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。
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数学代写|交换代数代写commutative algebra代考|Decomposition of Polynomials into Products of Irreducible Factors
8.14 Lemma Let $\mathbf{K}$ be a discrete field. The polynomials of $\mathbf{K}[X]$ can be decomposed into products of irreducible factors if and only if we have an algorithm to compute the zeros in $\mathbf{K}$ of an arbitrary polynomial of $\mathbf{K}[X]$.
D The second condition is a priori weaker since it amounts to determining the factors of degree 1 for some polynomial of $\mathbf{K}[X]$. Assume this condition is satisfied. To know whether there exists a decomposition $f=g h$ with $g$ and $h$ monic of fixed degrees $>0$, we apply Kronecker’s theorem. We obtain for each coefficient of $g$ and $h$ a finite number of possibilities (they are the zeros of monic polynomials that we can explicitly express according to the coefficients of $f$ ).
8.15 Proposition In $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ the polynomials can be decomposed into products of irreducible factors. A nonconstant polynomial of $\mathbb{Z}[X]$ is irreducible in $\mathbb{Z}[X]$ if and only if it is primitive and irreducible in $\mathbb{Q}[X]$.
D For $\mathbb{Q}[X]$ we apply Lemma 8.14. We must therefore show that we know how to determine the rational zeros of a monic polynomial $f$ with rational coefficients. We can even assume that the coefficients of $f$ are integral. The elementary theory of divisibility in $\mathbb{Z}$ shows then that if $a / b$ is a zero of $f, a$ must divide the leading coefficient and $b$ the constant coefficient of $f$; there is therefore only a finite number of tests to execute.
For $\mathbb{Z}[X]$, a primitive polynomial $f$ being given, we want to know if there exists a decomposition $f=g h$ with $g$ and $h$ of fixed degrees $>0$. We can assume $f(0) \neq 0$. We apply Kronecker’s theorem. A product $g_0 h_j$ for instance must be a zero in $\mathbb{Z}$ of a monic polynomial $q_{0, j}$ of $\mathbb{Z}[T]$ that we can compute. In particular, $g_0 h_j$ must divide $q_{0, j}(0)$, which only leaves a finite number of possibilities for $h_j$.
数学代写|交换代数代写commutative algebra代考|Rings of Integers of a Number Field
If $\mathbf{K}$ is a number field its ring of integers is the integral closure of $\mathbb{Z}$ in $\mathbf{K}$.
8.17 Proposition and definition (Discriminant of a number field) Let $\mathbf{K}$ be a number field and $\mathbf{Z}$ its ring of integers.
An element $y$ of $\mathbf{K}$ is in $\mathbf{Z}$ if and only if $\operatorname{Min}_{\mathbb{Q}, y}(X) \in \mathbb{Z}[X]$.
We have $\mathbf{K}=\left(\mathbb{N}^*\right)^{-1} \mathbf{Z}$.
Assume that $\mathbf{K}=\mathbb{Q}[x]$ with $x \in \mathbf{Z}$. Let $f(X)=\operatorname{Min}_{\mathbb{Q}, x}(X)$ be in $\mathbb{Z}[X]$ and $\Delta^2$ be the greatest square factor of $\operatorname{disc}_X f$.
Then, $\mathbb{Z}[x] \subseteq \mathbf{Z} \subseteq \frac{1}{\Delta} \mathbb{Z}[x]$
The ring $\mathbf{Z}$ is a free $\mathbb{Z}$-module of rank $[\mathbf{K}: \mathbf{Q}]$.
The integer $\operatorname{Disc}_{\mathbf{Z} / \mathbb{Z}}$ is well-defined. We call it the discriminant of the number field $\mathbf{K}$.
D 1. Results from Lemma 8.10 (Kronecker’s theorem).
Let $y \in \mathbf{K}$ and $g(X) \in \mathbb{Z}[X]$ be a nonzero polynomial that annihilates $y$. If $a$ is the leading coefficient of $g$, ay is integral over $\mathbb{Z}$.
Let $\mathbf{A}=\mathbb{Z}[x]$ and $n=[\mathbf{K}: \mathbf{Q}]$. Let $z \in \mathbf{Z}$, which we as $h(x) / \delta$ with $\delta \in \mathbb{N}^*$, $\langle\delta\rangle+\mathrm{c}(h)=\langle 1\rangle$ and $\operatorname{deg} h<n$. We have $\mathbf{A}+\mathbb{Z} z \subseteq \frac{1}{\delta} \mathbf{A}$ and it thus suffices to prove that $\delta^2$ divides $\operatorname{disc}X(f)$. The ring $\mathbf{A}$ is a free $\mathbb{Z}$-module of rank $n$, with the basis $\mathcal{B}_0=\left(1, x, \ldots, x^{n-1}\right)$. Proposition 5.10 gives $$ \operatorname{Disc}{\mathbf{A} / \mathbb{Z}}=\operatorname{disc}{\mathbf{A} / \mathbb{Z}}\left(\mathcal{B}_0\right)=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}_0\right)=\operatorname{disc}_X f
$$
交换代数代考
数学代写|交换代数代写commutative algebra代考|Decomposition of Polynomials into Products of Irreducible Factors
8.14引理设$\mathbf{K}$为离散域。当且仅当我们有一种算法来计算$\mathbf{K}[X]$的任意多项式在$\mathbf{K}$中的零时,$\mathbf{K}[X]$的多项式可以分解为不可约因子的乘积。
第二个条件是先验的较弱的,因为它相当于确定某个多项式的1次因子 $\mathbf{K}[X]$. 假设满足此条件。知道是否存在分解 $f=g h$ 有 $g$ 和 $h$ 固定度的Monic $>0$,我们应用克罗内克定理。的每个系数,我们得到 $g$ 和 $h$ 有限数量的可能性(它们是我们可以根据的系数显式表示的一元多项式的零) $f$ ).
