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$\require{AMScd}\DeclareMathOperator\Spec{Spec}$

Consider the following two commutative diagrams where $X':=X\times_{\operatorname{Spec} k, f_k} \operatorname{Spec} k$ is the pullback of $X$ along $f_k$ the absolute frobenius of $\operatorname{Spec} k$

$$\begin{CD} X' @>f>> X \\ @VVV @VVV \\ \operatorname{Spec} k @>f_k>> \operatorname{Spec}k \end{CD}$$

and

$$\begin{CD} S @>f_S>> S \\ @VVV @VVV \\ S @>f_S>> S \end{CD}$$

Prove that this diagram is commutative

$$\begin{CD} X'\times S @>(\pi_1,f_S)>> X\times S \\ @VVV @VVV \\ S @>f_S>> S \end{CD}$$ where $f_S$ the absolute frobenius of $S$ and $\pi_1$ is the first projection.


I am trying to find a solution using 2-fibre products but I don't know where to start, can we see a commutative square as an object and construct the fibered product of two squares?

Thank you for any hep.

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  • $\begingroup$ Is $k$ one of the fields $\mathbb{F}_q$ for $q = p^n$ and is $f_k$ the Frobenius on $k$, i.e., is $f_k(x) = x^{q}$ for $q = \lvert \mathbb{F}_q\rvert$? $\endgroup$ – Geoff May 16 '20 at 18:05
  • $\begingroup$ @Geoff indeed $k$ is a field of characteristic $p>0$, and for me I thought of the frobenius as the identity on topological spaces and the $p$-power on rings, so $f_k(x)=x^p$ for $x\in k$ and $f_S(x)=x^p$ for $x\in \mathcal O_S$ .Thank you! $\endgroup$ – PerelMan May 16 '20 at 18:23
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The Categorical Answer

There's nothing complicated here.

Just check commutativity directly. The vertical maps are the projections onto the second factor. By definition of $(\pi_1,f_S)$ (which only makes sense if I assume you really mean $\pi_1\times f_S$), we have $\pi_2(\pi_1,f_S)=f_S\pi_2$, which is what you get when you go around the other way.

The Algebraic Geometry Answer

If you want an even more explicit calculation, we have the following:

Without loss of generality you may assume that $X$ and $S$ are affine, since pullbacks (and products) are computed locally.

Then if $S=\newcommand\Spec{\operatorname{Spec}}\Spec A$, $X=\Spec B$, for $A$, $B$ $k$-algebras, and if $\phi_k : k\to k$ and $\phi_A : A\to A$ are the Frobenius maps, we have that $X' = \Spec (B\otimes_{\phi} k)$, and we need to show that the following diagram commutes: $$ \require{AMScd} \begin{CD} A @>\phi_A>> A\\ @Va\mapsto a \otimes 1 VV @VVa\mapsto a \otimes 1 \otimes 1V \\ A\otimes_k B @>>a\otimes b \mapsto \phi_A(a) \otimes b\otimes 1> A\otimes_k(B\otimes_\phi k)\\ \end{CD} $$

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  • $\begingroup$ Thank you for your help! Just one question about the categorical answer, do you consider that $\pi_2$ is the same map for both projections onto the second factor? $\endgroup$ – PerelMan May 17 '20 at 11:33
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    $\begingroup$ @PerelMan The two $\pi_2$s are different. They have different domains, so they can't be the same. $\endgroup$ – jgon May 17 '20 at 19:20
  • $\begingroup$ indeed, thank you very much for your help! $\endgroup$ – PerelMan May 17 '20 at 21:54

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