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数学代写|交换代数代写COMMUTATIVE ALGEBRA代写|MATH724 Polynomial extensions

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数学代写|交换代数代写COMMUTATIVE ALGEBRA代写|Polynomial extensions

In this part, one considers the extension $R \subset R[X]$, where $X$ is an indeterminate over $R$. The case in sight is when $R$ is a Noetherian ring since otherwise the results are more complicated. The treatment here is taken from [138, Chapitre III, (D)].

It is quite elementary to see that, for an arbitrary ring $R$, if $P \subset R$ is a prime ideal then its extension $P R[X]$ is also prime and its contraction to $R$ is $P$. The next simple result for polynomial extensions gives more information.

Lemma 2.5.29. Let $R$ be an arbitrary ring and let $Q \subsetneq Q^{\prime} \subset R[X]$ be distinct prime ideals having the same contraction to $R$. Then the extension to $R[X]$ of the common contraction is $Q$.

Proof. Set $P=Q \cap R=Q^{\prime} \cap R$. Passing to the domain $R / P$, one may assume that $P={0}$ (why?). Passing to the ring of fractions with respect to $S=R \backslash{0}$, may further assume that $R$ is a field (why?). Say, $R=k$. But for the principal ideal domain $k[X]$ the result is clear since $\operatorname{dim} k[X]=1$
As an immediate consequence, one gets the following.
Corollary 2.5.30. Let $R$ be an arbitrary ring. Then for any ideal $I \subset R$ and any minimal prime ideal $P$ of $I$, the extended ideal $P R[X]$ is a minimal prime of the extension $I R[X]$.
Proof. Suppose not. Then $I R[X] \subset Q \nsubseteq P R[X]$ for some prime ideal $Q$ of $R[X]$. Since $P R[X]$ contracts to $P$, Lemma 2.5.29 gives a contradiction.
Proposition 2.5.31. Let $R$ be a Noetherian ring.
(i) $h t(P)=h t(P R[X])$ for any prime ideal $P$ of $R$.
(ii) $\operatorname{dim} R[X]=\operatorname{dim} R+1$.

数学代写|交换代数代写COMMUTATIVE ALGEBRA代写|Arbitrary extensions

In this part, one considers the more encompassing case of an extension $R \subset S$ of domains.
The results are inspired from the following.
Proposition 2.5.36. Let $R \subset S$ be a finitely generated extension with $S$ a domain. Let $P \subset S$ denote a prime ideal such that $P \cap R={0} .$ Then
$$\text { height } P=\operatorname{trdeg}{R}(S)-\operatorname{trdeg}{R}(S / P) \text {, }$$
Proof. Since $P \cap R={0}$, passing to the localization $R_{{0}}$ of $R$ at its zero (prime) ideal preserves the primeness of $P S_{{0}}$; consequently, height $P=$ height $P S_{{0}}$. On the other hand, since $R_{{0}}$ is a field and $S_{{0}}$ is finitely generated over $R_{{0}}$, Theorem 2.3.7 yields
$$\operatorname{trdeg}{R}(S)=\operatorname{trdeg}{R_{{0}}}\left(S_{{0}}\right)=\operatorname{dim} S_{{0}}$$
and
$$\operatorname{trdeg}{R}(S / P)=\operatorname{trdeg}{R_{{0}}}\left(S_{{0}} / P S_{{0}}\right)=\operatorname{dim} S_{{0}} / P S_{{0}},$$
so the result follows from Proposition $2.5 .33$ as mentioned in Corollary 2.5.34.
The first important result is due to I. S. Cohen ([38]). It has been popularized in [108] as the dimension inequality.

数学代写|交换代数代写 COMMUTATIVE ALGEBRA代 GIPolynomial extensions

(一世) $h t(P)=h t(P R[X])$ 对于任何睋理想 $P$ 的 $R$.
(二) $\operatorname{dim} R[X]=\operatorname{dim} R+1$.

数学代写|交换代数代写COMMUTATIVE ALGEBRA代写|bitrary extensions

height $P=\operatorname{trdeg} R(S)-\operatorname{trdeg} R(S / P)$,

$$\operatorname{trdeg} R(S)=\operatorname{trdeg} R_{0}\left(S_{0}\right)=\operatorname{dim} S_{0}$$

$$\operatorname{trdeg} R(S / P)=\operatorname{trdeg} R_{0}\left(S_{0} / P S_{0}\right)=\operatorname{dim} S_{0} / P S_{0}$$

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