《基础代数学》作业 Week6-1
设 $R$ 是域 $k$ 上所有对角元相等的上三角矩阵全体 (固定阶数) 作成的集合. 证明 $R$ 是局部环.
注意到 $R$ 中的不可逆元为全体严格上三角矩阵, 显然是一个理想, 于是 $R$ 是局部环.
Let $R$ be an Artinian ring or a Noetherian ring. Then $M$ can be decomposed into a finite direct sum of indecomposable modules.
If $M$ is itself indecomposable, then we are done. Otherwise, we have a nontrivial decomposition $M=M_1⊕M_2$. If one of $M_1,M_2$ is indecomposable, we keep it and decompose the other one. If both of them are decomposable, we decompose both of them. Since $R$ is Artinian or Noetherian, this process must stop in finite steps, yielding a decomposition of $M$ into a finite direct sum of indecomposable modules.
Prove that the homomorphic image of a local ring is still a local ring.
Let $R$ be a local ring with unique maximal ideal $\mathfrak{m}$ (also the the set of all non-invertible elements of $R$), and let $f:R→S$ be a surjective ring homomorphism. We want to show that $S$ is also a local ring. Notice that $S≅R/I$ for some ideal $I$ of $R$, and we have $I⊆\mathfrak{m}$. It is obvious that $\mathfrak{m}/I$ is an ideal of $R/I$. Now we want to show that it is exactly the set of non-invertable elements in $R/I$.
First, any element in the ideal $\mathfrak{m}/I$ is non-invertible, since otherwise $\mathfrak{m}/I=R/I$.
Conversely, if an element $r+I∈R/I, r∉I$ is non-invertible, then $r$ is non-invertible in $R$, hence $r∈\mathfrak{m}$, so that $r+I∈\mathfrak{m}/I$, as desired.
Prove that a finite dimensional algebra $A$ is a local ring if and only if the left regular module $_AA$ is indecomposable.
Since $A$ is a finite dimensional, $A$ has composition series.
(⇒) If $_AA$ is decomposable, i.e., $_AA=M_1⊕M_2$ for some nontrivial submodules $M_1,M_2$, then consider the idempotent endomorphism $e:A→A$ defined by $e(a)=a_1$ where $a=a_1+a_2$ with $a_i∈M_i$. It is clear that $e$ is neither $0$ nor $1$, contradicting the fact that in a local ring, the only idempotent elements are $0$ and $1$.
(⇐) If $A$ is not a local ring, then there exists an idempotent element $e∈A$ such that $e≠0,1$. Consider the left ideals $Ae$ and $A(1-e)$. It is clear that $_AA=Ae⊕A(1-e)$, contradicting the indecomposability of $_AA$.