1. (a) $\begin{pmatrix} 0 &0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 &0 \\ 0 & 1 \end{pmatrix}$ which are the additive and multiplicative identities of $R$ are clearly in $M$. Take $\begin{pmatrix} a &b \\ b & a \end{pmatrix}, \begin{pmatrix} c &d \\ d & c \end{pmatrix} \in M$ then $\begin{pmatrix} a &b \\ b & a \end{pmatrix}+ \begin{pmatrix} c &d \\ d & c \end{pmatrix} = \begin{pmatrix} a+c & b+d \\ b+d & a+c \end{pmatrix} \in M$ which shows that $M$ is closed under matrix addition. Next $\begin{pmatrix} a &b \\ b & a \end{pmatrix} \begin{pmatrix} c &d \\ d & c \end{pmatrix} = \begin{pmatrix} ac+bd & ad+bc \\ bc+ad & bd+ac \end{pmatrix} \in M$ so $M$ is closed under matrix multiplication. Furthermore $\begin{pmatrix} a &b \\ b & a \end{pmatrix} + \begin{pmatrix} -a &-b \\ -b & -a \end{pmatrix} = \begin{pmatrix} 0& 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -a &-b \\ -b & -a \end{pmatrix}+ \begin{pmatrix} a &b \\ b & a \end{pmatrix}$ which shows that every element of $M$ has an additive inverse in $M$. So, using the criteria for subring from the lecture notes, this shows that $M$ is a subring of $R$. Lastly matrix multiplication on $M$ is commutative as shown $\begin{pmatrix} a &b \\ b & a \end{pmatrix} \begin{pmatrix} c &d \\ d & c \end{pmatrix} = \begin{pmatrix} ac+bd & ad+bc \\ bc+ad & bd+ac \end{pmatrix} = \begin{pmatrix} ca+db & cb+ad \\ da+cb & db+ca \end{pmatrix} = \begin{pmatrix} c &d \\ d & c \end{pmatrix}\begin{pmatrix} a &b \\ b & a \end{pmatrix}$so $M$ is a commutative subring of $R$. (b) The unit group of $R$ is the set of invertible $2 \times 2$ matrices which is given by $\text{GL}_{2}(\mathbb{R}) = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b,c,d \in \mathbb{R} \text{ and } ad-bc \neq 0 \right\} .$ 2. (a) Define the relation $E$ on $R$ by $aEb \iff av=b \quad \text{for some unit } v\in R.$Let $e$ denote the multiplicative identity in $R$. For any $r \in R$, $re=r$ so $rEr$ which shows that this relation is **reflexive** since $e$ is a unit ($e e=e$). Take $x,y\in R$ such that $xEy$ (i.e. $xv=y$ for some unit $v\in R$). Since $v$ is a unit, there exist $u \in R$ such that $uv=e$. So $(xv)u=yu \iff x(vu)=yu \iff yu=x$ (i.e. $yEx$) which shows that the relation is **symmetric**. Take $x,y,z \in R$ such that $xu_{1}=y$ and $yu_{2}=z$ for some units $u_{1},u_{2}\in R$ (i.e $xEy$ and $yEz$). There must exist $v_{1},v_{2} \in R$ such that $u_{1}v_{1}=e$ and $u_{2}v_{2}=e$. So $(yu_{2})v_{2}=zv_{2} \iff y(u_{2}v_{2})=zv_{2}\iff y=zv_{2}$. Substituting for $y$ gives $xu_{1}=zv_{2}\iff (xu_{1})u_{2}=(zv_{2})u_{2}\iff x(u_{1}u_{2})=z(v_{2}u_{2})=z$. Now $(u_{1}u_{2})(v_{2}v_{1})=u_{1}(u_{2}v_{2})v_{1}=u_{1}v_{1}=e$ so $(u_{1}u_{2})$ is a unit which shows that ($xEz$). Therefore $E$ is **transitive**. Hence $E$ is an equivalence relation since it is reflexive, symmetric and transitive. (b) The only unit in $\mathbb{Z}$ is 1. So $E$ is defined by $aEb \iff a=b$ so the for any $x \in \mathbb{Z}$, the equivalence class of $x$ is given by $[x]_{E} =\{ x \}$ so $\mathbb{Z}/E=\{ \{ x \} \mid x \in \mathbb{Z} \}$. 