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Consider the Fermat quartic surface $$F\!: x^4 + y^4 + z^4 + t^4 = 0$$ over an algebraically closed field $k$ of odd characterstics $p$. Shioda proved that for $p=3$ this surface is a generalized Zariski surface, i.e., there is a purely inseparable covering $F \dashrightarrow \mathbb{P}^2$ (or equivalently $\mathbb{P}^2 \dashrightarrow F$). Is this true for other primes $p$ such that $p \equiv 3 \ (\mathrm{mod} \ 4)$? How can a purely inseparable map be found explicitly?

Consider the Fermat quartic surface $$F\!: x^4 + y^4 + z^4 + t^4 = 0$$ over an algebraically closed field $k$ of odd characterstics $p$. Shioda proved that for $p=3$ this surface is a generalized Zariski surface, i.e., there is a purely inseparable covering $F \dashrightarrow \mathbb{P}^2$ (or equivalently $\mathbb{P}^2 \dashrightarrow F$). Is this true for other primes $p$ such that $p \equiv 3 \ (\mathrm{mod} \ 4)$?

Consider the Fermat quartic surface $$F\!: x^4 + y^4 + z^4 + t^4 = 0$$ over an algebraically closed field $k$ of odd characterstics $p$. Shioda proved that for $p=3$ this surface is a generalized Zariski surface, i.e., there is a purely inseparable covering $F \dashrightarrow \mathbb{P}^2$ (or equivalently $\mathbb{P}^2 \dashrightarrow F$). Is this true for other primes $p$ such that $p \equiv 3 \ (\mathrm{mod} \ 4)$? How can a purely inseparable map be found explicitly?

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Is the Fermat quartic surface a generalized Zariski surface?

Consider the Fermat quartic surface $$F\!: x^4 + y^4 + z^4 + t^4 = 0$$ over an algebraically closed field $k$ of odd characterstics $p$. Shioda proved that for $p=3$ this surface is a generalized Zariski surface, i.e., there is a purely inseparable covering $F \dashrightarrow \mathbb{P}^2$ (or equivalently $\mathbb{P}^2 \dashrightarrow F$). Is this true for other primes $p$ such that $p \equiv 3 \ (\mathrm{mod} \ 4)$?