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Suppose $G/S$ is a group scheme over $S$, $X/S$ is a scheme over $S$. $G$ acts on $X$ by the morphism $ \sigma : G \times_S X \to X$. Let $X'$ be a scheme over $X$. How to deine the group action on $X'$, i.e. $\sigma ' : G \times_S X' \to X'$ ?

Presumabely, this morphism should be the base extension of $\sigma$, i.e. $(G \times_S X) \times_X X'$. However, I don't know how to show this fibre product is $G \times_S X'$. (WLOG, one can assume $X,X'$ are affine schemes).

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Dear Li Zhan, I don't know what context this question came up in, but there is one natural way to base-change the $G$-action on $X$: if $S'$ is an $S$-scheme, and $G'$ and $X'$ are the base-changes of $G$ and $X$ over $S'$, then $G'$ acts naturally on $X$'. Regards, –  Matt E May 19 '12 at 2:07
    
@MattE, In Mumford's GIT p.4, he made the definition: Given an action $\sigma $ of $G/S$ on $X/S$, a pair $(Y,\phi)$ as above will be called a universal categorica quotient if, for all morphism $Y' \to Y$ we put $X'= X \times_Y Y'$ and let $\phi' : X' \to Y'$ denote $p_2$, then $(Y', \phi')$ is a categorical quotient of $X'$ by $G$. This definition let me to condider the extension of the group action. –  Li Zhan May 19 '12 at 8:27
    
Dear Li Zhan, I don't have a copy of GIT. What is "as above''? I doubt very much that there is any kind of error on p.4 of GIT (it would be very well-known if there were), so it is almost surely a problem with the interpretation. Regards, –  Matt E May 19 '12 at 11:34

3 Answers 3

up vote 2 down vote accepted

You can always make $G$ act trivially on $X'$. But presumbly you want the action be compatible with the action on $X$. Then it is not possible in general: just take for $X'$ a closed point of $X$ not stable by the action of $G$.

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Thank you so much! I understand now! –  Li Zhan May 18 '12 at 23:59

First, you forgot a "prime" in the $G\times_S X$ just before your last sentence.

However the one you want to prove is a general property of fibered products. One way is to draw the diagram where you put in a "tower" your two fiber products $G\times_SX$ and $(G\times_SX)\times_XX'$ (sorry I am not able to draw here). This will give you a morphism $G\times_SX'\to (G\times_SX)\times_XX'$ and you can prove it is an isomorphism. Note that if everything is affine your claim follows easily because you can write

\begin{equation} (M\otimes_BN)\otimes_AP\simeq M\otimes_B(N\otimes_AP) \end{equation}

whenever you have a ring homomorphism $B\to A$ and $N,P$ are $A$-modules, and $M$ is a $B$-module (and you are exactly in this situation on the level of morphisms of schemes).

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I don't think this works, because in the fibre product, $G \times_S X \to X$ is the projection, however, this is different from the group action $\sigma$, thus you cannot put $\sigma : G \times_S X \to X$ is a fibre digram. –  Li Zhan May 18 '12 at 10:12
    
Sorry, I do not understand your comment: in your original question you did define the group action $\sigma$ to be the projection morphism $G\times_SX\to X$. Also, you can put anything you want in a fiber diagram! It makes sense to construct a fiber product of any diagram of the shape $Z\to S\leftarrow W$. –  Brenin May 18 '12 at 17:04
    
First, $\sigma$ may not be the projection to $ X$. Second, because $\sigma$ is no longer the projection, the fibre product $(G \times_S X) \times_X X′$ may not be $G \times_S X′$ –  Li Zhan May 19 '12 at 0:01

edit: As Li Zhan points out, I misunderstood the question. I will leave what I had written here anyways.

Your last sentence should read "However, I don't know how to show this fibre product is $G \times_S X'$", i.e. it should be $X'$ instead of $X$.

The answer to your question follows from a more general fact, namely that towers of fibered squares are fibered squares (this is exercise 2.3.P in Vakil's FOAG, version of May, 16th).

Suppose that we have some category and two fibered squares

$W \longrightarrow Y$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$X \longrightarrow\,\, Z$

and

$W \longrightarrow Y$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$X \longrightarrow\,\, Z$

i.e. $W = Y\times_Z X$ and $U = W\times_X V$. Then if we put them toghether we get the diagram

$U \longrightarrow W \longrightarrow Y$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$V \longrightarrow X \longrightarrow\,\, Z$.

The claim is now that the outer rectangle

$U \longrightarrow Y$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$V \longrightarrow\,\, Z$

is again a fibered square (so the horizontal morphisms are the compositions of the morphisms in the single diagrams), i.e. $U = Y\times_Z V$.

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I think the argument you gave here is the same as the one given above by atricolf, thus because of the same reason I posted there, it seems doesn't work. –  Li Zhan May 18 '12 at 10:15

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