$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\s}{\sigma}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathbb{F}}$
example 1: $ R=\Z, I=n\Z$ with $f(r)=r+I$, ker f =I
example 2: Let $R=\R[x]$ and $\C$ be a ring. So there is an $i\in \C$ and a substitution homomorphism $\s=\s_{i}:\R[x] \rightarrow \C$ by $\s_{i}(A)=A(i)$. What is $I=ker \s$? Surely $x^{2}+1 \in I$, because $i^{2}+1=0$. It follows $(x^{2}+1)A \in I$ for every $A\in R$, so $(x^{2}+1)R\subset I$. But with $L=a+bx(a,b \in \R)$ it is : $\s(L)=a+bi\ne 0$. That gives us : $A=(x^{2}+1)Q+L$ with rest $L=a+bx$. It follows that $\tilde{\s}:R/I\rightarrow \C$ is a ringisomorphism. Especially $R/I = \R[x]/(x^{2}+1)\R[x]$ is a field, because $\C$ is a field.
example 3: $\Q[i]=\Q+\Q i$ (this is even a field)
example 4: $\Z[\frac{1}{2}]=\Z+\Z \frac{1}{2}+\Z(\frac{1}{2})^{2}+\cdots$
example 5: $a=2^{1/3}$, $\Z[a]=\Z+\Z a+ \Z a^{2}$
example 6: $Char(\F_{4}) = 2 $, because 1+1=0.
example 7 : $Char(\F_{2} \times \F_{2})= 2$
Example 1 : The $ker$ is everything that is mapped to 0, is it correct that this is always an ideal of R?
Example 2: I understand that $i^{2}+1=0$ in $\C$, but how can it be concluded from that, that $x^{2}+1 \in I$, because of the substitution homomorphism? How is the step(s??) from $\s(L) = a+bi$ to $A=(x^{2}+1)Q+L$ done?
Example 3: Apart from using the definition of a field, is there a way one can see immediately, that $\Q[i]$ is a field?
Example 4 and Example 5: Why isn't that the same as in Example 3?
Example 6: I think $\F _{4} = \{0,1,2,3\}$ but that is wrong according to this example...
Example 7: $\F _{2} \times \F_{2} = \{0,1\} \times \{0,1 \} = \{0,0\}, \{0,1\}, \{1,0\},\{1,1\}$
Characteristic is the littlest number so when the rest is 0 with r=mq+d, so when it is $\{0,0\}$; but why aren't there two characteristics even if it is the same ring (so the characteristic of this should be $2,2$ and not only 2?