= Solution
Let $D$ be a finite <integral domain>. For any $a\ne0$, multiplication $m_a:D\to D$ is injective: $ax=ay$ implies $a(x-y)=0$, hence $x=y$ because there are no zero divisors. An injective map of a finite set to itself is surjective, so $ab=1$ for some $b$. Every nonzero element is invertible; therefore \b[a finite integral domain is a field].
For an <irreducible polynomial> $f\in F[X]$, consider a nonzero class $[g]$ in the <quotient ring> $F[X]/(f)$. Since $f$ does not divide $g$, irreducibility gives $\gcd(f,g)=1$. The <Euclidean algorithm> in the <polynomial ring> gives polynomials $u,v$ with $uf+vg=1$. Modulo $(f)$, $[v][g]=1$. Thus \b[the quotient by an irreducible polynomial is a field].
For a <field with four elements>, use
$$
\boxed{\mathbb F_4=\mathbb F_2[X]/(X^2+X+1).}
$$
The quadratic has no root in $\mathbb F_2$, so is irreducible. Write $\alpha=[X]$. Polynomial division shows the elements are $0,1,\alpha,\alpha+1$, with $\alpha^2=\alpha+1$. Its multiplication table is
$$
\begin{array}{c|cccc}
\cdot&0&1&\alpha&\alpha+1\\ \hline
0&0&0&0&0\\
1&0&1&\alpha&\alpha+1\\
\alpha&0&\alpha&\alpha+1&1\\
\alpha+1&0&\alpha+1&1&\alpha
\end{array}
$$
For a <field with nine elements>, use
$$
\boxed{\mathbb F_9=\mathbb F_3[X]/(X^2+1).}
$$
The values of $X^2+1$ at $0,1,2$ are $1,2,2$, so the quadratic is irreducible. Every class is uniquely $a+b\beta$, $a,b\in\mathbb F_3$, with $\beta^2=-1=2$. These nine classes form a <finite field>; multiplication is $(a+b\beta)(c+d\beta)=(ac-bd)+(ad+bc)\beta$, with coefficients modulo three.
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