Unirationality of Fermat varieties in characteristic $p$

Question

In Shioda's famous paper "An Example of Unirational Surfaces in Characteristic $p$" (MSN), 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?

Answer 1

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.

Answer 2

Let $k$ be a field of characteristic $p > 2$. Fix a $\nu \in \mathbf{Z}_{\geq 1}$, and set $q := p^\nu$. Let $r \in \mathbf{Z}_{\geq 1}$ and let $$ X^{2r+1}_{q+1}\colon \{x_0^{q+1} + \cdots + x_{2r+1}^{q+1} = 0\} \subset \mathbf{P}^{2r+1}. $$ be the Fermat hypersurface of degree $q + 1$ in $\mathbf{P}^{2r+1}$. Our aim is to show the

Proposition. $X^{2r+1}_{q+1}$ is unirational.

As Shioda does in his article, the general statement that $X^{2r+1}_n$ is unirational whenever $q \equiv -1 \pmod{n}$ is reduced to this case by mapping $X_{q+1}^{2r+1}$ surjectively onto $X_n^{2r+1}$ via $x_i \mapsto x_i^{(q+1)/n}$.

Proof. We basically follow Shioda's method verbatim. Rearrange the defining equation and change coordinates by a root of $-1$ to obtain $$ (x_0^{q + 1} - x_{2r+1}^{q + 1}) + (x_{2r - 1}^{q + 1} - x_{2r}^{q+1}) = \sum_{i = 1}^{r - 1} (x_{2i-1}^{q + 1} - x_{2i}^{q+1}). $$ Making the change of variables $$ \begin{align*} y_0 & = x_{2r+1} + x_0, & y_{2i-1} & = x_{2i-1} + x_{2i}, \\ y_{2r+1} & = x_{2r+1} - x_0, & y_{2i} & = x_{2i-1} + x_{2i}, \end{align*} $$ the index $i$ ranging from $1$ to $r$, the equation above can be written as $$ y_0y_{2r+1}(y_0^{q - 1} + y_{2r+1}^{q-1}) + y_{2r - 1}y_{2r}(y_{2r - 1}^{q - 1} + y_{2r}^{q - 1}) = \sum_{i = 1}^{r - 1} y_{2i-1}y_{2i}(y_{2i-1}^{q - 1} + y_{2i}^{q - 1}). $$ Now pass to the affine chart with $y_0 \neq 0$, i.e. set $y_0 = 1$ in the equation above so that the function field $K = K(X_{q+1}^{2r+1})$ of $X_{q+1}^{2r+1}$ is $k(y_1,\ldots,y_{2r+1})$ in which the variables $y_i$ are related by the equation above.

Consider the field extension $L$ of $K$ obtained by adjoining a $q$th root $t$ of $y_{2r}$. The game now is to show $L$ is purely transcendental. To do this, consider the change of variables, in which a strangely convenient new variable $u$ is pervasive: $$ \begin{align*} y_{2r-1} & = uy_{2r}, \\ y_{2r+1} & = uv, \\ y_i & = uv_i\;\;\text{for}\;i = 1,\ldots,2r - 2. \end{align*} $$ With these new variables, $L$ is the field $k(t,u,v,v_1,\ldots,v_{2r-2})$ subject to the single relation $$ v(1 + u^{q - 1}v^{q - 1}) + t^{q(q + 1)}(u^{q-1} + 1) = u \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}) $$ where I have already cancelled a common factor of $u$. Now to isolate $v$. Throwing the term $t^{q(q+1)}$ to the right, $$ v + u^{q - 1}(v + t^{q+1})^q = -t^{q(q+1)} + \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}). $$ Make the change of variables $s = u(v + t^{q+1})$, so that $L \cong k(s,t,v,v_1,\ldots,v_{2r-2})$ subject to the relation $$ v + s^{q-1}(v + t^{q+1}) = -t^{q(q+1)} + \sum_{i = 1}^{r - 1} v_{2i-1}v_{2i}(v_{2i-1}^{q - 1} + v_{2i}^{q - 1}). $$ Rearranging one more time, we solve for $v$ in terms of the other $2r$ variables $$ v = \frac{-t^{q+1}(s^{q-1} + t^{q^2 - 1}) + \sum_{i = 1}^{r - 1}v_{2i-1}v_{2i}(v_{2i-1}^{q-1} + v_{2i}^{q-1})}{1 + s^{q-1}} $$ thereby eliminating the single relation imposed on the variables of our field. Thus $L \cong k(s,t,v_1,v_2,\ldots,v_{2r-2})$, as desired. $\square$