TUT ALG3 1

Problem 1

$(a)$: $R$ is clearly an abelian group under $+$. Associativity, distributivity, and existence of identity are easily verified.

$(b)$: $\alpha$ is a left zero divisor when $a=p$:

$$\left(\begin{array}{cc}p& \overline{0}\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& \overline{1}\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& \overline{0}\\ 0& 0\end{array}\right).$$

$\alpha$ is not a right zero divisor since right multiplication by $\alpha$ is injective (unless $a=0$)

$$\left(\begin{array}{cc}x& \overline{y}\\ 0& z\end{array}\right)\left(\begin{array}{cc}a& \overline{0}\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}xa& \overline{y}\\ 0& z\end{array}\right).$$

$(c)$: $\beta$ is a right zero divisor from $(b)$, and also from the fact that ${\beta }^2=0$ (which also makes it a left zero divisor).

Problem 2

$(a)$: It is easy to verify that $(\mathbb{H},+)$ is an abelian group.

Verifying associativity
Consider the injective map $\phi :\mathbb{H}\to M_2(\mathbb{C})$ defined by

$$\phi (a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k})=\left(\begin{array}{cc}a+b\mathbf{i}& c+d\mathbf{i}\\ -c+d\mathbf{i}& a-b\mathbf{i}\end{array}\right).$$

Clearly, $\phi$ respects addition: $\phi (q_1+q_2)=\phi (q_1)+\phi (q_2)$. Thus, $\phi$ is a group homomorphism $\phi :(\mathbb{H},+)\to (M_2(\mathbb{C}),+)$. Next, we see that $\phi$ respects multiplication too. It is easily verified that $\phi (1)=I$, $\phi (i{)}^2=\phi (j{)}^2=\phi (k{)}^2=-I$, and $\phi (i)\phi (j)=-\phi (j)\phi (i)=\phi (k)$. Since $\phi$ respects addition, this yields

$$\begin{array}{rl}& \phi ((a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k})(e+f\mathbf{i}+g\mathbf{j}+h\mathbf{k}))\\ & =\phi (a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k})\phi (e+f\mathbf{i}+g\mathbf{j}+h\mathbf{k}).\end{array}$$

Thus, $\phi$ respects multiplication. So,

$$\begin{array}{rl}& \phi ((q_1{q}_2)q_3)=\phi (q_1{q}_2)\phi (q_3)\\ & =\phi (q_1)\phi (q_2)\phi (q_3)=\phi (q_1)\phi (q_2{q}_3)=\phi (q_1(q_2{q}_3)).\end{array}$$

Since $\phi$ is injective, this implies $(q_1{q}_2)q_3=q_1(q_2{q}_3)$.

Division ring
For $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}\in \mathbb{H}$, let

$$q^{-1}\equiv \frac{a-b\mathbf{i}-c\mathbf{j}-d\mathbf{k}}{a^2+b^2+c^2+d^2}.$$

Then, $q{q}^{-1}=q^{-1}q=1$.

Center
For $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}\in \mathbb{H}$, $q\mathbf{i}-\mathbf{i}q=-2c\mathbf{k}+2d\mathbf{j}$. Thus, $q$ commutes with $\mathbf{i}$ iff $c=d=0$. Similarly, $q$ commutes with $\mathbf{j}$ iff $b=d=0$. Thus, $Z(\mathbb{H})\subseteq \mathbb{R}$. Since it is clear that $\mathbb{R}\subseteq Z(\mathbb{H})$, $Z(\mathbb{H})=\mathbb{R}$.

$(b)$: $2$ doesn’t have an inverse. The units are $\{\pm 1,\pm \mathbf{i},\pm \mathbf{j},\pm \mathbf{k}\}$. The center is $\mathbb{Z}$.

Problem 3

$(a)$:

$$\begin{array}{rl}& \left[\left(\sum a_i{x}^i\right)\left(\sum b_j{x}^j\right)\right]\left(\sum c_k{x}^k\right)\\ & =\left[\sum a_i{\sigma }^i(b_j)x^{i+j}\right]\left(\sum c_k{x}^k\right)\\ & =\sum a_i{\sigma }^i(b_j){\sigma }^{i+j}(c_k)x^{i+j+k}\\ & =\left(\sum a_i{x}^i\right)\left[\sum b_j{\sigma }^j(c_k)x^{j+k}\right]\\ & =\left(\sum a_i{x}^i\right)\left[\left(\sum b_j{x}^j\right)\left(\sum c_k{x}^k\right)\right]\end{array}$$

Distributivity trivially holds.

$(b)$: Let $\sum a_i{x}^i\in Z(R)$.

$$\begin{array}{rl}& \left[x,\sum a_j{x}^j\right]=0⟺a_j=\overline{a_j}\forall j⟺a_j\in R\forall j\\ & \left[b,\sum a_j{x}^j\right]=0⟺b{a}_j=a_j{\sigma }^j(b)\forall j⟺2|j\forall j.\end{array}$$

Thus, $Z(R)=\mathbb{R}[x^2]$.

$(c)$: Define a map $\phi :R/(R\cdot (x^2+1))\to \mathbb{H}$ by

$$a+bx+I\mapsto a+b(j)+I(j).$$