## 数学代写|交换代数代写commutative algebra代考|MAST90025

2022年9月26日

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## 数学代写|交换代数代写commutative algebra代考|Dedekind-Mertens Lemma

D First of all, notice that the products $f_i g_j$ are the coefficients of the polynomial $f(Y) g(X)$. Similarly, for some indeterminates $Y_0, \ldots, Y_m$, the content of the polynomial $f\left(Y_0\right) \cdots f\left(Y_m\right) g(X)$ is equal to $c(f)^{m+1} \mathrm{c}(g)$.

Let $h=f g$. Imagine that in the ring $\mathbf{B}=\mathbf{A}\left[X, Y_0, \ldots, Y_m\right]$ we are able to show the membership of the polynomial $f\left(Y_0\right) \cdots f\left(Y_m\right) g(X)$ in the ideal
$$\sum_{j=0}^m\left(h\left(Y_j\right) \prod_{k, k \neq j}\left(f\left(Y_k\right)\right\rangle\right) .$$
We would immediately deduce that $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.
This is more or less what is going to happen. We get rid of the denominators in Lagrange’s interpolation formula (we need at least $1+\operatorname{deg} g$ interpolation points):
In the ring $\mathbf{B}$, by letting $\Delta=\prod_{j \neq k}\left(Y_j-Y_k\right)$, we get:
$$\Delta \cdot g(X) \in \sum_{j=0}^m\left\langle g\left(Y_j\right)\right\rangle .$$
Thus by multiplying by $f\left(Y_0\right) \cdots f\left(Y_m\right)$ :
$$\Delta \cdot f\left(Y_0\right) \cdots f\left(Y_m\right) \cdot g(X) \in \sum_{j=0}^m h\left(Y_j\right) \prod_{k, k \neq j}\left\langle f\left(Y_k\right)\right\rangle .$$
If we show that for any $Q \in \mathbf{B}$ we have $\mathrm{c}(Q)=\mathrm{c}(\Delta \cdot Q)$, the previous membership gives $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.
Note that $\mathrm{c}\left(Y_i Q\right)=\mathrm{c}(Q)$ and especially that
$$\mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right) \subseteq \mathrm{c}\left(Q\left(Y_0, Y_1, \ldots, Y_m\right)\right) .$$
Therefore, by putting $Y_0=\left(Y_0 \pm Y_1\right) \mp Y_1, \mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right)=\mathrm{c}(Q)$. The following polynomials thus all have the same content:
$$\begin{gathered} Q, Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), Y_0 Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), \ \left(Y_0-Y_1\right) Q\left(Y_0, Y_1, \ldots, Y_m\right) . \end{gathered}$$

## 数学代写|交换代数代写commutative algebra代考|The Universal Splitting Algebra for a Monic Polynomial

Disclaimer In a context where we manipulate algebras, it is sometimes preferable to keep to the intuition that we want to have a field as the base ring, even if it is only a commutative ring. In which case we choose to give a name such as $\mathbf{k}$ to the base ring. This is what we are going to do in this section dedicated to the universal splitting algebra.
When we are truly dealing with a discrete field, we will use $\mathbf{K}$ instead.
We now proceed to the inverse operation to that which passes from the polynomial ring to the subring of symmetric polynomials.

In the presence of a monic polynomial $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ over a ring $\mathbf{k}$, we want to have at our disposal an extension of $\mathbf{k}$ where the polynomial is decomposed into linear factors. Such an extension can be constructed in a purely formal way. The result is called the universal splitting algebra.
4.1 Definition and notation Let $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ be a monic polynomial of degree $n$. We denote by Adu $\mathbf{k}{\mathbf{k}, f}$ the universal splitting algebra of $f$ over k defined as follows $$\mathrm{Adu}{\mathbf{k}, f}=\mathbf{k}\left[X_1, \ldots, X_n\right] / \mathcal{J}(f)=\mathbf{k}\left[x_1, \ldots, x_n\right]$$
where $\mathcal{J}(f)$ is the ideal of symmetric relators necessary to identify $\prod_{i=1}^n\left(T-x_i\right)$ with $f(T)$ in the quotient. Precisely, if $S_1, S_2, \ldots, S_n$ are the elementary symmetric functions of the $X_i$ ‘s, the ideal $\mathcal{J}(f)$ is given by
$$\mathcal{J}(f)=\left\langle S_1-s_1, S_2-s_2, \ldots, S_n-s_n\right\rangle$$
The universal splitting algebra $\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$ can be characterized by the following property.
4.2 Fact (Universal decomposition algebra, characteristic property)