8.15 In命题 $\mathbb{Z}[X]$ 和 $\mathbb{Q}[X]$ 多项式可以分解成不可约因子的乘积。的非常数多项式 $\mathbb{Z}[X]$ 是不可约的 $\mathbb{Z}[X]$ 当且仅当它是基本且不可约的 $\mathbb{Q}[X]$.
D对于$\mathbb{Q}[X]$,我们应用引理8.14。因此,我们必须证明我们知道如何确定具有有理数系数的单多项式$f$的有理数零。我们甚至可以假设$f$的系数是积分的。由$\mathbb{Z}$的可整除性的基本理论可知,如果$a / b$是$f, a$的零,则必须整除$f$的前导系数和$b$的常系数;因此,要执行的测试数量是有限的。
对于$\mathbb{Z}[X]$,给定一个原始多项式$f$,我们想知道是否存在一个分解$f=g h$与$g$和$h$的固定度$>0$。我们可以假设$f(0) \neq 0$。我们应用克罗内克定理。例如,产品$g_0 h_j$必须是我们可以计算的$\mathbb{Z}[T]$的单多项式$q_{0, j}$在$\mathbb{Z}$中的零。特别地,$g_0 h_j$必须除以$q_{0, j}(0)$,这就给$h_j$留下了有限的可能性。
数学代写|交换代数代写commutative algebra代考|Rings of Integers of a Number Field
如果$\mathbf{K}$是一个数字字段,那么它的整数环就是$\mathbf{K}$中$\mathbb{Z}$的整闭包。
8.17命题与定义(数域的判别式)设$\mathbf{K}$为一个数域,$\mathbf{Z}$为它的整数环。
当且仅当$\operatorname{Min}_{\mathbb{Q}, y}(X) \in \mathbb{Z}[X]$时,$\mathbf{K}$的元素$y$才会出现在$\mathbf{Z}$中。
我们有$\mathbf{K}=\left(\mathbb{N}^*\right)^{-1} \mathbf{Z}$。
假设$\mathbf{K}=\mathbb{Q}[x]$和$x \in \mathbf{Z}$。设$f(X)=\operatorname{Min}_{\mathbb{Q}, x}(X)$在$\mathbb{Z}[X]$中,$\Delta^2$是$\operatorname{disc}_X f$的最大平方因子。
然后, $\mathbb{Z}[x] \subseteq \mathbf{Z} \subseteq \frac{1}{\Delta} \mathbb{Z}[x]$
戒指$\mathbf{Z}$是一个免费的$\mathbb{Z}$ -模块,排名$[\mathbf{K}: \mathbf{Q}]$。
整数$\operatorname{Disc}_{\mathbf{Z} / \mathbb{Z}}$是定义良好的。我们称它为数字域$\mathbf{K}$的判别式。
解析:选D。引理8.10(克罗内克定理)的结果。
设$y \in \mathbf{K}$和$g(X) \in \mathbb{Z}[X]$是一个非零多项式,它湮没了$y$。如果$a$是$g$的前系数,ay是对$\mathbb{Z}$的积分。
让$\mathbf{A}=\mathbb{Z}[x]$和$n=[\mathbf{K}: \mathbf{Q}]$。让$z \in \mathbf{Z}$,我们将其与$\delta \in \mathbb{N}^*$, $\langle\delta\rangle+\mathrm{c}(h)=\langle 1\rangle$和$\operatorname{deg} h<n$表示为$h(x) / \delta$。我们有$\mathbf{A}+\mathbb{Z} z \subseteq \frac{1}{\delta} \mathbf{A}$,因此足以证明$\delta^2$能除$\operatorname{disc}X(f)$。环$\mathbf{A}$是一个免费的$\mathbb{Z}$ -模块,等级为$n$,基为$\mathcal{B}_0=\left(1, x, \ldots, x^{n-1}\right)$。提案5.10给出 $$ \operatorname{Disc}{\mathbf{A} / \mathbb{Z}}=\operatorname{disc}{\mathbf{A} / \mathbb{Z}}\left(\mathcal{B}_0\right)=\operatorname{disc}{\mathbf{K} / \mathbb{Q}}\left(\mathcal{B}_0\right)=\operatorname{disc}_X f
$$
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