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# 数学代写|交换代数代写Commutative Algebra代考|Classical specialization

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## 数学代写|交换代数代写Commutative Algebra代考|Classical specialization

Here, one assumes a more particular case of specialization, following the classical procedure of specializing variables of a ground polynomial ring. Let $A$ denote a Noetherian ring of finite Krull dimension-in the applications it will mostly be the ground ring of variables to be specialized-and let $R$ denote a finitely generated $A$-algebra. Let $\mathfrak{n} \subset A$ denote a maximal ideal such that the extended ideal $\mathfrak{n} R$ is prime, and $\mathcal{I} \subset R$ any ideal not contained in $\mathfrak{n} R$. Set $I:=(\mathcal{I}, \mathfrak{n}) / \mathfrak{n} R$ and $k:=A / \mathfrak{n}$.
Proposition 7.3.50. Consider the specialization homomorphism
$$\mathfrak{s}: \mathcal{R}R(\mathcal{I}) \otimes_A k \rightarrow \mathcal{R}{R / \mathfrak{n} R}(I)$$
Then:
(1) $\operatorname{ker}(\mathfrak{s})$ is a minimal prime ideal of $\mathcal{R}R(\mathcal{I}) \otimes_A k$ and, for any minimal prime $\mathfrak{Q}$ of $\mathcal{R}_R(\mathcal{I}) \otimes_A$ k other than $\operatorname{ker}(\mathfrak{s})$, one has that $\mathfrak{Q}$ corresponds to a minimal prime of $\operatorname{gr}{\mathcal{I}}(R) \otimes_A k \simeq\left(\mathcal{R}R(\mathcal{I}) \otimes_A k\right) / \mathcal{I}\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right)$ and so $$\operatorname{dim}\left(\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right) / \mathfrak{Q}\right) \leq \operatorname{dim}\left(\operatorname{gr}{\mathcal{I}}(R) \otimes_A k\right)$$
In particular,
$$\operatorname{dim}\left(\mathcal{R}R(\mathcal{I}) \otimes_A k\right)=\max \left{\operatorname{dim}(R / \mathfrak{n} R)+1, \operatorname{dim}\left(\operatorname{gr}{\mathcal{I}}(R) \otimes_A k\right)\right}$$
(2) Let $t \geq 0$ be an integer such that $\ell\left(\mathcal{I}{\mathfrak{P}}\right) \leq \operatorname{ht}(\mathfrak{P} / \mathrm{n} R)+t$ for every prime ideal $\mathfrak{P} \in$ $\operatorname{Spec}(R)$ containing $(\mathcal{I}, \mathfrak{n})$. Then $$\operatorname{dim}\left(\operatorname{gr}{\mathcal{I}}(R) \otimes_A k\right) \leq \operatorname{dim}(R / \mathrm{n} R)+t$$
(3) $\operatorname{dim}(\operatorname{ker}(\mathfrak{s})) \leq \operatorname{dim}\left(\operatorname{gr}_{\mathcal{I}}(R) \otimes_A k\right)$

Proof. (1) Let $P \in \operatorname{Spec}(R)$ be a prime ideal not containing $\mathcal{I}$. Localizing the surjection $\mathfrak{s}: \mathcal{R}R(\mathcal{I}) \otimes_A k \rightarrow \mathcal{R}{R / \mathrm{n} R}(I)$ at $R \backslash P$, one easily sees that it becomes an isomorphism. It follows that some power of $\mathcal{I}$ annihilates $\operatorname{ker}(\mathfrak{s})$, i.e.,
$$\mathcal{I}^l \cdot \operatorname{ker}(\mathfrak{s})=0$$
for some $l>0$. Since $I \neq 0$, then $\mathcal{I} \nsubseteq \operatorname{ker}(\mathfrak{s})$. Thus, any minimal prime ideal of $\mathcal{R}R(\mathcal{I}) \otimes_A k$ contains either the prime ideal $\operatorname{ker}(\mathfrak{s})$ or the ideal $\mathcal{I}$. Thus, $\operatorname{ker}(\mathfrak{s})$ is a minimal prime and any other minimal prime $\mathfrak{Q}$ of $\mathcal{R}_R(\mathcal{I}) \otimes_A k$ contains $\mathcal{I}$. Clearly, then any such $\mathfrak{Q}$ is a minimal prime of $\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right) / \mathcal{I}\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right) \simeq \operatorname{gr}{\mathcal{I}}(R) \otimes_A k$. Since $\operatorname{dim}\left(\mathcal{R}_{R / \mathrm{n} R}(I)\right)=$ $\operatorname{dim}(R / \mathfrak{n} R)+1$, the claim follows.

