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I'm trying to undersand the various frobenuis morphism we can consider on a scheme. Those are concepts I just touched, so I'm not sure about some stuff I'm writing and things became clearer while I develop the question; any comment or precisation will then be appreciated, in addition to the answers to the various questions that arises in what follows.


If $A$ is a ring with characteristic $p$ then there is a Frobenius morphism $F:A\to A$ which sends $a$ to $F(a)=a^p$; this morphism is injective iif there are no nilpotent elements and in general is not surjective.

From now on let $K$ be a field of characteristic $p$, and $X$ a $K$-scheme.

Absolute Frobenius

Whave the absolute Frobenius morphism $F^{abs}:X\to X$, which is defined on an affine open subset $U\cong \operatorname{Spec}(A)$ by the Frobenius on $A$.

Since we are looking for the q-th roots of elements of prime ideals, $F^{abs}$ is the identity on the points of $X$, while sends a section in its q-th power as pointed in this answer.

Unless $K=\mathbb F_p$, the absolute Frobenius is not a morphism of $K$-schemes , but it makes the following diagram commute \begin{array}{rcl} X&\overset{F^{abs}}{\longrightarrow}&X\\ \downarrow&&\downarrow&\\ Y&\overset{F^{abs}}{\longrightarrow}&Y \end{array} for any morphism $X\to Y$. This will be useful to define the relative version of this morphism.

Frobenius Twist

The Frobenius twist of $X$ is defined as the fibered product $X^{(p)}:= X\times_{\operatorname{Spec}(K)}\operatorname{Spec}(K)$, where $\operatorname{Spec}(K)$ is regarded as a $K$-scheme whith the absolute Frobenius as structure morphism.

+++Here come the first questions: if $K=\mathbb F_p$ then $X^{(p)}\cong X$ since we are extending via the identity, but in general I cannot understand its structure. Wiki gives the explicit answer in the affine case (there is called extension by scalars), but it's not clear to me how it works and furthermore I don't understand it geometrically: which are the points of $X^{(p)}$? Which are his $K$-points? Are the $\mathbb F_p$-points of $X$ and $X^{(p)}$ the same? In particular, what can we say if the field K is perfect (so F is an automorphism)? I expect them to be isomorphic, even though not naturally.

I'm not very familiar with tensor product of $K$-algebras, but trying to think in the affine case $X\cong\operatorname{Spec}(A)$, it seems to me that $X^{(p)}$ should correspond to the tensor product algebra $A\otimes K$, where $v\otimes\lambda$ is identified to $\lambda^pv\otimes 1$, and where $K$ acts as $\alpha.v\otimes\lambda=\alpha v\otimes\lambda=v\otimes\alpha^p\lambda$

Relative Frobenius

For what we've said, we find that there is a universal morphism $F^{rel}:X\to X^{(p)}$ which is called relative Frobenius.

This is now a morphism of $K$-schemes: in the affine case it sends $v\otimes\lambda$ to $\lambda v^p$ and it's worth noting that the $K$-algebra structure on $X$ is given by $\alpha.v=\alpha^pv$

+++And here come more questions: If $K=\mathbb F_p$ then the relative and the absolute Frobenius coincide, modulo the iusomorphism we said before, and from the affine case I can have a hint of what happens in general to the structure sheaf, but what about the points? Is it surjective/injective? I would like to understand it at least in the case where $X\cong X^{(p)}$.


For the moment it's all. I will ask other questions about the geometric and aritmetic Frobenius, for which we need to define first what is a $k$-structure on a $K$-scheme. I'm pondering if opening a new post for it, or expanding this one.

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