6.1 Algebras and ideals

中文

An algebra consists of a vector space $\mathcal{A}$ over a field $\mathbb{K}$ together with a law of composition or product of vectors, $\mathcal{A}\times \mathcal{A}\to \mathcal{A}$ , denoted

$$(A,B)\mapsto AB\in \mathcal{A}(A,B\in \mathcal{A}),$$

which satisfies a pair of distributive laws:

$$A(aB+bC)=aAB+bAC,(aA+bB)C=aAC+bBC\text{\hspace{1em}(6.1)}$$

for all scalars $a$, $b\in \mathbb{K}$ and vectors $A$, $B$ and $C$. In the right-hand sides of Eq. (6.1) quantities such as $aAB$ are short for $a(AB)$ ; this is permissible on setting $b=0$ , which gives the identities $aAB=(aA)B=A(aB)$ . In Section 3.2 it was shown that $0A=A0=O$ for all $A\in \mathcal{A}$. , taking careful note of the difference between the zero vector $O$ and the zero scalar 0. Hence

$$OA=(0A)A=0(AA)=O,AO=A(0A)=0AA=O.$$

We have used capital letters $A$, $B$, etc. to denote vectors because algebras most frequently arise in spaces of linear operators over a vector space $V$. There is, however, nothing in principle to prevent the more usual notation $u$, $v,..$. for vectors and to write $uv$ for their product. The vector product has been denoted by a simple juxtaposition of vectors, but other notations such as $A\times B,A\otimes B,A\wedge B$ and $[A,B]$ may arise, depending upon the context. The algebra is said to be associative if $A(BC)=(AB)C$ for all $A,B,C\in \mathcal{A}$. It is called commutative if $AB=BA$ for all $A$, $B\in \mathcal{A}$

Example 6.1

On the vector space of ordinary three-dimensional vectors ${\mathbb{R}}^3$ define the usual vector product $\mathbf{u}\times \mathbf{v}$

$$(\mathbf{u}\times \mathbf{v}{)}_i=\sum _{j=1}^3\sum _{k=1}^3{ϵ}_{ijk}{u}_j{v}_k$$

where

$${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{ if any pair of indices }i,j,k\text{ are equal },\\ 1& \text{ if }ijk\text{ is an even permutation of }123,\\ -1& \text{ if }ijk\text{ is an odd permutation of }123.\end{array}$$

The vector space ${\mathbb{R}}^3$ with this law of composition is a non-commutative, non-associative algebra. The product is non-commutative since

$$\mathbf{u}\times \mathbf{v}=-\mathbf{v}\times \mathbf{u},$$

and it is non-associative as

$$(\mathbf{u}\times \mathbf{v})\times \mathbf{w}-\mathbf{u}\times (\mathbf{v}\times \mathbf{w})=(\mathbf{u}\cdot \mathbf{v})\mathbf{w}-(\mathbf{v}\cdot \mathbf{w})\mathbf{u}$$

does not vanish in general.

Example 6.2

The vector space $L(V,V)$ of linear operators on a vector space $V$ forms an associative algebra where the product $AB$ is defined in the usual way,

$$(AB)u=A(Bu).$$

The distributive laws Eq. (6.1) follow trivially and the associative law $A(BC)=(AB)C$ holds for all linear transformations. It is, however, non-commutative as $AB\ne BA$ in general.

Similarly the set of all $n\times n$ real matrices ${\mathcal{M}}_n$ forms an algebra with respect to matrix multiplication, since it may be thought of as being identical with $L({\mathbb{R}}^n,{\mathbb{R}}^n)$ where ${\mathbb{R}}^n$ is the vector space of $n\times 1$ column vectors. If the field of scalars is the complex numbers, we use ${\mathcal{M}}_n(\mathbb{C})$ to denote the algebra of $n\times n$ complex matrices.