## 数学代写|交换代数代写Commutative Algebra代考|Combinatorial preliminaries

The interest here lies in the set of functions $\mathbb{Z} \rightarrow \mathbb{Z}$, which has a structure of an Abelian group $\mathfrak{Z}$ under the natural addition of functions. Now, a polynomial $p=$ $p(X) \in \mathbb{Q}[X]$ induces a function $\mathbb{Q} \rightarrow \mathbb{Q}$ by evaluation and, since $\mathbb{Q}$ is infinite, can be identified with the latter. One is interested in the subset of those $p$ such that $p(n) \in \mathbb{Z}$ for every $n \in \mathbb{Z}$. This set, identified with the corresponding subset of functions from $\mathbb{Z}$ to $\mathbb{Z}$, is a subgroup $\mathfrak{Q} \subset \mathfrak{Z}$. Its elements will be called polynomial functions-always keeping in mind that as polynomials they have rational coefficients.

The focus will be on the nongroup theoretic equivalence relation that identifies two functions $f, g: \mathbb{Z} \rightarrow \mathbb{Z}$ provided $f(n)=g(n)$ for all $n \gg 0$. The elements of $Z$ that belong to $\mathfrak{Q}$ up to this equivalence relation are very interesting; the Hilbert function will be one such, as one will see.

The (first) difference operator is the endomorphism $\Delta$ of $\mathfrak{Z}$ such that $\Delta(f)(n)=f(n+$ 1) $-f(n)$, for $f \in \mathcal{Z}$ and $n \in \mathbb{Z}$. For $l \geq 1$, the lth iterated $\Delta^l$ of $\delta$ is naturally defined by $\Delta^l=\Delta\left(\Delta^{l-1}\right)$, while $\Delta^0$ is defined by $\Delta^0(f)=f$, for every $f \in \mathcal{Z}$.

The reason to understand $\mathfrak{Z}$ in terms of its subgroup $\mathfrak{Q}$ is that the latter has a very simple structure.
Proposition 7.4.1. With the above notation, one has:
(a) (Polya, Ostrowski) Let $\left(\begin{array}{l}X \ i\end{array}\right)$ denote the polynomial function $n \mapsto\left(\begin{array}{l}n \ i\end{array}\right)$. Then $\mathfrak{Q}$ is the free Abelian subgroup with basis the set $\left{\left(\begin{array}{l}X \ i\end{array}\right) \mid i=0,1, \ldots\right}$.
(b) The following conditions are equivalent for an element $f \in \mathfrak{Z}$ :
(i) $f$ is a polynomial function of degree $\leq d$
(ii) There exists an integer $d \geq 1$ such that $\Delta^{d+1} f(n)=0$ for all $n \in \mathbb{Z}$.

## 数学代写|交换代数代写COMMUTATIVE ALGEBRA代考|CLASSICAL SPECIALIZATION

$$\mathfrak{s}: \mathcal{R} R(\mathcal{I}) \otimes_A k \rightarrow \mathcal{R} R / \mathrm{n} R(I)$$

$1 \operatorname{ker}(\mathfrak{s})$ 是的最小素理想 $\mathcal{R} R(\mathcal{I}) \otimes_A k$ 并且，对于任何最小素数 $\mathfrak{Q}$ 的 $\mathcal{R}R(\mathcal{I}) \otimes_A \mathrm{k}$ 除了ker $(\mathfrak{s})$,一个有那个 $\mathfrak{Q}$ 对应于的最小素数 $\operatorname{gr} \mathcal{I}(R) \otimes_A k \simeq\left(\mathcal{R} R(\mathcal{I}) \otimes_A k\right) / \mathcal{I}\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right)$ 所以 $$\operatorname{dim}\left(\left(\mathcal{R}_R(\mathcal{I}) \otimes_A k\right) / \mathfrak{Q}\right) \leq \operatorname{dim}\left(\operatorname{gr} \mathcal{I}(R) \otimes_A k\right)$$ 尤其， 2 让 $t \geq 0$ 是一个整数，使得 $\ell(\mathcal{I P}) \leq \mathrm{ht}(\mathfrak{P} / \mathrm{n} R)+t$ 对于每个素理想 $\mathfrak{P} \in \operatorname{Spec}(R)$ 含有 $(\mathcal{I}, \mathfrak{n})$. 然后 $$\operatorname{dim}\left(\operatorname{gr} \mathcal{I}(R) \otimes_A k\right) \leq \operatorname{dim}(R / \mathrm{n} R)+t$$ $$3 \operatorname{dim}(\operatorname{ker}(\mathfrak{s})) \leq \operatorname{dim}\left(\operatorname{gr}{\mathcal{I}}(R) \otimes_A k\right)$$

$$\mathcal{I}^l \cdot \operatorname{ker}(\mathfrak{s})=0$$

## 数学代写|交换代数代写COMMUTATIVE ALGEBRA代考|COMBINATORIAL PRELIMINARIES

a Polya, Ostrowski让 $(X i)$ 表示多项式函数 $n \mapsto(n i)$. 然后Q是基于集合的自由阿贝尔子群
$b$ 以下条件对于一个元素是等价的 $f \in Z:$
$i f$ 是次数的多项式函数 $\leq d$
$i i$ 存在一个整数 $d \geq 1$ 这样 $\Delta^{d+1} f(n)=0$ 对全部 $n \in \mathbb{Z}$.

## Matlab代写

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