Let $S$ be an $\mathbb{F}_p$-scheme for a given prime $p$. Let $X$ be a scheme over $S$. Then, we can consider the Frobenius morphism of $X$ relative to $S$, defined by taking the unique morphism given by universal property from $X$ to $X^{(p)}:=X\times_{S,\phi} S$, where $\phi$ is the absolute Frobenius of $S$. Now, if we have another scheme $Y$, over $X^{(p)}$, and we consider its base change along Frobenius, this defines a scheme $Y\times_{X^{(p)}}X$, and a morphism $\psi:Y\times_{X^{(p)}}X\rightarrow Y$. Is it true that $Y\cong (Y\times_{X^{(p)}}X)^{(p)}$ and that the morphism $\psi$ is the Frobenius of $Y\times_{X^{(p)}}X$ relative to $S$? Thank you for any kind of suggestion!

  • $\begingroup$ What is the morphism $X\to X^{(p)}$? $\endgroup$ – KReiser Nov 4 '18 at 5:50
  • $\begingroup$ It's the Frobenius morphism. You start with $X$ over $S$, where $S$ is an $\mathbb{F}_p$-scheme. You take the absolute Frobenius over $S$, i.e. the morphism which is the identity on the topological space and sends $x$ to $x^p$ on sections. Then you pullback $X$ over $S$ along this absolute Frobenius. This defines $X^{(p)}$. Then you show that there are two maps, one is the absolute Frobenius over $X$ and the other one is the structure morphism over $S$ which makes the pullback diagram commute. This, via universal property, induces a unique map $X\rightarrow X^{(p)}$, which is the relative Frob $\endgroup$ – Zariski93 Nov 4 '18 at 8:28

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