If $\mathcal{A}$ is a finite dimensional algebra and $E_1,E_2,\dots ,E_n$ any basis, then let $C_{ij}^k$ be a set of scalars defined by

$$E_i{E}_j=C_{ij}^k{E}_k.\text{\hspace{1em}(6.2)}$$

The scalars $C_{ij}^k\in \mathbb{K}$. , uniquely defined as the components of the vector $E_i{E}_j$ with respect to the given basis, are called the structure constants of the algebra with respect to the basis $\{E_i\}$ . This is a common way of defining an algebra for, once the structure constants are specified with respect to any basis, we can generate the product of any pair of vectors $A=a^i{E}_i$ and $B=b^j{E}_j$ by the distributive law Eq. (6.1),

$$AB=(a^i{E}_i)(b^j{E}_j)=a^i{b}^j{E}_i{E}_j=\left(a^i{b}^j{C}_{ij}^k\right)E_k.$$

Exercise

Show that an algebra is commutative iff the structure constants are symmetric in the subscripts, $C_{ij}^k=C_{ji}^k$

Let $\mathcal{A}$ and $\mathcal{B}$ be any pair of algebras. A linear map $\phi :\mathcal{A}\to \mathcal{B}$ is called an algebra homomorphism if it preserves products, $\phi (AB)=\phi (A)\phi (B)$

Exercise

Show that for any pair of scalars $a$, $b$ and vectors $A$, $B$, $C$

$$\phi (A(aB+bC))=a\phi (A)\phi (B)+b\phi (A)\phi (C).$$

A subalgebra $\mathcal{B}$ of $\mathcal{A}$ is a vector subspace that is closed under the law of composition,

$$A\in \mathcal{B},B\in \mathcal{B}⟹AB\in \mathcal{B}.$$

Exercise

Show that if $\phi :\mathcal{A}\to \mathcal{B}$ is an algebra homomorphism then the image set $\phi (\mathcal{A})\subseteq \mathcal{B}$ is a subalgebra of $\mathcal{B}$.

A homomorphism $\phi$ is called an algebra isomorphism if it is one-to-one and onto; the two algebras $\mathcal{A}$ and $\mathcal{B}$ are then said to be isomorphic.

Example 6.3

On the vector space ${\mathbb{R}}^{\infty }$ define a law of multiplication

$$(a_0,a_1,a_2,\dots )(b_0,b_1,b_2,\dots )=(c_0,c_1,c_2,\dots )$$

where

$$c_p=a_0{b}_p+a_1{b}_{p-1}+\cdots +a_p{b}_0.$$

Setting $A=(a_0,a_1,a_2,\dots ),B=(b_0,b_1,b_2,\dots )$ , it is straightforward to verify Eq. (6.1) and the commutative law $AB=BA$ . Hence with this product law, ${\mathbb{R}}^{\infty }$ is a commutative algebra. Furthermore this algebra is associative,

$$\begin{array}{l}(A(BC){)}_p={\sum }_{i=0}^p{a}_i(bc{)}_{p-i}\\ ={\sum }_{i=0}^p{\sum }_{j=0}^{p-i}{a}_i{b}_j{c}_{p-i-j}\\ ={\sum }_{i+j+k=p}{a}_i{b}_j{c}_k\\ ={\sum }_{j=0}^p{\sum }_{i=0}^{p-j}{a}_j{b}_{p-i-j}{c}_i\\ =((AB)C{)}_p.\end{array}$$

The infinite dimensional vector space of all real polynomials $\mathcal{P}$ is also a commutative and associative algebra, whereby the product of a polynomial $f(x)=a_0+a_{1}x+a_2{x}^2+\cdot \cdot \cdot +a_n{x}^n$ of degree $n$ and $g(x)=b_0+b_{1}x+b_2{x}^2+\cdot \cdot \cdot +b_m{x}^m$ of degree $m$ results in a polynomial $f(x)g(x)$ of degree $m+n$ in the usual way. On explicitly carrying out the multiplication of two such polynomials it follows that the map $\phi :\mathcal{P}\to {\mathbb{R}}^{\infty }$ defined by

$$\phi (f(x))=\left(a_0,a_1,a_2,\dots ,a_n,0,0,\dots \right)$$

is an algebra homomorphism. In Example 3.10 it was shown that the map $\phi$ (denoted $S$ in that example) establishes a vector space isomorphism between $\mathcal{P}$ and the vector space ${\hat{\mathbb{R}}}^{\infty }$ of sequences having only finitely many non-zero terms. If $A\in {\hat{\mathbb{R}}}^{\infty }$ let us call its length the largest natural number $p$ such that $a_p\ne 0$. From the law of composition it follows that if $A$ has length $p$ and $B$ has length $q$ then $AB$ is a vector of length $\le p+q$ . The space ${\hat{\mathbb{R}}}^{\infty }$ is a subalgebra of ${\mathbb{R}}^{\infty }$ , and is isomorphic to the algebra $\mathcal{P}$

