# 数学代写|同调代数代写Homological Algebra代考|MMATH4301 Radicals and Jordan–Hölder Conditions

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## 数学代写|同调代数代写Homological Algebra代考|Radicals and Jordan–Hölder Conditions

Ordinary representation theory of finite groups, i.e. over fields of characteristic 0 , can be done using the following three basic theorems. First, Maschke’s theorem shows that the group ring of a finite group over a ground field is semisimple whenever the group order is invertible in the ground field. Second, Wedderburn’s theorem determines the structure of semisimple finite dimensional algebras over a ground field, and third Krull-Schmidt’s theorem proves that a direct sum decomposition of finite dimensional modules into indecomposables is unique up to permutation and isomorphism of factors.

We have seen that the hypothesis on the group order being invertible is indeed a non-trivial hypothesis and that not all algebras are semisimple. What can we say about these more general algebras?

Working with Noetherian modules almost always includes the use of a particularly important result, namely Nakayama’s lemma, which is shown below. This involves the notion of a radical.

Definition 1.6.1 Let $A$ be an algebra and let $M$ be an $A$-module. The Jacobson radical $\operatorname{rad}(M)$ of $M$ is the intersection of all maximal $A$-submodules of $M$. We sometimes call the Jacobson radical the radical if no confusion may occur. Moreover, $\operatorname{rad}(A)$ is the radical of the regular $A$-module.

Example 1.6.2 The radical of $\mathbb{Z}$ is 0 , as it is the intersection of all prime ideals. Indeed, an element is in the radical of $\mathbb{Z}$ if it is divisible by all primes.
The radical of a semisimple artinian ring is 0 by Lemma 1.4.28.
Lemma 1.6.3 Let $A$ be a ring. Then $\operatorname{rad}(A)$ is a two-sided ideal of $A$.
Proof By definition, $\operatorname{rad}(A)$ is a left ideal of $A$. If $m$ is a maximal ideal of $A$, then $A / m$ is a simple $A$-module. If $S$ is a simple $A$-module, then by Remark 1.4.25 we get that $S=A \cdot s$ for any $s \in S \backslash{0}$ and $S \simeq A / \operatorname{ann}(s)$ where $\operatorname{ann}(s):={a \in A \mid a \cdot s=0} .$ Moreover, $\operatorname{ann}(s)$ is a maximal ideal. Indeed, if $\operatorname{ann}(s)<\mathfrak{m}$, then $A / \mathfrak{m}$ is a quotient of $A / \operatorname{ann}(s) \simeq S$. Now, $\operatorname{ann}(S):=\bigcap_{s \in S \backslash{0}} \operatorname{ann}(s)$ is a two-sided ideal of $A$ since $S$ is an $A$-module. Therefore $\operatorname{rad}(A)$ is included in the intersection of the two-sided ideals $\operatorname{ann}(S)$ for all simple $A$-modules $S$. If $S$ is a simple $A$-module, then $S \simeq A / \mathrm{m}$ for some maximal ideal $m$. If $a \in \operatorname{ann}(S)$, then $a \cdot A \subseteq \mathfrak{m}$ and hence $a \in \mathfrak{m}$. Therefore $\operatorname{ann}(S) \leq \mathfrak{m}$ and the intersection of $\operatorname{ann}(S)$ over all simple $A$-modules is included in $\operatorname{rad}(A)$. Hence we get that the intersection of $\operatorname{ann}(S)$, the intersection taken over all simple $A$-modules, equals $\operatorname{rad}(A)$. But these annihilators $\operatorname{ann}(S)$ are two-sided ideals of $A$.

## 数学代写|同调代数代写Homological Algebra代考|The Jordan–Hölder Theorem

We have already seen the fundamental role of the Krull-Schmidt theorem. In general, however, this does not give us full information. Indeed, given an algebra $A$, the KrullSchmidt theorem shows that given a Noetherian and artinian $A$-module $M$ then $M$ can be decomposed in an essentially unique way into indecomposable modules. In the semisimple situation we have seen that indecomposable modules are simple, and the classification of simple modules implies the knowledge of any given module. If $A$ is not semisimple, indecomposable modules can be constructed from two copies of one simple module in various ways to obtain a fairly complicated structure. We give a first example.

