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Let $f:X\to Y$ be a flat surjective morphism with reduced fibers between affine varieties over an alg. closed field of char. zero, k. Let G be a reductive group acting on X fiberwise. How do I show that all the G-modules $k[f^{-1}(y)]$ are isomorphic?

Edit: I am adding the original setting of the article: Let S be a reductive monoid with unit group G (that is, S in an affine algebraic variety with a structure of a monoid and G is reductive). Let $G_0$ be the commutator subgroup of G. Since $G\times G$ acts on S, we have $k[S]\hookrightarrow k[G]$ (if it is not obvious, you can just believe me). Consider the $G_0\times G_0$ action on S, and set $A=S//(G_0\times G_0)$ (good quotient). Let $f:S\to A$ (it is surjective). Assume that f is flat with integral fibers. Why are all the $G_0\times G_0$ modules $k[f^{-1}(a)]$ isomorphic?

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This isn't true with the hypotheses you've stated: what if $G = 1$, for instance? The fibers of a flat morphism of affine varieties are generally not isomorphic... –  Akhil Mathew Aug 29 '12 at 1:12
    
I'm sorry, I forgot two more conditions. This is a step in an article I'm reading which I don't fully get. f is also surjective with reduced fibers. I am guessing this will have to do something with faithfully flatness... –  Yatir Aug 29 '12 at 5:09
    
@Yatir, you are still missing conditions on $f$. –  user18119 Aug 29 '12 at 10:31
    
@QiL, are you saying this because you have figured out what other conditions I need? The author of the article doesn't say why the fibers are isomorphic as G-modules. I can tell you what else I know about f and maybe you can help. f is dominant and arises from a good quotient of X ($Y=X//G_0$, where $G_0=(G,G)$ is the commutator). –  Yatir Aug 29 '12 at 11:39
    
@Yatir, a trivial sufficient condition would be $X=Y\times G$. But you didn't give any condition on $f$ related to the action of $G$, that was the problem. Now if $Y$ is a quotient of $X$ by $G_0$, then likewise $X$ is a torsor under $G_0$ over $Y$, so the fibers are isomorphic to $G_0$. But I don't see how $G$ acts on the fibers of $f$ is $G_0\ne G$ ! It would act on $Y$, so doesn't preserve the fibers of $f$. –  user18119 Aug 29 '12 at 14:30

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