Ideals and factor algebras

A vector subspace $\mathcal{B}$ of $\mathcal{A}$ is a subalgebra if it is closed with respect to products, a property that may be written $\mathcal{BB}\subseteq \mathcal{B}$ . A vector subspace $\mathcal{L}$ of $\mathcal{A}$ is called a left ideal if

$$L\in \mathcal{L},A\in \mathcal{A}⟹AL\in \mathcal{L},$$

or, in the above notation, $\mathcal{A}\mathcal{L}\subseteq \mathcal{L}$ . Similarly a right ideal $\mathcal{R}$ is a subspace such that

$$\mathcal{R}\mathcal{A}\subseteq \mathcal{R}.$$

A two-sided ideal or simply an ideal is a subspace $\mathcal{I}$ that is both a left and right-sided ideal. An ideal is always a subalgebra, but the converse is not true.

Ideals play a role in algebras parallel to that played by normal subgroups in group theory (see Section 2.5). To appreciate this correspondence let $\phi :\mathcal{A}\to \mathcal{B}$ be an algebra homomorphism between any two algebras. As in Section 3.4, define the kernel $\ker \phi$ of the linear map $\phi$ to be the vector subspace of $\mathcal{A}$ consisting of those vectors that are mapped to the zero element $O^{\prime }$ of $\mathcal{B}$, namely $\ker \phi ={\phi }^{-1}(O^{\prime })$ .

Theorem 6.1 · Kernels of algebra homomorphisms and factor algebras

The kernel of an algebra homomorphism $\phi :\mathcal{A}\to \mathcal{B}$ is an ideal of $\mathcal{A}$. Conversely, if $\mathcal{I}$ is an ideal of $\mathcal{A}$ then there is a natural algebra structure defined on the factor space $\mathcal{A}/\mathcal{I}$ such that the map $\phi :\mathcal{A}\to \mathcal{A}/\mathcal{I}$ whereby $A\mapsto [A]\equiv A+\mathcal{I}$ is a homomorphism with kernel $\mathcal{I}$.

Proof

The vector subspace $\ker \phi$ is a left ideal of $\mathcal{A}$, for if $B\in \ker \phi$ and $A\in \mathcal{A}$ then $AB\in \ker \phi$ , for

$$\phi (AB)=\phi (A)\phi (B)=\phi (A)O^{\prime }=O^{\prime }.$$

Similarly $\ker \phi$ is a right ideal.

If $\mathcal{I}$ is an ideal of $\mathcal{A}$, denote the typical elements of $\mathcal{A}/\mathcal{I}$ by the coset $[A]=A+\mathcal{I}$ and define an algebra structure on $\mathcal{A}/\mathcal{I}$ by setting $[A][B]=[AB]$ . This product rule is ‘natural’ in the sense that it is independent of the choice of representative from [$A$] and [$B$], for if $[A^{\prime }]=[A]$ and $[B^{\prime }]=[B]$ then $A^{\prime }\in A+\mathcal{I}$ and $B^{\prime }\in B+\mathcal{I}$ . Using the fact that $\mathcal{I}$ is both a left and right ideal, we have

$$A^{\prime }{B}^{\prime }\in (A+\mathcal{I})(B+\mathcal{I})=AB+A\mathcal{I}+\mathcal{I}B+\mathcal{II}=AB+\mathcal{I}.$$

Hence $[A^{\prime }][B^{\prime }]=[A^{\prime }{B}^{\prime }]=[AB]=[A][B]$ . The map $\phi :\mathcal{A}\to \mathcal{A}/\mathcal{I}$ defined by $\phi (A)=[A]$ is clearly a homomorphism, and its kernel is ${\phi }^{-1}([O])=\mathcal{I}$