Proof The unique morphism is given as $$ \begin{align*} \phi(n)&=\overset{n\text{ times}}{\overbrace{1+1+\ldots+1}} \text{ for }n\geq 0\\ \phi(n)&=\underset{-n\text{ times}}{\underbrace{-1-1-\ldots-1}} \text{ for }n\leq 0\\ \end{align*} $$ Thus, $\Ker\phi$ is of the form $n\Z$, and it is an ideal in $\Z$ (since, $\Ker$ of a morphism is an ideal).

Proof (by contradiction): Let $n=\Char k$ and suppose $n=p_1p_2$ where $p_i\geq 2$ is prime, then $0=\phi(n)=\phi(p_1)\phi(p_2)$ implying atleast one of $\phi(p_i)=0$. This is a contradiction because $2\leq p_i< n$, and thus not in $n\Z$ which is the $\Ker\phi$.

Examples

1. $\Q,\Z,\R,\C$ have Characteristic zero.
2. $\Z/n\Z$ has Characteristic $n$, for $n=p$ a prime this is a field.

Proof Let $\phi:\Z\ra k$ be the unique morphism of rings, then $\phi$ factorizes through the kernel $n\Z$, giving a canonical injection $\Z/n\Z\ra k$. If $n=0$, then $k$ contains $\Ima \phi\simeq\Z$, and the subfield of $k$ is generated by the $\Ima\phi$ which is isomorphic to $\Q$. If $n=p$ a prime number, then $k$ contains $\Ima\phi\simeq\Z/p\Z.$

Proof

As $k$ is finite, it is a finite dimensional vector space over $\F_p$. The choice of a base defines a vector space isomorphism $\F_p^n\simeq k$. Since the cardinality of $\F_p^n$ is $p^n$ we are done.
Since the order of $\tio{k}$ is $q-1$ we get $x^{q-1}=1$ for all $x\neq 0$ or $x^q=x$ for all $x\in k$. Since, the polynomial $X^q-X$ admits atmost $q$ roots ( also the cardinality of $k$, with all its elements satisfying the equation), we deduce that $k$ is the set of roots of $X^q-X$.
Let $k$ and $k'$ have the same cardinality $q=p^n$. Pick $\Omega$ as algebraically closed field of $\Char p$ and cardinality $q$, it thus contains $\F_p$ (which is embedded in $\Omega$ via an identity map). The identity map of $\F_p$ can be extended to $\phi$ and $\phi'$ to give embeddings of $k$ and $k'$ in $\Omega$ respectively. But, $\Omega$ is a unique subfield of cardinality $q$ (uniqueness coming from part 2 above), we have $\phi(k)=\phi'(k')=\Omega$ (all three have of cardinality $q$ and uniqueness from part 2 gives equality). Hence, $\phi^{-1}\circ \phi'$ gives the isomorphism between $k'$ and $k$.
The frobenius map in $\Omega$ is given as $F:x\mapsto x^p$ and thus $F^n:x\ra x^q$ (where $q=p^n$). Since, the field has $\Char p$ the binomial theorem gives $$(x\pm y)^q=F^n(x\pm y)=F^n(x)\pm F^n(y)=x^q\pm y^q $$ and $(xy)^q=x^qy^q$ giving the stability of $\F_q$ by product, multiplication and inverse, hence making $\F_q$ a subfield.
 We now show that $\F_q$ is necessarily the set of roots of $X^q-X$. The roots are simple, if atleast one of them was a double root it would negate $(X^q-X)'=-1$ (A polynomial $f$ has repeated root iff $f'=0$). Thus, all the roots are in $\F_q$ and there are precisely $q$ of them.
If $n|m$ the roots of $X^{p^n}-X$ are roots of $X^{p^m}-X$ and we have the inclusion $\F_{p^n}\subset \F_{p^m}$. Inversely, if $\F_{p^n}\subset\F_{p^m}$ then ${F}^{\times}_{p^n}\subset {F}^{\times}_{p^n}$ and thus $(p^n-1)|(p^m-1)$ (theorem of lagrange on order of groups). Writing the euclidean division $m=an+r,0\leq r< n$. We then have $$p^m-1=p^{an}p^r-1=(p^{an}-1)p^r+p^r-1 $$ But, according to the formula of partial sums of a geometric series $(p^n-1)|(p^{an}-1)$. Hence, $p^n-1|p^r-1<p^n-1$, which is never possible hence $r=0$, giving the desired result of $n|m$.
  Let $q=p^n$, for $N\geq 1$, the elements of $\F_{q^{N!}}$ are algebraic over $\F_q$ and thus lie in the algebraic closure $\F_{q^{N!}}\subset\wbr{\F}_q$.
Now let $x\in\wbr{\F}_q$, then there is a non trivial polynomial $P(X)\in\F_q[X]$ with $x$ as a root, that is $P(x)=0$. If $F^n$ denotes the frobenius map applied $n$ times then $P(F^{sn}(x))=0$ for all $s\geq 0$ (recall $q=p^n$). Hence, $\{F^{sn}(x)\}_{s\geq 0}$ is contained within the finite set of roots of $P$. Thus there is a $t>s\geq 0$ such that $F^{tn}(x)=F^{sn}x$ which implies $F^{(t-s)n}(x)=x$ and thus $x\in\F_{q^{t-s}}$. Since every $x$ lives in some $\F_{q^w}$, the algebraic closure has to be increasing union of these.
$\square$
Note that for $n|m$, if $d$ is the dimension of $\F_{p^m}$ over $\F_{p^n}$, then as vector spaces $\F_{p^m}\simeq(\F_{p^n})^d$. Comparing the cardinality one obtains $p^m=(p^n)^d$ or $d=m/n$.
 $\F_q$ is a splitting field of $X^q-X$ over $\F_p$ (in $\Omega$).Thus, when talking about a finite field $\F_q$, in general it means that an algebraically closed field of $\Char p$ was chosen.

