Let \(F\) be a finite field. Suppose the subfield generated by \[ \{x^3:x\in F\} \] is a proper subfield of \(F\). Prove that \[ |F|=4. \]
Solution
Write \[ q\coloneqq\abs F=p^n \] with \(p\) prime, and let \(K\) be the subfield of \(F\) generated by \(\{x^3:x\in F\}\).
1. The characteristic \(p\) is either \(2\) or \(3\).
Suppose that \(p\ne 2,3\), and let \(a\in F\). Since \(K\) contains every cube and contains the prime field, \[ (a+1)^3+(a-1)^3-2a^3=6a \] belongs to \(K\). Because \(6\) is invertible in the prime field, it follows that \(a\in K\). Thus \(F=K\), contrary to the hypothesis.
2. The characteristic is \(p=2\).
If \(p=3\), then the Frobenius map \[ F\longrightarrow F,\qquad x\longmapsto x^3 \] is an automorphism of the finite field \(F\). Hence every element of \(F\) is a cube, so again \(K=F\), contrary to the hypothesis. Step 1 therefore leaves only \(p=2\).
3. The integer \(n\) is even, and the set of nonzero cubes hascardinality\[ \frac{2^n-1}{3}. \]
By step 2, \(q=2^n\). The group \(F^\times\) is cyclic of order \(2^n-1\). If cubing were surjective on \(F^\times\), then every nonzero element of \(F\) would be a cube and hence lie in \(K\), forcing \(K=F\). Thus the endomorphism \[ F^\times\longrightarrow F^\times,\qquad x\longmapsto x^3 \] is not surjective. For a cyclic group of order \(2^n-1\), this happens exactly when \(3\mid 2^n-1\). Since \(2\equiv-1\pmod 3\), this divisibility forces \(n\) to be even. Its kernel then has order \(3\), so its image, the set of nonzero cubes, has cardinality \((2^n-1)/3\).
4. One has\[ \frac{2^n-1}{3}\le 2^{n/2}-1. \]
Because \(K\) is a proper subfield of \(F=\FF_{2^n}\), there is a proper divisor \(d\) of \(n\) such that \[ \abs K=2^d. \] Hence \(d\le n/2\). Every nonzero cube belongs to \(K^\times\), so step 3 gives \[ \frac{2^n-1}{3}\le \abs{K^\times} =2^d-1 \le 2^{n/2}-1. \]
5. The field has cardinality\[ \boxed{\abs F=4}. \]
Set \(m\coloneqq 2^{n/2}\). By step 3, \(n\) is a positive even integer, so \(m\ge2\). Step 4 becomes \[ \frac{m^2-1}{3}\le m-1. \] Since \(m-1>0\), cancelling this factor gives \[ m+1\le3. \] Thus \(m=2\), so \(n=2\) and \(\abs F=2^2=4\).
6. Q.E.D.
Step 5 proves the required cardinality.