$\newcommand{\nfrac}[2]{\frac{\displaystyle{#1}}{\displaystyle{#2}}}$
Notes, Theorem 15.2.1
iv
Take arbitrary $a' \in \varphi(A)$ and $s' \in S$. Then we have $\varphi(s) = s'$ and $\varphi(a) = a'$ where $as, sa \in A$. It follows $a's' = \varphi(a) \varphi(s) = \varphi(as) \in \varphi(A)$. Similarly, $s'a' \in \varphi(A)$.
viii
Fact. $\varphi(0) = 0$.
Observe $\varphi(0) = \varphi(0 + 0) = \varphi(0) + \varphi(0)$, implying by additive cancellation $0 = \varphi(0)$.
Theorem. Main problem.
$(\rightarrow)$ We know $\varphi(0) = 0$ for any homomorphism. For any $r \in Ker \; \varphi$, by definition $\varphi(r) = 0$, implying by injectivity $r = 0$.
$(\leftarrow)$ Take $\varphi(r) = \varphi(r')$. Then
$$\begin{aligned} \varphi(r) - \varphi(r') &= 0 \\ \varphi(r - r') &= 0 \end{aligned}$$Implying $r - r' = 0$, i.e $r = r'$.
ix
Clearly $\varphi^{-1}$ is a bijection. It remains to show operations are preserved.
For arbitrary $a', b' \in S$, we know there are $a, b \in R$, such that $\varphi(a) = a'$ and $\varphi(b) = b'$. It follows for the operation $*$, interpreted as either addition $+$ or multiplication $\cdot$:
$$\begin{aligned} & & \varphi^{-1}(a' * b') &= \varphi^{-1}(a') * \varphi^{-1}(b') \\ & \longleftrightarrow & \varphi^{-1}(\varphi(a) * \varphi(b)) &= (\varphi^{-1} \circ \varphi)(a) * (\varphi^{-1} \circ \varphi)(b) \\ & \longleftrightarrow & \varphi^{-1}(\varphi(a * b)) &= a * b \\ & \longleftrightarrow & a * b &= a * b \end{aligned}$$Gallian (9th ed), Chapter 15
6
Call the map $\varphi$. Observe
$$\begin{aligned} \varphi(3 \cdot 3) &= \phi(1) = 3 \cdot 1 = 3 \\ \varphi(3) \cdot \varphi(3) &= (3 \cdot 3) \cdot (3 \cdot 3) = 9 \cdot 9 = 9 \end{aligned}$$8
Denote $\varphi(1) = a$. Clearly,
$$\begin{aligned} \varphi(x) &= \varphi(\underbrace{1 + \dots + 1}_{x \text{ times}}) = \underbrace{\varphi(1) + \dots + \varphi(1)}_{x \text{ times}} = x \cdot \varphi(1) = ax \\ a^2 &= \varphi(1) \cdot \varphi(1) = \varphi(1 \cdot 1) = \varphi(1) = a \end{aligned}$$10 (a)
Let $\varphi$ be an arbitrary homomorphism.
$$\begin{aligned} \varphi(4) = \varphi(2) + \varphi(2) &= \varphi(2) \cdot \varphi(2) \\ 3x + 3x &= 3x \cdot 3x \end{aligned}$$By injectivity $x \neq 0$.
Claim. no such integer $x$ satisfies that. Observe
$$\begin{aligned} 3x + 3x &= 3x \cdot 3x \\ x + x &= 3x^2 \\ 2x &= 3x^2 \\ 2 &= 3x \end{aligned}$$17 (compare with thm 15.2.1 (vii) of the notes)
Yes. We show a more general statement, that the following $\varphi$ is a ring homomorphism:
$$\varphi : \mathbb{Z}_n \longrightarrow \mathbb{Z}_{kn}, \qquad \varphi(x) = kx, \quad \text{where } k = mn + 1.$$Lemma. Addition is preserved, i.e $\varphi(a + b) = \varphi(a) + \varphi(b)$.
