如果你也在 怎样代写交换代数Commutative Algebra 这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。交换代数Commutative Algebra是计划的局部研究中的主要技术工具。对不一定是换元的环的研究被称为非换元代数;它包括环理论、表示理论和巴拿赫代数的理论。
交换代数Commutative Algebra换元代数本质上是对代数数论和代数几何中出现的环的研究。在代数理论中,代数整数的环是Dedekind环,因此它构成了一类重要的换元环。与模块化算术有关的考虑导致了估值环的概念。代数场扩展对子环的限制导致了积分扩展和积分封闭域的概念,以及估值环扩展的公理化概念。
statistics-lab™ 为您的留学生涯保驾护航 在代写交换代数commutative algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写交换代数commutative algebra代写方面经验极为丰富,各种代写交换代数commutative algebra相关的作业也就用不着说。
数学代写|交换代数代写commutative algebra代考|Definition of the Discriminant of a Monic Polynomial
We define the discriminant of a univariate monic polynomial $f$ over a commutative ring A starting with the case where $f$ is the generic monic polynomial of degree $n$ :
$$
f(T)=T^n-S_1 T^{n-1}+S_2 T^{n-2}+\cdots+(-1)^n S_n \in \mathbb{Z}\left[S_1, \ldots, S_n\right][T]=\mathbb{Z}[S][T] .
$$
We can write $f(T)=\prod_i\left(T-X_i\right)$ in $\mathbb{Z}\left[X_1, \ldots, X_n\right]$ (Corollary 1.6), and we set
$$
\operatorname{disc}T(f)=(-1)^{n(n-1) / 2} \prod{i=1}^n f^{\prime}\left(X_i\right)=\prod_{1 \leqslant i<j \leqslant n}\left(X_i-X_j\right)^2 .
$$
As this polynomial in the $X_i$ ‘s is clearly variable permutation invariant, there exists a unique polynomial in the $S_i$ ‘s, $D_n\left(S_1, \ldots, S_n\right) \in \mathbb{Z}[S]$, which is equal to $\operatorname{disc}T(f)$. In short, the auxiliary variables $X_i$ can indeed vanish. Then, for a “concrete” polynomial $$ g(T)=T^n-s_1 T^{n-1}+s_2 T^{n-2}+\cdots+(-1)^n s_n \in \mathbf{A}[T], $$ we define $\operatorname{disc}_T(g)=D_n\left(s_1, \ldots, s_n\right)$. Naturally, if it happens that $g(T)=\prod{i=1}^n\left(T-b_i\right)$ in a ring $\mathbf{B} \supseteq \mathbf{A}$, we would then $\operatorname{obtain~}^{\operatorname{disc}T}(g)=\prod{1 \leqslant i<j \leqslant n}\left(b_i-b_j\right)^2$ by evaluating the formula (1). In particular, by using the universal splitting algebra we could directly define the discriminant by this formula.
A monic polynomial is said to be separable when its discriminant is invertible.
Diagonalization of the Matrices on a Ring
Let us first recall that if $f \in \mathbf{A}[T]$, a zero of $f$ in an $\mathbf{A}$-algebra $\mathbf{B}$ (given by a homomorphism $\varphi: \mathbf{A} \rightarrow \mathbf{B})$ is a $y \in \mathbf{B}$ which annihilates the polynomial $f^{\varphi}$, the image of $f$ in $\mathbf{B}[T]$.
In addition, the zero $y$ is said to be simple if $f^{\prime}(y) \in \mathbf{B}^{\times}$(we also say that it is a simple root of $f$ ).
Here, we are interested in the diagonalizations of matrices on an arbitrary commutative ring, when the characteristic polynomial is separable.
First of all, we have the classical “Kernels’ Lemma” II-4.8.
Next is a generalization of the theorem which states (in the discrete field case) that a simple zero of the characteristic polynomial defines a proper subspace of dimension 1.
数学代写|交换代数代写commutative algebra代考|The Generic Matrix is Diagonalizable
Consider $n^2$ indeterminates $\left(a_{i, j}\right){i, j \in \llbracket 1 . . n \rrbracket}$ and let $A$ be the corresponding matrix (it has coefficients in $\left.\mathbf{A}=\mathbb{Z}\left[\left(a{i, j}\right)\right]\right)$.
5.3 Proposition The generic matrix $A$ is diagonalizable over a ring $\mathbf{B}$ containing $\mathbb{Z}\left[\left(a_{i, j}\right)\right]=\mathbf{A}$.
Det $f(T)=T^n-s_1 T^{n-1}+\cdots+(-1)^n s_n$ be the characteristic polynomial of $A$. Then the coefficients $s_i$ are algebraically independent over $\mathbb{Z}$. To realize this, it suffices to specialize $A$ as the companion matrix of a generic monic polynomial.
In particular, the discriminant $\Delta=\operatorname{disc}(f)$ is nonzero in the integral ring $\mathbf{A}$. Then consider the ring $\mathbf{A}1=\mathbf{A}[1 / \Delta] \supseteq \mathbf{A}$ and the universal splitting algebra $\mathbf{C}=\operatorname{Adu}{\mathbf{A}1, f}$. Let the $x_i$ be the elements of $\mathbf{C}$ such that $f(T)=\prod_i\left(T-x_i\right)$. Finally, apply Proposition 5.2. If we want to obtain a diagonalizable matrix, we invert for instance $a=\prod_i \operatorname{det}\left(\left(A-x_i \mathbf{I}_n\right){1 . . n-1,1 . . n-1}\right)$. This is an element of $\mathbf{A}$ and it suffices to convince ourselves that it is nonzero by exhibiting a particular matrix, for example the companion matrix of the polynomial $X^n-1$.
