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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|Finding All Simple Modules

The Jordan-Hölder theorem shows that every module which has a composition series can be built from simple modules. Therefore, it is a fundamental problem of representation theory to understand what the simple modules of a given algebra are.
Recall from Example $2.25$ the following notion. Let $A$ be a $K$-algebra and $V$ an $A$-module. Then for every $v \in V$ we set $\operatorname{Ann}{A}(v)={a \in A \mid a v=0}$, and call this the annihilator of $v$ in $A$. We have seen in Example $2.25$ that for every $v \in V$ there is an isomorphism of $A$-modules $A / A{A} n_{A}(v) \cong A v$. In the context of simple modules this takes the following form, which we restate here for convenience.

Lemma 3.18. Let $A$ be a $K$-algebra and $S$ a simple A-module. Then for every non-zero $s \in S$ we have that $S \cong A / A_{A}(s)$ as A-modules.

Proof. As in Example $2.25$ we consider the A-module homomorphism $\psi: A \rightarrow S, \psi(a)=a s$. Since $S$ is simple and $s$ non-zero, this map is surjective by Lemma 3.3, and by definition the kernel is $\operatorname{Ann}{A}(s)$. So the isomorphism theorem yields $A / \operatorname{Ann}{A}(s) \cong \operatorname{im}(\psi)=A s=S$.

This implies in particular that if an algebra has a composition series, then it can only have finitely many simple modules:

Theorem 3.19. Let A be a $K$-algebra which has a composition series as an $A$ module. Then every simple A-module occurs as a composition factor of A. In particular, there are only finitely many simple A-modules, up to isomorphism.

Proof. By Lemma $3.18$ we know that if $S$ is a simple $A$-module then $S \cong A / I$ for some $A$-submodule $I$ of $A$. By Proposition $3.17$ there is some composition series of $A$ in which $I$ is one of the terms. Since $A / I$ is simple there are no further $A$ submodules between $I$ and $A$ (see Lemma 3.4). This means that $I$ can only appear as the penultimate entry in this composition series, and $S \cong A / I$, so it is a composition factor of $A$.
For finite-dimensional algebras we have an interesting consequence.

## 数学代写|表示论代写Representation theory代考|Simple Modules for Factor Algebras of Polynomial

We will now determine the simple modules for an algebra $A$ of the form $K[X] / I$ where $I$ is a non-zero ideal with $I \neq K[X]$; hence $I=(f)$ where $f$ is a polynomial of positive degree. Note that this does not require us to know a composition series of $A$, in fact we could have done this already earlier, after Lemma 3.4.
Proposition 3.23. Let $A=K[X] /(f)$ with $f \in K[X]$ of positive degree.
(a) The simple A-modules are up to isomorphism precisely the A-modules $K[X] /(h)$ where $h$ is an irreducible polynomial dividing $f$.
(b) Write $f=f_{1}^{a_{1}} \ldots f_{r}^{a_{r}}$, with $a_{i} \in \mathbb{N}$, as a product of irreducible polynomials $f_{i} \in K[X]$ which are pairwise coprime. Then A has precisely $r$ simple modules. up to isomorphism, namely $K[X] /\left(f_{1}\right), \ldots, K[X] /\left(f_{r}\right)$.

Proof. (a) First, let $h \in K\lfloor X\rfloor$ be an irreducible polynomial dividing $f$. Then $K[X] /(h)$ is an $A$-module, by Exercise $2.23$, with $A$-action given by
$$\left(g_{1}+(f)\right)\left(g_{2}+(h)\right)=g_{1} g_{2}+(h) .$$
Since $h$ is irreducible, the ideal $(h)$ is maximal, and hence $K[X] /(h)$ is a simple A-module, by Lemma 3.4.

