# Pullback of a scheme along the Frobenius morphism

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!

• What is the morphism $X\to X^{(p)}$? – KReiser Nov 4 '18 at 5:50
• 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 – Zariski93 Nov 4 '18 at 8:28