Section 24 Noncommutative Examples

Rings of Endomorphisms

NOTE

Let $A$ be any abelian group. A homomorphism of A into itself is an endomorphism of A.

Let the set of all endomorphisms of A be $End(A)$. We have $\varphi ,\psi \in End(A)$ , and any $a\in A$ $(\varphi +\psi )(a)=\varphi (a)+\psi (a)$ $(\varphi +\psi )(a)=(\psi +\varphi )(a)$ The homomorphism 0 is defined by $0(a)=e$, where e is the additive identity of A. And we define $(-\varphi )(a)=-\varphi (a)$ is in $End(A)$, and $\varphi +(-\varphi )=0$.

Thus $⟨End(A),+⟩$ is an abelian group

It can be also proven that $⟨End(A),+,\cdot ⟩$ is a ring under homomorphism addition and homomorphism multiplication (function composition).

Since function composition is not commutative, $⟨End(A),+,\cdot ⟩$ is also not commutative. An example is $⟨End(⟨\mathbb{Z}\times \mathbb{Z},+⟩),+,\cdot ⟩$

24.3 Weyl Algebra

Define two Endomorphism $X,Y\in End(⟨F[x],+⟩)$ (F is a field of characteristic zero, see characteristic of ring)

• X: multiply the polynomial by x
• Y: the derivative of the polynomial $YX-XY=1$ $x^2+1$ → $2x$ → $2x^2$ $x^2+1$→ $x^3+x$ → $3x^2+1$ $3x^2+1-2x^2=x^2+1$

The subring generated by X and Y and multiplications by elements of $F$ (multiply by constant) is the Weyl Algebra. Important in quantum mechanics.

24.5 Definition Group Ring

$RG$ is defined as ${\sum }_{i\in I}{a}_i{g}_i$

p.223

The ring $RG$ defined above is the group ring of $G$ over $R$. If $F$ is a field, then $FG$ is the group algebra of G over F.

Think G as the variables $\{x,y,z,\cdots \}$ and R as the coefficients $\{a_0,a_1,a_2,\cdots \}$ with every $a_i\in R$ And $⟨RG,+⟩$ is a abelian group with additive identity ${\sum }_{i\in I}0g_i$

24.6 Example

$Z_{2}G$ with $G=\{e,a\}$ The elements of $Z_{2}G$ are

• 0e+0a
• 0e+1a
• 1e+0a
• 1e+1a or $\{0,a,e,e+a\}$

The Multiplication of two elements of $RG$ is defined by:

$\left({\sum }_{i\in I}{a}_i{g}_i\right)\left({\sum }_{i\in I}{b}_i{g}_i\right)={\sum }_{i\in I}\left({\sum }_{g_j{g}_k=g_i}{a}_j{b}_k\right)g_i$ For example: $r\cdot s=(1e+1a)(1e+0a)=Coef(e)e+Coef(a)a=(r_e{s}_e+r_a{s}_a)e+(r_e{s}_a+r_a{s}_e)a$ $=(1\cdot 1+1\cdot 0)e+(1\cdot 0+1\cdot 1)a=e+a$ It’s literally the same as calculating a normal polynomial multiplication.

Quaternions

The quaternions of Hamilton are the standard example of a strictly skew field or a noncommutative division ring (see Definition of a strictly skew field).

The Set $\mathbb{H}$ defined on $\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}$.

• 1=(1,0,0,0)

• i=(0,1,0,0)

• j=(0,0,1,0)

• k=(0,0,0,1) We define

• $1a=a1=a$ for $a\in \mathbb{H}$

• $i^2=j^2=k^2=-1$

• $ij=k,jk=i,ki=j,ji=-k,kj=-i,ik=-j$

• Excercise 19 Quaternions are isomorphic to a subring of $M_2(\mathbb{C})$ The map $\varphi :\mathbb{H}\to M_2(\mathbb{C})$ is defined by

a\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} +b\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} +c\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} +a\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$$ Note that $a+bi+cj+dk$ can be rewritten as $a+bi+(c+di)j$ or $z_1+z_2j$ see [[Section 13 Homomorphisms#how-to-show--phi-g-rightarrow-g-is-an-isomorphism|How to prove isomorphism]] ## 24.10 Wedderburn's Theorem > [!PDF|note] [[A First Course in Abstract Algebra.pdf#page=231&selection=85,0,85,36&color=note|p.226]] > > > Every finite division ring is a field. Elegant