Let \(K\) and \(L\) be finite fields. Show that \(K\) is isomorphic to a subfield of \(L\) if and only if \(\#K=p^r\) and \(\#L=p^s\) for the same prime \(p\), with \(r\mid s\). Here \(r,s\) are positive integers.
The inequality \(r\leq s\) is not sufficient: \(\mathbb F_4\) does not embed in \(\mathbb F_8\), because the tower law would require \(2\mid 3\). If both fields lie in one algebraic closure of \(\mathbb F_p\), the criterion \(r\mid s\) also characterizes the containment \(K\subseteq L\).
Solution
Every finite field has prime characteristic and cardinality a positive power of that prime [@DF04]. When an embedding \(K\hookrightarrow L\) is given, identify \(K\) with its image.
1. Forward direction (\(K\hookrightarrow L\implies \operatorname{char} K = \operatorname{char} L = p\) and \(r \mid s\)):
1.1. If \(K\) is a subfield of \(L\), then \(L\) is a finite-dimensional vector space over \(K\).Let \(d = [L : K] \ge 1\) denote the degree of the extension.
Multiplication \(K\times L\to L\) restricted from \(L\) makes \(L\) a \(K\)-vector space, and \(L\) is finite.
1.2. Since \(K\) has characteristic \(p\), \(L\) must also have characteristic \(p\).The cardinality of \(L\) is related to the cardinality of \(K\) by:\[ \# L = (\# K)^d = (p^r)^d = p^{rd}. \]
A \(d\)-dimensional vector space over a field with \(q\) elements contains \(q^d\) elements.
1.3. Since \(\# L = p^s\), we have \(s = rd\), which implies \(r \mid s\) (and therefore \(r \le s\)).
Unique factorization in \(\mathbb Z\) turns \(p^s = p^{rd}\) into \(s=rd\).
2. Reverse direction (\(\#K = p^r, \, \#L = p^s\) with \(r \mid s \implies K \hookrightarrow L\)):
2.1. If \(r \mid s\), then \((p^r - 1) \mid (p^s - 1)\) because:\[ x^k - 1 = (x - 1)(x^{k-1} + \dots + 1) \quad \text{applied to } x = p^r, \, s = kr. \]
Evaluate the displayed factorization of \(x^k-1\) at the integer \(x=p^r\).
2.2. The multiplicative group \(L^\times\) is a cyclic group of order \(p^s - 1\).Since \((p^r - 1) \mid (p^s - 1)\), \(L^\times\) contains a unique cyclic subgroup \(H\) of order \(p^r - 1\).
The multiplicative group of a finite field is cyclic, and a cyclic group has exactly one subgroup for each divisor of its order [@DF04].
2.3. The subset \(K' = H \cup \{0\} \subseteq L\) consists precisely of all roots of the polynomial \(f(x) = x^{p^r} - x \in \mathbb{F}_p[x]\) in \(L\).Since the map \(\phi(x) = x^{p^r}\) is an automorphism of \(L\) (the \(r\)-th power of the Frobenius automorphism), the fixed set:\[ K' = \operatorname{Fix}(\phi) = \{\alpha \in L \mid \alpha^{p^r} = \alpha\} \]is a subfield of \(L\).
The Frobenius map is an automorphism of a finite field, and the fixed elements of any field automorphism form a subfield [@DF04].
2.4. The subfield \(K'\) has cardinality \(\# K' = p^r = \# K\).Since any two finite fields with \(p^r\) elements are isomorphic, \(K \cong K' \subseteq L\), so \(K\) is isomorphic to a subfield of \(L\).
Finite fields of the same cardinality are isomorphic [@DF04].
3. Conclusion:\(K\) embeds in \(L\) if and only if \(\#K=p^r\) and \(\#L=p^s\)for the same prime \(p\) with \(r\mid s\).
2. Inside a common algebraicclosure, any subfield with \(p^r\) elements is exactly theroot set of \(T^{p^r}-T\): all its elements are roots byLagrange’s theorem and the polynomial has at most \(p^r\)roots. Thus \(K\) equals the constructed \(K'\) in thatsetting, proving the final assertion of the remark.
Solution
1. An embedding forces the stated characteristic and divisibility.
Every finite field has prime characteristic \(p\) and is a finite-dimensional vector space over its prime subfield \(\mathbb F_p\) [@DF04]. Its size is therefore \(p^r\), where \(r\) is that positive dimension. An embedding preserves the identity and hence the characteristic. Identify \(K\) with its image in \(L\). If \(d=[L:K]\), the tower law gives \[ s=[L:\mathbb F_p]=[L:K][K:\mathbb F_p]=dr. \] Consequently \(r\mid s\). This also proves the four-versus-eight obstruction in the remark.
2. If \(r\mid s\), then \(L\) contains a subfield of size \(p^r\).
Write \(s=rd\) with \(d\geq1\), and put \(h(T)=T^{p^r}-T\). In \(\mathbb F_p[T]/(h)\) the class \(t\) satisfies \(t^{p^r}=t\). Iterating this identity \(d\) times gives \(t^{p^s}=t\). Hence \(h\) divides \(T^{p^s}-T\) in \(\mathbb F_p[T]\).
Lagrange’s theorem in \(L^\times\) shows that every \(a\in L\) satisfies \(a^{p^s}=a\), including zero. There are \(p^s\) such elements, so the monic polynomial of that degree factors as \[ T^{p^s}-T=\prod_{a\in L}(T-a). \] Its divisor \(h\) thus splits in \(L\). Its derivative is \(-1\), so its roots are distinct and number exactly \(p^r\). Let \(K'\) be their set.
It contains zero and one. The characteristic-\(p\) binomial identity, iterated \(r\) times, gives \((a-b)^{p^r}=a^{p^r}-b^{p^r}\). Thus for \(a,b\in K'\) the difference \(a-b\) lies in \(K'\). Multiplicativity gives \((ab)^{p^r}=ab\), and for \(a\ne0\) it gives \((a^{-1})^{p^r}=a^{-1}\). Hence \(K'\) is a subfield of \(L\), of the claimed cardinality.
3. The field \(K\) embeds in this subfield \(K'\).
Every element of \(K\) is a root of \(h\), again by Lagrange’s theorem in \(K^\times\). Thus \(K\) is a splitting field of \(h\) over \(\mathbb F_p\). The field \(K'\) is another: its elements are all the roots. Uniqueness of splitting fields up to base-field isomorphism gives \(K\cong K'\) [@DF04]. Composing this isomorphism with \(K'\subseteq L\) gives the required embedding.
Inside one algebraic closure, a field of size \(p^r\) is exactly the root set of \(h\): it contains \(p^r\) roots and a degree-\(p^r\) polynomial has no others. Thus in that setting \(K=K'\), which proves the literal-containment assertion of the remark.