Proof Let $G\subset\tio{k}$ be a finite subgroup of finite order $|G|=n$, then $x^n=1$ for all $x\in G$ (by theorem of Lagrange). Hence, $X^n-1$ splits in $k[X]$ with distinct roots, the roots being exactly the $n$ elements of $G$. Now for $d$ an integer that divides $n$, we get $(X^d-1)|(X^n-1)$, hence $X^d-1$ splits with distinct roots.
  For example, let $p$ be a prime number that divides $n$ and $r\geq 1$ maximal such that $d=p^r$ divides $n$, then $X^d-1$ has $d$ distinct roots in $k$. Roots of order different from $d$ are the roots of $X^{p^{r-1}}-1$, hence there are at most $p^{r-1}$. As $p^r>p^{r-1}$ there is atleast $x\in k$ root of $X^d-1$ that is not the root of $X^{p^{r-1}}-1$, and thus $x$ has order $p^r$. Now let $x_1,\ldots, x_N$ be obtained for each prime number that divides $n$ and let $y=x_1\cdots x_N$. Since, $G$ is commutative and for $i\neq j$ the order of $x_i$ is prime with that of $x_j$, the order of $y$ is the product of orders of $x_1,\ldots, x_N$, that is $n$. Hence, $G$ is cyclic generated by $y$.

Remark If we know the structure of the finite Abelian groups, this result is evident. Indeed, we know that $\tio{k}$ is isomorphic to a product $$\Pi=\prod_{i=1}^d\Z/n_i\Z $$ with $1<n_1|\cdots|n_d$ (note the multiplicative law on $\tio{k}$, whereas on the right side the law is additive with identity $0$ ). The number of solutions of $X^{n_1}-1$ is atmost $n_1$. In $\Pi$, they correspond to the solution of the equation $n_1\Pi=0$. If $d>1$ there are atleast $2n_1$, namely the elements of $\Z/n_1\Z$ and those from $(n_2/n_1)\cdot(\Z/n_2\Z)\simeq \Z/n_1\Z$, a contradiction to the product form of abelian group.

Proof Let $x$ be the generator of the cyclic group $\F^\times_{q^m}$. Since $[\F_{q^m}:\F_q]=[\F_q[x]:\F_q]=m$, the minimal polynomial $P$ of $x$ on $\F_q$ is of degree $m$. A morphism $\sigma\in G=\Aut_{\F_q}(\F_{q^m})$ leaves $P$ invariant so that $\sigma(x)$ is a root of $P$, which are at most $m$ in $\F_{q^m}$. As $x$ generates $\F_{q^m}^\times$, the morphism $\sigma$ is determined by $\sigma(x)$ so that $\card G\leq m$. Furthermore, $\varphi_q$ is of order $m$.If not, there exists $0<d<m$ such that $F^d=\Ida$, and thus $x^{q^d}=x$ contradicting that order of $x$ is $q^m-1$. Or, $\varphi_q$ is an automorphism since it is an injective map (as all morphisms of fields) between finite subsets of same cardinality.

The field extension $\F_{p^n}/\F_p$ is Galois, since $\F_{p^n}$ is the splitting field of the separable polynomial $X^{p^n}-X\in\F_p[X]$. The theorem above asserts that $\gal(\F_{p^n}/\F_p)$ is generated by $\varphi_p$ given as $\varphi(x)=x^p$.

If $P(X)\in\F_p[X]$ is an irreducible polynomial, the the field $\F_p[X]/P(X)$ is finite over $\F_p[X]$ and we have already shown in Section 1.2 above that it is thus of the form $\F_{p^d}$ for $d$ as the degree of $P(X)$. If $\alpha$ is the root of the polynomial we have $\F_p(\alpha)=\F_p[X]/P(X)=\F_{p^d}$. The Galois group is generated by $\varphi_p$ given as $\varphi(x)=x^p$, and hence all the $d$ roots of $P(X)$ are $\alpha,\alpha^p, \alpha^{p^2},\ldots, \alpha^{p^{d-1}}$ (note that $\alpha^{p^d}=\alpha$). All these roots are distinct since $P(X)$ is separable over $\F_p[X]$.

Examples
1. Consider irreducible polynomial $X^3-2\in\F_7[X]$, and $\F_7(\alpha)=\F_7[X]/(X^3-2) $ then $\gal(\F_7(\alpha)/\F_7)=\{\alpha\mapsto\alpha, \alpha\mapsto \alpha^7,\alpha\mapsto\alpha^9\}$. The three distinct roots of $X^3-2$ are $\alpha,\alpha^7,\alpha^9$ in $\F_7(\alpha)$.
2. Consider irreducible polynomial $X^3+X^2+1\in\F_2[X]$, and $\F_2(\alpha)=\F_2[X]/(X^3+X^2+1) $ then $\gal(\F_2(\alpha)/\F_2)=\{\alpha\mapsto\alpha, \alpha\mapsto \alpha^2,\alpha\mapsto\alpha^4\}$. The three distinct roots of $X^3+X^2+1$ are $\alpha,\alpha^2,\alpha^4$ in $\F_7(\alpha)$.