The fibers of a flat quotient morphism. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:29:47Z http://mathoverflow.net/feeds/question/106018 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106018/the-fibers-of-a-flat-quotient-morphism The fibers of a flat quotient morphism. yatir 2012-08-31T07:01:10Z 2012-09-02T05:34:39Z <p>Let $G$ be an affine algebraic group, $G_0$ be its derived subgroup, and $S$ be an algebraic monoid with $G$ as its unit group. It can be shown that $k[S]\hookrightarrow k[G]$ as $G\times G$ modules. Let $A=S//(G_0\times G_0)$ be the good quotient ($A=Speck[S]^{G_0\times G_0}$), and let $\pi:S\to A$ (it is surjective). Further, assume that $\pi$ is flat with integral fibers. In an article I'm reading, the author claims that all $k[\pi^{-1}(a)]$ (for $a\in A$) are isomorphic as $G_0\times G_0$ modules. Does anybody know how to prove this?</p> <p><strong>A direction I was thinking of:</strong> The first thing I tried, is to show that in each fiber, each irreducible module shows up in the same multiplcity. That is, if $$k[S]=\bigoplus V_i$$ (decomposition as irreducilbe $G_0\times G_0$-modules), then $\dim_k Hom_{G_0\times G_0}(V,k[S]\otimes_{k[A]} k(a))$ has to be constant, where $k(a)$ ($a\in A$) is the residue field (constant=independent of a).</p> <p>This, I believe, can be reduced to showing that $\dim_k V\otimes_{k[A]} k(a)$ is constant, for each $V$.</p> <p>Since $V$ is flat over $k[A]$, $k[A]$ is an integral domain, and $V$ is finitely generated over $k[A]$, then I know that $\dim_{k(a)} V\otimes_{k[A]} k(a)$ is constant. </p> <p>Could that be used to show the result I need? Maybe a different approach?</p>