Let $k$ be a field of characteristic $p$, and define an embedding of fields $f:k\to k$ by $f(\lambda)=\lambda^p$. Then, given any scheme over $k$, $X$, define the Frobenius twist of $X$ as
$$X^{(1)}:=X\times_{f}\operatorname{spec}k$$
where we consider $\operatorname{spec}k$ as a scheme over $k$ via the map $f$. The paper I'm reading says there is a natural morphism of schemes $X\to X^{(1)}$, which I'm trying to define. It seems to me I should use the universal property of the fiber product of schemes, so I need to find a natural map $\varphi:X\to X$ so that the maps
$$X\xrightarrow{\varphi} X\to\operatorname{spec}k$$
and
$$X\to\operatorname{spec}k\xrightarrow{f^*}\operatorname{spec}k$$
are equal. How might I define $\varphi$? Or is there another way to view this natural map $X\to X^{(1)}$?
Note: In the affine case, I believe the comorphism of $\varphi$ is
$$\begin{array}{rcl}
\varphi^*:k[X^{(1)}]&\longrightarrow&k[X]\\
a\otimes\lambda&\longmapsto&\lambda\cdot a^p
\end{array}$$
but I'd like to have a description of $\varphi$ independent of any affine covering.
Best Answer
The morphism $\phi$ you are looking for is called the absolute Frobenius morphism of $X$. It is the identity on points, and on regular functions it is given by $h\mapsto h^p$. The same for the arrow that you denote $f^\ast:\textrm{Spec }k\to \textrm{Spec } k$, which is the comorphism of $f:\lambda\mapsto\lambda^p$. You can verify easily that the absolute Frobenius commutes with every morphism of schemes over $k$, so if you call $\sigma$ the structure morphism of $X/k$, you get $f^\ast\circ\sigma=\sigma\circ \phi$. The universal property of $X^{(p)}$ gives you the morphism $X\to X^{(p)}$, called the relative Frobenius morphism of $X$.
If your variety $X$ is defined over $\mathbb F_p$, then the absolute and relative Frobenius coincide. (Indeed, the drawback of the absolute Frobenius is that it does not respect the usual $k$-algebra structure in general, unless $k=\mathbb F_p$; that's why we make the base change by $f$, and define the $k$-structure of $X^{(p)}$ through the second projection to $\textrm{Spec }k$, where this $k$ has the "modified" $k$-algebra structure, namely $\alpha.x:=\alpha^px$.)
If $X=\textrm{Spec }A$ is affine, then the relative Frobenius is the $k$-morphism corresponding to the morphism you mention at the end, namely $a\otimes\lambda\mapsto \lambda a^p$.