= Solution
A <finite field> is a <field> with finitely many elements. Its characteristic is the least positive integer $p$ for which $p\cdot1=0$; such an integer exists because the additive group is finite. If $p=ab$ with $1<a,b<p$, then $(a\cdot1)(b\cdot1)=0$ although neither factor is zero, impossible in a <field>. Hence $p$ is prime, and the multiples of $1$ form its <prime field> $\mathbb F_p$. The whole <field> is a finite-dimensional <vector space> over this subfield, say of <dimension> $s\geq1$, so its cardinality is $p^s$. Thus \b[$q=p^s$ and its characteristic is $p$].
For existence, let $K$ be a <splitting field> over $\mathbb F_p$ of $P(T)=T^{p^s}-T$. Since $P'(T)=-1$, it has exactly $p^s$ distinct roots. Its root set $F$ contains $0,1$ and is closed under addition and multiplication, because iterated <Frobenius endomorphism> gives
$$
(a+b)^{p^s}=a^{p^s}+b^{p^s},\qquad (ab)^{p^s}=a^{p^s}b^{p^s}.
$$
It is also closed under additive inverses, and for $a\ne0$, $(a^{-1})^{p^s}=(a^{p^s})^{-1}=a^{-1}$. Thus $F$ is a <field> with exactly $p^s$ elements. Since it contains every root, it is the entire <splitting field>. This proves existence.
For uniqueness, let $L$ be any <field> with $p^s$ elements. Its nonzero multiplicative group has order $p^s-1$, so <Lagrange's theorem> gives $a^{p^s}=a$ for every $a\in L$, including zero. Hence $L$ is a <splitting field> of the same <polynomial>. Here is the embedding argument giving uniqueness, rather than merely invoking it by name. Extend the identity on $\mathbb F_p$ from one root at a time. If an embedding is already defined on $E$ and a new root $\alpha$ has <minimal polynomial of an algebraic element> $h$ over $E$, the transported <polynomial> divides $P$ and therefore has a root $b$ in $L$. Evaluation at $b$ induces an embedding $E[T]/(h)\to L$, extending the existing embedding to $E(\alpha)$. Repeating for the finitely many roots embeds the constructed <field> $F$ in $L$. They have the same finite cardinality, so this embedding is an <isomorphism>. Thus \b[the <field> of order $p^s$ is unique up to <isomorphism>].
We prove cyclicity of the multiplicative group, also establishing the more general <finite multiplicative subgroup of a field is cyclic> result. Let $H$ be a finite subgroup of a <field>'s multiplicative group, and let $e$ be the least common multiple of all element orders. For every prime $\ell$ dividing $e$, choose an element whose order has the maximal $\ell$-power $\ell^{a_\ell}$; raising it to the factor of its order coprime to $\ell$ gives an element of order $\ell^{a_\ell}$. Since the group is abelian, the product of these elements has order $\prod_\ell\ell^{a_\ell}=e$: for commuting elements of coprime orders, an equation $(ab)^t=1$ forces $a^t=b^{-t}$ into the intersection of their cyclic subgroups, which is trivial, so both orders divide $t$. Every member of $H$ is a root of $T^e-1$. The <Lagrange root bound over a field> gives $|H|\leq e$, while the element just constructed has order $e$ and gives $e\leq|H|$. Thus equality holds and that element generates $H$. In particular
$$
\boxed{\mathbb F_{p^s}^{\times}\cong\mathbb Z/(p^s-1)\mathbb Z.}
$$
For the <field with nine elements>, use $\mathbb F_3[u]/(u^2+1)$. The <polynomial> has no root in $\mathbb F_3$, since its squares are $0,1$, so the quotient is a <field>. In the <basis> $(1,u)$, a vector $(a,b)$ represents $a+bu$, and
$$
(a,b)+(c,d)=(a+c,b+d),\qquad(a,b)(c,d)=(ac-bd,ad+bc)
$$
with coefficients reduced modulo three. Put $\beta=1+u$. Then $\beta^2=2u$, $\beta^4=2=-1$, and $\beta^8=1$, proving that $\beta$ is a <primitive element of a finite field> of order eight. Its complete power/vector table is
$$
\begin{array}{c|c|c}
\text{label}&\text{element}&\text{vector in }(1,u)\\\hline
a_0&0&(0,0)\\
a_1&\beta^0=1&(1,0)\\
a_2&\beta&(1,1)\\
a_3&\beta^2&(0,2)\\
a_4&\beta^3&(1,2)\\
a_5&\beta^4&(2,0)\\
a_6&\beta^5&(2,2)\\
a_7&\beta^6&(0,1)\\
a_8&\beta^7&(2,1)
\end{array}
$$
For completeness, the full addition and multiplication tables in these labels are displayed below. The subscripts are element labels; they are not the scalar residues of $\mathbb F_3$.
$$
\begin{array}{c|rrrrrrrrr}
+ & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8\\\hline
a_0 & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8\\
a_1 & a_1 & a_5 & a_8 & a_4 & a_6 & a_0 & a_3 & a_2 & a_7\\
a_2 & a_2 & a_8 & a_6 & a_1 & a_5 & a_7 & a_0 & a_4 & a_3\\
a_3 & a_3 & a_4 & a_1 & a_7 & a_2 & a_6 & a_8 & a_0 & a_5\\
a_4 & a_4 & a_6 & a_5 & a_2 & a_8 & a_3 & a_7 & a_1 & a_0\\
a_5 & a_5 & a_0 & a_7 & a_6 & a_3 & a_1 & a_4 & a_8 & a_2\\
a_6 & a_6 & a_3 & a_0 & a_8 & a_7 & a_4 & a_2 & a_5 & a_1\\
a_7 & a_7 & a_2 & a_4 & a_0 & a_1 & a_8 & a_5 & a_3 & a_6\\
a_8 & a_8 & a_7 & a_3 & a_5 & a_0 & a_2 & a_1 & a_6 & a_4
\end{array}
$$
$$
\begin{array}{c|rrrrrrrrr}
\times & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8\\\hline
a_0 & a_0 & a_0 & a_0 & a_0 & a_0 & a_0 & a_0 & a_0 & a_0\\
a_1 & a_0 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8\\
a_2 & a_0 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8 & a_1\\
a_3 & a_0 & a_3 & a_4 & a_5 & a_6 & a_7 & a_8 & a_1 & a_2\\
a_4 & a_0 & a_4 & a_5 & a_6 & a_7 & a_8 & a_1 & a_2 & a_3\\
a_5 & a_0 & a_5 & a_6 & a_7 & a_8 & a_1 & a_2 & a_3 & a_4\\
a_6 & a_0 & a_6 & a_7 & a_8 & a_1 & a_2 & a_3 & a_4 & a_5\\
a_7 & a_0 & a_7 & a_8 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6\\
a_8 & a_0 & a_8 & a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7
\end{array}
$$
A nonzero element $\alpha=\beta^j$ satisfies $\alpha^4=1$ exactly when $8\mid4j$, so $j$ is even. Thus, with $e=1$,
$$
\boxed{\{\alpha:\alpha^4=e\}=\{(1,0),(2,0),(0,1),(0,2)\}.}
$$
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