= Solution
Map the <integers> into a <field> by $n\mapsto n1$. Its <kernel (algebra)> is a <prime ideal> of $\mathbb Z$, since the <image> has no <zero divisors>. Thus the <kernel (algebra)> is either zero or $(p)$ for a <prime> $p$. In the zero case, the map extends injectively to the <rational numbers> by taking quotients of nonzero <integers>. In the other case, its <image> is the <field> $\mathbb F_p$. This proves the prime-field assertion.
A <finite field> cannot contain the infinite rational <field>, so it has <characteristic> $p$. It is a <finite-dimensional vector space> over its <prime field> $\mathbb F_p$, say of <dimension> $n\ge1$, and choosing a <basis> counts its elements as
$$
\boxed{|F|=p^n.}
$$
For existence, put $q=p^n$ and take a <splitting field> of $x^q-x$ over $\mathbb F_p$. Its <derivative> is $-1$, so it has exactly $q$ distinct <polynomial roots>. The <polynomial roots> are closed under <addition> and <multiplication> because the Frobenius identity gives $(u+v)^q=u^q+v^q$ and $(uv)^q=u^qv^q$. They include $0,1$, are closed under <additive inverses>, and for a nonzero <polynomial root> $u$, $(u^{-1})^q=(u^q)^{-1}=u^{-1}$. They therefore form a <field> of exactly $q$ elements. Indeed this <field> is the <splitting field> because it contains all <polynomial roots> and is generated by them.
In order 4, <characteristic> is 2. For $\omega\ne0,1$, the multiplicative <group> has order 3, so $\omega^3=1$ and $\omega^2\ne1,\omega$. Factorization of $x^3-1$ gives $\omega^2+\omega+1=0$. The quadratic $x^2+x+1$ has no <polynomial root> in $\mathbb F_2$ and is the <minimal polynomial>. Write $s=\omega^2=\omega+1$. The complete tables are
$$
\begin{array}{c|cccc}
+&0&1&\omega&s\\\hline
0&0&1&\omega&s\\
1&1&0&s&\omega\\
\omega&\omega&s&0&1\\
s&s&\omega&1&0
\end{array}
\qquad
\begin{array}{c|cccc}
\cdot&0&1&\omega&s\\\hline
0&0&0&0&0\\
1&0&1&\omega&s\\
\omega&0&\omega&s&1\\
s&0&s&1&\omega
\end{array}.
$$
For an element $\xi\ne0,1$ in order 8, the <tower law> makes $[\mathbb F_2(\xi):\mathbb F_2]$ divide3. It is not1, so its <minimal polynomial> is an irreducible monic cubic. Its <constant term> must be1. A cubic $x^3+ax^2+bx+1$ has no <polynomial root> at 1 exactly when $a+b=1$, and absence of a <polynomial root> is equivalent to cubic irreducibility. Consequently the two possibilities are
$$
\boxed{x^3+x+1\quad\text{and}\quad x^3+x^2+1.}
$$
Both occur: together with $x$ and $x+1$ they factor $x^8-x$ over $\mathbb F_2$, whose distinct <polynomial roots> are exactly the eight elements of the <field>. Each cubic contributes its three <polynomial roots>.
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