# 数学代写|表示论代写Representation theory代考|Subalgebras, Ideals and Factor Algebras

## 数学代写|表示论代写Representation theory代考|Subalgebras, Ideals and Factor Algebras

In analogy to the ‘subspaces’ of vector spaces, and ‘subgroups’ of groups, we should define the notion of a ‘subalgebra’. Suppose $A$ is a $K$-algebra, then, roughly speaking, a subalgebra $B$ of $A$ is a subset of $A$ which is itself an algebra with respect to the operations in $A$. This is made precise in the following definition.

Definition 1.14. Let $K$ be a field and $A$ a $K$-algebra. A subset $B$ of $A$ is called a $K$-subalgebra (or just subalgebra) of $A$ if the following holds:
(i) $B$ is a $K$-subspace of $A$, that is, for every $\lambda, \mu \in K$ and $b_1, b_2 \in B$ we have $\lambda b_1+\mu b_2 \in B$
(ii) $B$ is closed under multiplication, that is, for all $b_1, b_2 \in B$, the product $b_1 b_2$ belongs to $B$.
(iii) The identity element $1_A$ of $A$ belongs to $B$.
Exercise 1.3. Let $A$ be a $K$-algebra and $B \subseteq A$ a subset. Show that $B$ is a $K$-subalgebra of $A$ if and only if $B$ itself is a $K$-algebra with the operations induced from $A$.

Remark 1.15. Suppose $B$ is given as a subset of some algebra $A$, with the same addition, scalar multiplication, and multiplication. To decide whether or not $B$ is an algebra, there is no need to check all the axioms of Definition 1.1. Instead, it is enough to verify conditions (i) to (iii) of Definition 1.14 .
We consider several examples.
Example 1.16. Let $K$ be a field.
(1) The $K$-algebra $M_n(K)$ of $n \times n$-matrices over $K$ has many important subalgebras.
(i) The upper triangular matrices
$$T_n(K):=\left{a=\left(a_{i j}\right) \in M_n(K) \mid a_{i j}=0 \text { for } i>j\right}$$
form a subalgebra of $M_n(K)$. Similarly, one can define lower triangular matrices and they also form a subalgebra of $M_n(K)$.

## 数学代写|表示论代写Representation theory代考|Algebra Homomorphisms

As with any algebraic structure (like vector spaces, groups, rings) one needs to define and study maps between algebras which ‘preserve the structure’.

Definition 1.22. Let $A$ and $B$ be $K$-algebras. A map $\phi: A \rightarrow B$ is a $K$-algebra homomorphism (or homomorphism of $K$-algebras) if
(i) $\phi$ is a $K$-linear map of vector spaces,
(ii) $\phi(a b)=\phi(a) \phi(b)$ for all $a, b \in A$,
(iii) $\phi\left(1_A\right)=1_B$.
The map $\phi: A \rightarrow B$ is a $K$-algebra isomorphism if it is a $K$-algebra homomorphism and is in addition bijective. If so, then the $K$-algebras $A$ and $B$ are said to be isomorphic, and one writes $A \cong B$. Note that the inverse of an algebra isomorphism is also an algebra isomorphism, see Exercise 1.14.
Remark 1.23.
(1) To check condition (ii) of Definition 1.22, it suffices to take for $a, b$ any two elements in some fixed basis. Then it follows for arbitrary elements of $A$ as long as $\phi$ is $K$-linear.
(2) Note that the definition of an algebra homomorphism requires more than just being a homomorphism of the underlying rings. Indeed, a ring homomorphism between $K$-algebras is in general not a $K$-algebra homomorphism.

For instance, consider the complex numbers $\mathbb{C}$ as a $\mathbb{C}$-algebra. Let $\phi: \mathbb{C} \rightarrow \mathbb{C}, \phi(z)=\bar{z}$ be the complex conjugation map. By the usual rules for complex conjugation $\phi$ satisfies axioms (ii) and (iii) from Definition 1.22, that is, $\phi$ is a ring homomorphism. But $\phi$ does not satisfy axiom (i), since for example
$$\phi(i i)=\phi\left(i^2\right)=\phi(-1)=-1 \neq 1=i(-i)=i \phi(i) .$$
So $\phi$ is not a $\mathbb{C}$-algebra homomorphism. However, if one considers $\mathbb{C}$ as an algebra over $\mathbb{R}$, then complex conjugation is an $\mathbb{R}$-algebra homomorphism. In fact, for $r \in \mathbb{R}$ and $z \in \mathbb{C}$ we have
$$\phi(r z)=\overline{r z}=\bar{r} \bar{z}=r \bar{z}=r \phi(z) .$$
We list a few examples, some of which will occur frequently. For each of these, we recommend checking that the axioms of Definition 1.22 are indeed satisfied.

# 表示论代考

## 数学代写|表示论代写Representation theory代考|Subalgebras, Ideals and Factor Algebras

(i) $B$ 是一个 $K$-子空间 $A$ ，也就是说，对于每个 $\lambda, \mu \in K$ 和 $b_1, b_2 \in B$ 我们有 $\lambda b_1+\mu b_2 \in B$
(二) $B$ 在乘去下是封闭的，也就是说，对于所有 $b_1, b_2 \in B$ ，产品 $b_1 b_2$ 属于 $B$.
(iii) 身份要素 $1_A$ 的 $A$ 属于 $B$.

(1) 的 $K$-代数 $M_n(K)$ 的 $n \times n$-矩阵超过 $K$ 有许多重要的子代数。
(i) 上三角矩阵
$T_{-} n(K):=\backslash$ left ${a=\backslash$ left(a_{i $} \backslash \backslash$ right $) \backslash$ in $M _n(K) \backslash$ mid $a_{-}{i$ i $}=0 \backslash$ text ${$ for $\left.} i>j \backslash r i g h t\right}$

## 数学代写|表示论代写Representation theory代考|Algebra Homomorphisms

(i) $\phi$ 是一个 $K$-向量空间的线性映射，
(ii) $\phi(a b)=\phi(a) \phi(b)$ 对全部 $a, b \in A$ ，
(iii) $\phi\left(1_A\right)=1_B$.

(1) 要检查定义 1.22 的条件 (ii)，只需考虑 $a, b$ 某个固定基础上的任意两个元素。然后它 遵循任意元素 $A$ 只要 $\phi$ 是 $K$ – 线性。
(2) 请注意，代数同态的定义不仅仅需要底层环的同态。实际上，环同态之间 $K$-代数一 般不是 $K$-代数同态。

$$\phi(i i)=\phi\left(i^2\right)=\phi(-1)=-1 \neq 1=i(-i)=i \phi(i) .$$所以 $\phi$ 不是一个 $C$-代数同态。但是，如果考虑 $\mathbb{C}$ 作为代数 $\mathbb{R}$, 那么复共轭是 $\mathbb{R}$-代数同态。 事实上，对于 $r \in \mathbb{R}$ 和 $z \in \mathbb{C}$ 我们有
$$\phi(r z)=\overline{r z}=\bar{r} \bar{z}=r \bar{z}=r \phi(z)$$

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