3. The ideals in $(\mathbb{Z}/6\mathbb{Z},+_{6},\times_{6})$ are $\{ [0]_{6} \}$, $\{ [0]_{6}, [1]_{6},[2]_{6},[3]_{6}, [4]_{6}, [5]_{6} \}$, $\{ [0]_{6}, [2]_{6}, [4]_{6}\}$ and $\{ [0]_{6}, [3]_{6}\}$. 4. (a) First show that $(aR,+)$ is a subgroup of $(R,+)$. Note that $0=a\times 0 \in aR$ (i.e. $aR$ contains the additive identity of $R$). Take $x,y \in R$ then $x+y \in R$ since $(R,+)$ is a group. So $ax+ay=a(x+y) \in aR$, from distributivity, which shows that $aR$ is closed under addition. Also for any $x \in R$, its additive inverse $-x \in R$ (since $R$ is a group) so $ax+a(-x)=a(x+(-x))=a(0)=0$. So every element of $aR$ has an inverse in $aR$. Therefore $aR$ is a subgroup of $R$. Next take $x \in aR$ and $r \in R$. Then $x=as$ for some $s \in R$ so $rx=xr = (as)r= a(sr) \in aR$ using commutativity and associativity. So $aR$ is a two-sided ideal of $R$. 5. (a) $(\mathbb{Z}/10\mathbb{Z})^{*}=\{ [1]_{10}, [3]_{10}, [7]_{10}, [9]_{10} \}$. (b) $(\mathbb{Z}/12\mathbb{Z})^{*}=\{ [1]_{12},[5]_{12}, [7]_{12}, [11]_{12} \}$. (c) Clearly $\langle [7]_{10} \rangle= \{ [1]_{10}, [3]_{10}, [7]_{10}, [9]_{10} \}$. Now $[1]_{12}$ has order 1 while $[5]_{12}, [7]_{12}, [11]_{12}$ have order 2 so $(\mathbb{Z}/12\mathbb{Z})^{*}$ is not cyclic since none of its elements have order 4. 6. Using Fermat's little theorem, $5^{22} \equiv 1 \pmod{23}$. Note that $\phi(22)=22 \times \frac{1}{2} \times \frac{10}{11}=10$. So by Euler's theorem $5^{10} \equiv 1\pmod{22}$. Now $5^{21} = (5^{10})^{2} \times 5 \equiv 5 \pmod{22}$. Hence $5^{21} = 22m+5$ for some $m\in \mathbb{Z}$. Therefore $\begin{align} 5^{5^{21}} &= 5^{22m} 5^{5} \\ &\equiv 5^{5} \pmod{23} \\ &\equiv 5^{2} \times 5^{2} \times 5 \pmod{23} \\ &\equiv 2 \times 2 \times 5 \pmod{23} \\ & \equiv 20 \pmod{23}.\end{align}$ 7. (a) First show that $I$ is a subgroup of $R$. Note that the zero polynomial, the additive identity of $\mathbb{R}[x]$, is clearly in $I$. Take $f(x),g(x)\in I$. Then $0=0+0=f(1)+g(1)=(f+g)(1)$ so $(f+g)(x) \in I$ which shows that $I$ is closed under polynomial addition. Furthermore $0=-f(0)$ so $-f(x) \in I$ which shows that every element of $I$ has an additive inverse in $I$. Therefore $I$ is a subgroup of $R$. Take $g(x) \in R$ and $f(x) \in I$ then $0=f(0)\times g(0)= fg(0)$ so $fg(x)\in I$ so which shows that $I$ is an ideal of $R$. Now the multiplicative identity of $R$ is given by the constant polynomial $f(x)=1$. Note that $f(0)=1\neq 0$ so $f(x)\not \in I$ which shows that $I\neq R$. (b) Suppose $J$ is an ideal of $R$ such that $I$ is proper subset of $J$. So for all $i(x) \in I$, we have $i(x) \in J$. Also $\exists j(x) \in J$ such that $j(x) \not \in I$. Using division algorithm, $j(x)=(x-1)g(x)+c$ for some $g(x) \in R, c\neq 0$. Note that $(x-1)g(x) \in I \subsetneq J$ hence using closure $c=j(x)-(x-1)g(x)\in J$. Since this constant polynomial is a unit, we must have that $J=R$.