Ultimately, consider $\mathbf{A}2=\mathbf{A}[1 /(a \Delta)] \supseteq \mathbf{A}$ and take $\mathbf{B}=\mathrm{Adu}{\mathbf{A}_2, f} \supseteq \mathbf{A}_2$.
The strength of the previous result, “which makes life considerably easier” is illustrated in the following two subsections.
交换代数代考
数学代写|交换代数代写commutative algebra代考|Definition of the Discriminant of a Monic Polynomial
我们定义了交换环a上的单变量单多项式$f$的判别式,从$f$是次为$n$的一般单多项式开始:
$$
f(T)=T^n-S_1 T^{n-1}+S_2 T^{n-2}+\cdots+(-1)^n S_n \in \mathbb{Z}\left[S_1, \ldots, S_n\right][T]=\mathbb{Z}[S][T] .
$$
我们可以在$\mathbb{Z}\left[X_1, \ldots, X_n\right]$(推论1.6)中写$f(T)=\prod_i\left(T-X_i\right)$,我们设置
$$
\operatorname{disc}T(f)=(-1)^{n(n-1) / 2} \prod{i=1}^n f^{\prime}\left(X_i\right)=\prod_{1 \leqslant i<j \leqslant n}\left(X_i-X_j\right)^2 .
$$
由于$X_i$’s中的这个多项式明显是变量置换不变量,所以在$S_i$’s中存在一个唯一的多项式$D_n\left(S_1, \ldots, S_n\right) \in \mathbb{Z}[S]$,它等于$\operatorname{disc}T(f)$。简而言之,辅助变量$X_i$确实可以消失。然后,对于一个“具体的”多项式$$ g(T)=T^n-s_1 T^{n-1}+s_2 T^{n-2}+\cdots+(-1)^n s_n \in \mathbf{A}[T], $$,我们定义$\operatorname{disc}_T(g)=D_n\left(s_1, \ldots, s_n\right)$。自然地,如果在环$\mathbf{B} \supseteq \mathbf{A}$中发生$g(T)=\prod{i=1}^n\left(T-b_i\right)$,那么我们将通过计算公式(1)来$\operatorname{obtain~}^{\operatorname{disc}T}(g)=\prod{1 \leqslant i<j \leqslant n}\left(b_i-b_j\right)^2$。特别是,通过使用全称分裂代数,我们可以直接用这个公式定义判别式。
当一个单多项式的判别式可逆时,我们就说它是可分离的。
环上矩阵的对角化
让我们首先回忆一下,如果$f \in \mathbf{A}[T]$,在$\mathbf{A}$ -代数$\mathbf{B}$(由同态$\varphi: \mathbf{A} \rightarrow \mathbf{B})$给出)中$f$的零是$y \in \mathbf{B}$,它湮灭了多项式$f^{\varphi}$, $f$在$\mathbf{B}[T]$中的像。
此外,如果$f^{\prime}(y) \in \mathbf{B}^{\times}$,零点$y$被认为是简单的(我们也说它是$f$的简单根)。
这里,我们对任意交换环上的矩阵的对角化感兴趣,当特征多项式是可分的。
首先,我们有经典的“核引理”II-4.8。
接下来是定理的推广,该定理指出(在离散域的情况下)特征多项式的简单零定义了维数为1的固有子空间。
数学代写|交换代数代写commutative algebra代考|The Generic Matrix is Diagonalizable
考虑$n^2$不确定$\left(a_{i, j}\right){i, j \in \llbracket 1 . . n \rrbracket}$,并设$A$为对应的矩阵(它在$\left.\mathbf{A}=\mathbb{Z}\left[\left(a{i, j}\right)\right]\right)$中有系数)。
5.3命题一般矩阵$A$在含有$\mathbb{Z}\left[\left(a_{i, j}\right)\right]=\mathbf{A}$的环$\mathbf{B}$上是可对角的。
Det $f(T)=T^n-s_1 T^{n-1}+\cdots+(-1)^n s_n$为$A$的特征多项式。那么系数$s_i$在$\mathbb{Z}$上是代数无关的。为了实现这一点,将$A$专门化为一般单多项式的伴随矩阵就足够了。
特别地,在积分环$\mathbf{A}$中,判别式$\Delta=\operatorname{disc}(f)$是非零的。然后考虑环$\mathbf{A}1=\mathbf{A}[1 / \Delta] \supseteq \mathbf{A}$和普遍分裂代数$\mathbf{C}=\operatorname{Adu}{\mathbf{A}1, f}$。让$x_i$成为$\mathbf{C}$的元素,这样$f(T)=\prod_i\left(T-x_i\right)$。最后,应用命题5.2。如果我们想要得到一个可对角化的矩阵,我们反演例如$a=\prod_i \operatorname{det}\left(\left(A-x_i \mathbf{I}_n\right){1 . . n-1,1 . . n-1}\right)$。这是$\mathbf{A}$的一个元素,通过展示一个特定的矩阵,例如多项式$X^n-1$的伴随矩阵,它足以使我们确信它是非零的。
最后,考虑$\mathbf{A}2=\mathbf{A}[1 /(a \Delta)] \supseteq \mathbf{A}$和$\mathbf{B}=\mathrm{Adu}{\mathbf{A}_2, f} \supseteq \mathbf{A}_2$。
前一个结果的强度“使生活变得相当容易”将在以下两个小节中说明。
统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。