesis $a M \subset I M$, obtaining a certain square matrix $\mathcal{A}$. Then use the idea in Remark $2.2 .2$ to $\operatorname{derive} \operatorname{det}(\mathcal{A}) M=0$ and then apply the remaining hypothesis to get an equation of integral dependence of the form $(a \operatorname{det}(\mathcal{A}))^{l}=0$ for some $l$.

The set of all elements of $R$ integral over $I$ is called the integral closure (in $R$ ) of $I$, here denoted $\tilde{I}$. By extension, any intermediate ideal $I \subset J \subset \tilde{I}$ will be said to be integral over $I$.

## 数学代写|交换代数代写commutative algebra代考|Krull dimension and Noether normalization

The basic element of this part is a finite chain of prime ideals
$$P_{0} \subsetneq P_{1} \subsetneq \cdots \subsetneq P_{n}$$
The length of the chain (2.3.0.1) is the integer $n$. The next definition mimics somewhat the version of the dimension of a vector space by means of chains of subspaces. Yet, its impact is far greater allowing for introducing new invariants not found in the case of vector spaces.

Definition 2.3.1. The height of a prime ideal $P \subset R$ is the maximum of the lengths of chains of prime ideals whose top is $P$ :
$$P_{0} \subsetneq P_{1} \subsetneq \cdots \subsetneq P_{n}-P$$
One denotes this maximum by ht $(P)$ or ht $P$. A chain of primes as above is said to be saturated if no prime ideal can be properly inserted in the chain so as to increase its length. There is no offthand reason for ht $(P)$ to be actually a (finite) number. There are two types of obstruction: first, in principle a given chain may not be extended to a (finite) saturated one; second, there may be saturated chains of ever increasing length. Later one will show that for certain rings-Noetherian rings $-\mathrm{ht}(P)<\infty$ for any prime, but this is by no means a trivial result.

An alternative terminology is often used: the codimension of $P(\operatorname{denoted} \operatorname{cod} P)$. This is a slight abuse inherited from the special case where ht $P$ is a true codimension in the ambient ring $R$ and will mainly be used in those cases.
By a similar token, one can introduce the following notion.
Definition 2.3.2. The (Krull) dimension of the ring $R$ is the maximum of the heights of its prime ideals.

The dimension of $R$ will be denoted $\operatorname{dim} R$. Like for height, there is even less chance that an arbitrary ring have finite Krull dimension, as prime ideals might turn out to admit ever increasing heights. There are even examples of Noetherian rings with infinite Krull dimension (see Example 2.5.9).
This notion can be extended to an arbitrary ideal $I \subset R$ by setting
$$\text { ht } I:=\min {h t P \mid P \supset I \text { a prime ideal }} .$$
Clearly, one may restrict to the minimal prime overideals of $I$ in $R$-when $R$ is Noetherian, these will be finitely many (Proposition 2.5.20).

## 数学代写|交换代数代写commutative algebra代考|Noether normalization and the dimension theorem

The following result is due to E. Noether (see History $2.8 .3$ below).
Theorem 2.3.5 (Noether normalization). Let $R$ denote a finitely generated algebra over a field $k$. Then there exists a finite algebraically independent subset $\mathfrak{2}$ of $R$ such that $R$ is integral over the $k$-subalgebra $k[\mathfrak{A}]$.

Proof. The proof given here only works in the case where $k$ has an infinite number of elements $-$ the case where $k$ is a finite field will be left to the curious reader.

Write $R=k\left[x_{1}, \ldots, x_{n}\right]$ and induct on $n$. If $n=0$, take $\mathfrak{A}=\emptyset$. Assume that $n \geq 1$. If the set $\left{x_{1}, \ldots, x_{n}\right}$ is algebraically independent over $k$ (i. $e ., R$ is a polynomial ring over $k$, then one is done with $\mathfrak{A}=\left{x_{1}, \ldots, x_{n}\right}$. Thus, assume that $\left{x_{1}, \ldots, x_{n}\right}$ is algebraically dependent over $k$.

Take a nonzero polynomial $F$ in the polynomial ring $k\left[X_{1}, \ldots, X_{n}\right]$ (note the capital $X^{\prime}$ s) such that $F\left(x_{1}, \ldots, x_{n}\right)=0$. One may assume that the term in $X_{n}$ does not vanish.

Claim. Let $\mathrm{a}{i}=x{i}-c_{i} x_{n}$, for $i=1, \ldots, n-1$ and $c_{i} \in k$. Then, for suitable choice of the $c_{i}$ ‘s, $R$ is integral over its subalgebra $k\left[a_{1}, \ldots, a_{n-1}\right]$.

Note that, by the inductive assumption, the content of the claim is all that is needed. Thus, one proceeds to prove the claim. Note that $x_{i}=\mathfrak{a}{i}+c{i} x_{n}$, so substituting upon $F\left(x_{1}, \ldots, x_{n}\right)=0$ gives $F\left(a_{1}+c_{1} x_{n}, \ldots, a_{n-1}+c_{n-1} x_{n}, x_{n}\right)=0$. Let $f\left(X_{1}, \ldots, X_{n}\right)$ denote the term of $F$ of highest degree in $X_{n}$. Then, expanding the left-side as a polynomial in $x_{n}$, the term of highest degree has the form $x_{n}^{r} f\left(c_{1}, \ldots, c_{n-1}, 1\right)$, for some integer $r \geq 1$. Therefore, since $f\left(c_{1}, \ldots, c_{n-1}, 1\right) \in k$, provided this coefficient does not vanish, one gets an equation of integral dependence of $x_{n}$ over the subalgebra $k\left[a_{1}, \ldots, a_{n-1}\right]$, as required in the claim.

In order to make sure that $f\left(c_{1}, \ldots, c_{n-1}, 1\right) \neq 0$ for suitable choice of the $c_{j}$ ‘s is where the assumption that $k$ is infinite comes in. Indeed, suppose that $f\left(c_{1}, \ldots, c_{n-1}, 1\right)$ vanishes for any choice of $\left(c_{1}, \ldots, c_{n-1}\right) \in k^{n-1}$. Thus, one is saying that the nonzero polynomial $g=f\left(X_{1}, \ldots, X_{n-1}, 1\right)$ in $n-1$ variables vanishes for every $\left(c_{1}, \ldots, c_{n-1}\right) \in k^{n-1}$, an infinite vector space. From this, one can derive the existence of a nonzero polynomial in one variable over the same $k$ with an infinite set of roots, which is absurd.

## 数学代写|交换代数代写commutative algebra代考|Integral closure of ideals

X米+一个米−1X米−1+⋯+一个1X+一个0

(i)一个∈我~.
(ii) 存在一个有限生成的R-模块米这样一个∈我米:R米并且这样0:R米⊂0:R一个

(二)⇒(i) 应用与证明 2.2.1 相同的方法，(iii)⇒(i), 到有限的一组生成器米, 使用假设一个米⊂我米，得到某个方阵一个. 然后使用 Remark 中的想法2.2.2至派生⁡这⁡(一个)米=0然后应用剩余的假设得到形式的积分依赖方程(一个这⁡(一个))l=0对于一些l.

## 数学代写|交换代数代写commutative algebra代考|Krull dimension and Noether normalization

H T 我:=分钟H吨磷∣磷⊃我 一个首要理想 .

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