Let
$$a + b = nq + r, \qquad r \in \{0,1,\dots,n-1\},$$then
$$k(a + b) = knq + kr, \qquad kr \in \{0,\dots,kn-k\}.$$Note $kr = \varphi(a + b)$. Observe also
$$(k\cdot a) + (k\cdot b) = knq + kr, \qquad ka, kb \in \{0,\dots,kn-k\}.$$Note $kr = \varphi(a) + \varphi(b)$.
Lemma. Multiplication is preserved, i.e $\varphi(a \cdot b) = \varphi(a)\cdot\varphi(b)$.
Let
$$a \cdot b = nq + r, \qquad r \in \{0,1,\dots,n-1\},$$then
$$\begin{aligned} k (a \cdot b) &= knq + kr & &\text{where } kr \in \{0,\dots,kn-k\} \end{aligned}$$Note $kr = \varphi(a \cdot b).$
$$\begin{aligned} (k\cdot a)(k\cdot b) &= (k\cdot k\cdot n)q + (k\cdot k)r & &\text{where } ka,kb \in \{0,\dots,kn-k\}, \\ &= (k\cdot k\cdot n)q + kmnr + kr & &\text{since } k = mn + 1, \\ &= kn(kq + mr) + kr. \end{aligned}$$Note $kr = \varphi(a)\cdot\varphi(b).$
Corollary. $\varphi : \mathbb{Z}_5 \longrightarrow \mathbb{Z}_{30}$ is a ring homomorphism.
21
Theorem. All homomorphisms from $\mathbb{Z}_6$ to $\mathbb{Z}_6$.
For any homomorphism $\varphi$, observe $\varphi(x) = x\varphi(1) = x \cdot_{Z_6} \varphi(1)$, since $x \in Z_6$.
So $\varphi(1)$ decides $\varphi$. Thereby, all homomorphisms correspond to $\varphi(1) = 0, 1, \dots, 5$.
Theorem. All homomorphisms from $\mathbb{Z}_{20}$ to $\mathbb{Z}_{30}$.
Any $x \in \mathbb{Z}_{20}$ is also in $\mathbb{Z}_{30}$. So similarly, all homomorphisms correspond to $\varphi(1) = 0, 1, \dots, 29$.
34
Note. Up against the stated question we show $\varphi$ is a ring homomorphism, regardless of the divisibility condition — I am looking for your feedback ya Ibrahim.
Lemma. Addition
$$\begin{aligned} &\phi(a + b) = \phi(a) + \phi(b) \\ \Longleftrightarrow &(a + b) \bmod n = \bigl[(a \bmod n) + (b \bmod n)\bigr] \bmod n \end{aligned}$$Observe
$$\begin{aligned} a + b &= nq + r,\quad 0 \le r < n \\ a &= nq_0 + r_0 \\ b &= nq_1 + r_1 \\ a + b &= n(q_0 + q_1) + (r_0 + r_1) \end{aligned}$$Take $r_0 + r_1 = nq' + r'$. Then
$$a + b = n(q_0 + q_1 + q') + r',\quad 0 \le r' < n$$by uniqueness $r = r'$.
Lemma. Multiplication
$$\begin{aligned} &\phi(a \cdot b) = \phi(a) \cdot \phi(b) \\ \Longleftrightarrow &(a \cdot b) \bmod n = \bigl[(a \bmod n)\cdot(b \bmod n)\bigr] \bmod n \end{aligned}$$Observe
$$\begin{aligned} a \cdot b &= nq + r,\quad 0 \le r < n \\ a &= nq_0 + r_0 \\ b &= nq_1 + r_1 \\ a \cdot b &= (nq_0 + r_0)(nq_1 + r_1) \\ &= n^2 q_0 q_1 + nq_0 r_1 + nq_1 r_0 + r_0 r_1 \end{aligned}$$Take $r_0 r_1 = nq' + r'$, $0 \le r' < n$.
$$a \cdot b = n\bigl(nq_0 q_1 + q_0 r_1 + q_1 r_0 + q'\bigr) + r'$$By Euclid, $r = r'$.