Example 1.6.23 Let $K$ be a field and let $K\langle X, Y\rangle$ be the free algebra with two generators $X$ and $Y$ (cf Definition 1.4.30 for a precise definition) and let $A$ be the quotient $K\left\langle e_1, e_2, X, Y\right\rangle / I$ where $I$ is the smallest two-sided ideal of $K\langle X, Y\rangle$ containing the set
\begin{aligned} &\left{X^2, Y^2, X Y, Y X, e_1^2-e_1, e_2^2-e_2, e_1 e_2, e_2 e_1\right. \ &\left.e_1 X-X e_2, e_1 Y-Y e_2, e_1 X-X, e_1 Y-Y, e_2 X, e_2 Y, X e_1, X e_2\right} \end{aligned}
More economically and suggestively we write $A$ as a quiver algebra as follows.

## 数学代写|同调代数代写Homological Algebra代考| radical and Jordan-Hölder Conditions

1.6.1设$A$为代数，设$M$为$A$ -module。$M$的Jacobson根式$\operatorname{rad}(M)$是$M$的所有极大子模块$A$的交集。如果没有混淆的话，我们有时称雅各布森根号为根号。此外，$\operatorname{rad}(A)$是常规的$A$ -module的根

$\operatorname{rad}(A)$ 左理想是什么 $A$。如果 $m$ 最大的理想是什么 $A$，那么 $A / m$ 是一个简单的 $A$-module。如果 $S$ 是一个简单的 $A$-module，然后通过注释1.4.25我们得到它 $S=A \cdot s$ 对于任何 $s \in S \backslash{0}$ 和 $S \simeq A / \operatorname{ann}(s)$ 哪里 $\operatorname{ann}(s):={a \in A \mid a \cdot s=0} .$ 此外， $\operatorname{ann}(s)$ 是一种极大的理想。的确，如果 $\operatorname{ann}(s)<\mathfrak{m}$，那么 $A / \mathfrak{m}$ 是的商 $A / \operatorname{ann}(s) \simeq S$。现在， $\operatorname{ann}(S):=\bigcap_{s \in S \backslash{0}} \operatorname{ann}(s)$ 是双面理想的吗 $A$ 自从 $S$ 是一个 $A$-module。因此 $\operatorname{rad}(A)$ 都包含在双面理想的交集中 $\operatorname{ann}(S)$ 为了一切简单 $A$-modules $S$。如果 $S$ 是一个简单的 $A$-module, then $S \simeq A / \mathrm{m}$ 为了某种极大的理想 $m$。如果 $a \in \operatorname{ann}(S)$，那么 $a \cdot A \subseteq \mathfrak{m}$ 因此 $a \in \mathfrak{m}$。因此 $\operatorname{ann}(S) \leq \mathfrak{m}$ 和交叉点 $\operatorname{ann}(S)$ 总的来说很简单 $A$-modules包含在 $\operatorname{rad}(A)$。因此我们得到 $\operatorname{ann}(S)$，路口接管了一切简单 $A$-modules， = $\operatorname{rad}(A)$。但是这些湮灭者 $\operatorname{ann}(S)$ 双面理想是什么 $A$.

## 数学代写|同调代数代写Homological Algebra代考|The Jordan-Hölder定理

\begin{aligned} &\left{X^2, Y^2, X Y, Y X, e_1^2-e_1, e_2^2-e_2, e_1 e_2, e_2 e_1\right. \ &\left.e_1 X-X e_2, e_1 Y-Y e_2, e_1 X-X, e_1 Y-Y, e_2 X, e_2 Y, X e_1, X e_2\right} \end{aligned}

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