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Let $S$ be a polynomial ring which carries the action of a semi-simple linear algebraic group $G$ (I'm interested in a product of $GL$'s). Take $S$ and $G$ to be over an algebraically closed field.

Say I have an ideal $I_0 \subset S$ that is generated by a finite set of highest weight vectors for $G$.

If $I_0$ is a radical ideal then does it follow that $I=G\cdot I_0$ is too?

I'm trying to exploit the fact that I know something definite about the highest weight vectors of a family of ideals. I want to show that the ideals in my family are all radical. I can't compute a Gröbner basis for $I$, but I think I can for $I_0$ and thus show that it is radical. I'm hoping this buys me something.

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    $\begingroup$ May I assume $I_0$ homogeneous, and $G$ acting linearly? I'm just trying to understand the geometry so far. ${\mathbb P}V(I_0)$ is a $B$-invariant projective variety, and you're looking at the intersection of all its $G$-translates. That closed subscheme will contain some closed $G$-orbits, and it's enough to know that it's reduced along those... I doubt this is helping. $\endgroup$
    – Allen Knutson
    May 13, 2013 at 3:45
  • $\begingroup$ Yes assume $I_0$ is homogeneous and that $G$ acts linearly. For me, $I$ is actually a polynomial representation of $G$. $\endgroup$
    – J. Knecht
    May 13, 2013 at 5:28
  • $\begingroup$ Your question still seems hard to sort out, so anything you can add to illustrate or motivate it would be useful. For instance, what happens if your group just has rank 1? Aside from that, I'm guessing you want to work only in characteristic 0. Otherwise the representation theory gets far more complicated and involves serious unknowns. $\endgroup$
    – Jim Humphreys
    May 27, 2013 at 23:41

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