Subfield of finite field generated by cubed elements

Question

So assume we have a finite field $F$ and we form the subfield $L$ generated by all elements of the form $x^3$, where $x\in F$. If $L\neq F$ then $L$ must have exactly four elements.

Now my thought was to examine the multiplicative subgroup of $F$, which is cyclic to derive this result, but I'm not sure how to carry through with the argument. There is probably some number theory involved, but I wasn't sure. Can someone help with this.

Answer 1

If $F$ has $q=p^k$ elements, $F^\times$ is cyclic of order $q-1$. The (non-zero) cubes are a subgroup of this, either of index $3$ or the full group. In the latter case we are done, hence assume that the non-zero cubes are a subgroup of index $3$ in $F^\times$. We have to look at addition, but as soon as we find just one non-cube that is the sum of two cubes, $L^\times$ must be strictly larger than the group of non-zero cubes, hence must be all of $F^\times$ and we are done. Hence we may assume that the sum of cubes is always a cube. Then the cubes form a non-trivial additive subgroup of $F$, which must be of order $p^l$ for some $0<l\le k$.

Combining the multiplicative and the additive result, we have $$p^l-1=\frac13(p^k-1),$$ i.e., $$(3-p^{k-l})\cdot p^l=2.$$ As $l>0$, we conclude $p=2$, and then $l=1$ and $k-l=1$.

Answer 2

Hagen's argument is nice. We can also do this with a cardinality argument only involving the multiplicative groups. The idea is to use known information about cardinalities of the proper subfields.

Assume that $F$ has $p^k$ elements. The group $F^*$ is cyclic of order $p^k-1$, so the cubes either fill up all of $F^*$ or they form a subgroup of of order $(p^k-1)/3$. If $F$ has a proper subfield, we must have $k\ge2$.

In the former case obviously $L=F$. In the latter case $L^*$ has at least $(p^k-1)/3$ elements. But, if $L$ is a proper subfield we have $|L^*|=p^\ell-1$ for some proper divisor $\ell\mid k$. In particular, $\ell\le(k/2)$. We can conclude that the index of the group $L^*$ in $F^*$ is thus $$ [F^*:L^*]=\frac{p^k-1}{p^\ell-1}\ge\frac{p^k-1}{p^{k/2}-1}=p^{k/2}+1. $$ On the other hand we saw that $[F^*:L^*]=3.$ Hence $$ 3\ge p^{k/2}+1. $$ Recalling that $k\ge2$ we are left with the alternative $p=k=2$. Therefore $|F|=p^k=4.$

For the sake of completeness we need to check that the conclusion holds when $F=\Bbb{F}_4$. In that case all the non-zero element of $F$ are cube roots of $1$, so in this case $L=\{0,1\}=\Bbb{F}_2$ is a proper subfield.