1. Let $\mathbf{C}$ be a $\mathbf{k}$-algebra such that $f(T)$ is decomposed into a product of factors $T-z_i$. Then, there exists a unique homomorphism of $\mathbf{k}$-algebras of $\mathbf{A}$ to $\mathbf{C}$ which sends the $x_i$ ‘s to the $z_i$ ‘s.
2. This characterizes the universal splitting algebra $\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$, up to unique isomorphism.
3. Moreover, if $\mathbf{C}$ is generated (as a $\mathbf{k}$-algebra) by the $z_i$ ‘s, the universal splitting algebra is isomorphic to a quotient of $\mathbf{A}$.

# 交换代数代考

## 数学代写|交换代数代写交换代数代考|Dedekind-Mertens引理

$$\sum_{j=0}^m\left(h\left(Y_j\right) \prod_{k, k \neq j}\left(f\left(Y_k\right)\right\rangle\right) .$$我们马上就能推断出来 $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$这或多或少就是将要发生的事情。我们去掉了拉格朗日插值公式中的分母(我们至少需要 $1+\operatorname{deg} g$ 插补点):

$$\Delta \cdot g(X) \in \sum_{j=0}^m\left\langle g\left(Y_j\right)\right\rangle .$$

$$\Delta \cdot f\left(Y_0\right) \cdots f\left(Y_m\right) \cdot g(X) \in \sum_{j=0}^m h\left(Y_j\right) \prod_{k, k \neq j}\left\langle f\left(Y_k\right)\right\rangle .$$
$Q \in \mathbf{B}$ 我们有 $\mathrm{c}(Q)=\mathrm{c}(\Delta \cdot Q)$，以前的会员给予 $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.

$$\mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right) \subseteq \mathrm{c}\left(Q\left(Y_0, Y_1, \ldots, Y_m\right)\right) .$$

$$\begin{gathered} Q, Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), Y_0 Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), \ \left(Y_0-Y_1\right) Q\left(Y_0, Y_1, \ldots, Y_m\right) . \end{gathered}$$

## 数学代写|交换代数代写交换代数代考|一个Monic多项式的通用分裂代数

4.1定义和符号Let $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ 是次的一多项式 $n$。我们用阿杜表示 $\mathbf{k}{\mathbf{k}, f}$ 的普适分裂代数 $f$ 除以k，定义如下 $$\mathrm{Adu}{\mathbf{k}, f}=\mathbf{k}\left[X_1, \ldots, X_n\right] / \mathcal{J}(f)=\mathbf{k}\left[x_1, \ldots, x_n\right]$$
where $\mathcal{J}(f)$ 是否有必要确定理想的对称关系式 $\prod_{i=1}^n\left(T-x_i\right)$ 用 $f(T)$ 在商中。没错，如果 $S_1, S_2, \ldots, S_n$ 初等对称函数 $X_i$ ’s，理想 $\mathcal{J}(f)$
$$\mathcal{J}(f)=\left\langle S_1-s_1, S_2-s_2, \ldots, S_n-s_n\right\rangle$$

1. 设$\mathbf{C}$为$\mathbf{k}$ -代数，使$f(T)$分解为因子$T-z_i$的乘积。然后，$\mathbf{A}$到$\mathbf{C}$的$\mathbf{k}$ -代数存在一个唯一同态，它将$x_i$ ‘s发送到$z_i$ ‘s。
2. 这表征了普遍分裂代数$\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$，直到唯一同态。

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