Conversely, let $S$ be any simple $A$-module. By Lemmas $3.18$ and $3.4$ we know that $S$ is isomorphic to $A / U$ where $U$ is a maximal submodule of $A$. By the submodule correspondence, see Theorem $2.28$, we know $U=W /(f)$ where $W$ is an ideal of $K[X]$ containing $(f)$, that is, $W=(h)$ where $h \in K[X]$ and $h$ divides $f$. Applying the isomorphism theorem yields
$$A / U=(K[X] /(f)) /(W /(f)) \cong K[X] / W .$$
Isomorphisms preserve simple modules (see Exercise 3.3), so with $A / U$ the module $K[X] / W$ is also simple. This means that $W=(h)$ is a maximal ideal of $K[X]$ and then $h$ is an irreducible polynomial.
(b) By part (a), every simple A-module is isomorphic to one of $K[X] /\left(f_{1}\right), \ldots, K[X] /\left(f_{r}\right)$ (use that $K[X]$ has the unique factorization property, hence $f_{1}, \ldots, f_{r}$ are the unique irreducible divisors of $f$, up to multiplication by units). On the other hand, these $A$-modules are pairwise non-isomorphic: suppose $\psi: K[X] /\left(f_{i}\right) \rightarrow K[X] /\left(f_{j}\right)$ is an $A$-module homomorphism, we show that for $i \neq j$ it is not injective. Write $\psi\left(1+\left(f_{i}\right)\right)=g+\left(f_{j}\right)$ and consider the coset $f_{j}+\left(f_{i}\right)$ Since $f_{i}$ and $f_{j}$ are irreducible and coprime, this coset is not the zero element in $K[X] /\left(f_{i}\right)$. But it is in the kernel of $\psi$, since
\begin{aligned} \psi\left(f_{j}+\left(f_{i}\right)\right) &=\psi\left(\left(f_{j}+\left(f_{i}\right)\right)\left(1+\left(f_{i}\right)\right)\right)=\left(f_{j}+\left(f_{i}\right)\right) \psi\left(1+\left(f_{i}\right)\right) \ &=\left(f_{j}+\left(f_{i}\right)\right)\left(g+\left(f_{j}\right)\right)=f_{j} g+\left(f_{j}\right) \end{aligned}
which is the zero element in $K[X] /\left(f_{j}\right)$. In particular, $\psi$ is not an isomorphism.

## 数学代写|表示论代写Representation theory代考|Finding All Simple Modules

Jordan-Hölder 定理表明，每个具有组合序列的模块都可以由简单的模块构建。因此，理解给定代数的简单模块是什么是表示论的一个基 本问题。
$\mathrm{~ 从 示 例 中 回 忆 2 . 2 5 下 面 的 概 。}$ 称其为殀灭者 $v$ 在 $A$. 我们已经在示例中看到 $2.25$ 那对于每个 $v \in V$ 有一个同构 $A$-模块 $A / A A n_{A}(v) \cong A v$. 在简单模块的上下文中，它采 用以下形式，为方便起见，我们在此重述。

## 数学代写|表示论代写Representation theory代考|Simple Modules for Factor Algebras of Polynomial

(a) 简单 $\mathrm{A}$ 模完全符合同构 $\mathrm{A}$ 模 $K[X] /(h)$ 在哪里 $h$ 是不可约多项式除法 $f$.
(b) 写 $f=f_{1}^{a_{1}} \ldots f_{r}^{a_{r}}$ ，和 $a_{i} \in \mathbb{N}$ ，作为不可约多项式的乘积 $f_{i} \in K[X]$ 它们是成对互质的。那么 $\mathrm{A}$ 正好 $r$ 简单的模块。直到同构，即 $K[X] /\left(f_{1}\right), \ldots, K[X] /\left(f_{r}\right)$

$$\left(g_{1}+(f)\right)\left(g_{2}+(h)\right)=g_{1} g_{2}+(h) .$$

$$A / U=(K[X] /(f)) /(W /(f)) \cong K[X] / W \text {. }$$

(b) 根据 (a) 部分，每个简单的 A 模都同构于 $K[X] /\left(f_{1}\right), \ldots, K[X] /\left(f_{r}\right)$ (使用那个 $K[X]$ 具有唯一的因式分解性质，因此 $f_{1}, \ldots, f_{r}$ 是 唯一的不可约因数 $f$ ，最多乘以单位 $)$ 。另一方面，这些 $A$-模块是成对非同构的：假设 $\psi: K[X] /\left(f_{i}\right) \rightarrow K[X] /\left(f_{j}\right)$ 是一个 $A$-模同态， 我们证明对于 $i \neq j$ 它不是单射的。写 $\psi\left(1+\left(f_{i}\right)\right)=g+\left(f_{j}\right)$ 并考虑陪集 $f_{j}+\left(f_{i}\right)$ 自从 $f_{i}$ 和 $f_{j}$ 是不可约且互质的，这个陪集不是 $K[X] /\left(f_{i}\right)$. 但它在内核中 $\psi$ ，自从
$$\psi\left(f_{j}+\left(f_{i}\right)\right)=\psi\left(\left(f_{j}+\left(f_{i}\right)\right)\left(1+\left(f_{i}\right)\right)\right)=\left(f_{j}+\left(f_{i}\right)\right) \psi\left(1+\left(f_{i}\right)\right) \quad=\left(f_{j}+\left(f_{i}\right)\right)\left(g+\left(f_{j}\right)\right)=f_{j} g+\left(f_{j}\right)$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师