In Shioda's famous paper "[An Example of Unirational Surfaces in Characteristic $p$](https://link.springer.com/article/10.1007%2FBF01350715)" ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=374149)), the author proved that the Fermat surface over the characteristic-$p$ field $k$
$$
x\_1^n +x\_2^n + x\_3^n+x\_4^n =0
$$
is unirational if $q \equiv -1 \pmod n$ for some $p$-power $q$.


At end of this paper, he mentioned that he proved it in the higher-dimension case with similar method, *i.e.*,
$$ x\_1^n +x\_2^n + \cdots +x\_{m+1}^n =0$$
is unirational if $m \geq 3$ is odd and $q \equiv -1 \pmod n$ for some $p$-power $q$. But I can't prove it with his method in the surface case. How does his method generalize?



Unirationality of Fermat varieties in characteristic $p$
Let $F^r$ be the Fermat hypersurface of fixed degree $n$ in $\mathbb{P}^{r+1}$, so that $\dim(F^r)=r$ (in any characteristic). There is a rational dominant map $\varphi :F^r\times F^s \cdots\!\!\twoheadrightarrow F^{r+s}$: if you write the equation of $F^r$ in affine coordinates $x\_1^n+\ldots +x^n\_{r+1}=-1$, and that of $F^s$ $\ y\_1^n+\ldots +y^n\_{s+1}=-1$, put $\varphi (x,y)= (x\_1,\ldots ,x\_{r+1},\lambda y\_1,\ldots ,\lambda y\_{s+1})$ in $\mathbb{P}^{r+s+1}$, with $\lambda ^n=-1$. Thus once you know that $F^2$ is unirational, the same holds for $F^r$ for any $r$ even. I don't know if this is what Shioda had in mind, but he certainly used a lot that trick in his subsequent work.


