Exercise 10.17

Proof. Let ${\omega }_1,{\omega }_2,\dots ,{\omega }_m$ be a basis for $F_{q^m}$ over $F_q$.

Consider

Then $f$ is a form of degree $m$ in $m$ variables.

By definition, $f\in {𝔽}_{q^m}(x_1,\dots ,x_m)$. We show first that $f\in {𝔽}_q[x_1,\dots ,x_m]$.

Let $f$ be the Frobenius automorphism on ${𝔽}_{q^m}$, defined by

By Corollary 1 of Proposition 7.1.1, for every $\alpha \in {𝔽}_{q^m}$, $\alpha \in {𝔽}_q$ if and only if $F(\alpha )=\alpha$. If $p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^d{a}_{i_1,\dots ,i_m}{x}_1^{i_1}\cdots x_m^{i_m}\in {𝔽}_{q^m}[x_1,\dots ,x_m]$, define $F\cdot p={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^{d}F(a_{i_1,\dots ,i_m})x_1^{i_1}\cdots x_m^{i_m}$. Then $F\cdot p\in {𝔽}_{q^m}[x]⟺F\cdot p=p$ and $F\cdot (\mathit{pq})=(F\cdot p)(F\cdot q)$ for all $p,q\in {𝔽}_{q^m}[x_1,\dots ,x_m]$.

Then, using this last property,

Therefore $f\in {𝔽}_q[x_1,\dots ,x_m]$.

Now we prove that $f$ has no non trivial zero $\bar{a}=({\alpha }_1,\dots ,{\alpha }_m)\in {𝔽}_q^m\setminus \{(0,\dots ,0)\}$. If $f$ had such a zero $({\alpha }_1,\dots ,{\alpha }_m)\in {𝔽}_q^m$, then

Then for some $i\in \text{[[}0,m-1\text{]]}$,

Applying $F^{m-i}$ to this equality, and using $F({\alpha }_i)={\alpha }_i$, we obtain

Since ${\omega }_i^{q^m}={\omega }_i,i=1,\dots ,m$, this gives

Since ${\omega }_1,{\omega }_2,\dots ,{\omega }_m$ is a basis for $F_{q^m}$ over $F_q$, this proves

So $f$ has no non trivial zero.

Note : this proves that we cannot extend the Chevalley’s Theorem to the forms of degree $m$ in $m$ varibles. □