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Extra info for Actions of Linearly Reductive Groups on Affine Pi Algebras

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This shows that some additional assumption about the conductor is necessary, if "B is affine" should imply "A is affine". 13 below. PROOF. Denote by NA and NB the nil radicals of A and 5 , respectively. 6, it is sufficient to show that A/NA is affine. Since NB D i i s a nilpotent ideal of A, it is contained in N^. And since the extension A C B is centralizing, the ideal generated by NA in B is nilpotent. Hence NA C iV#, in fact NA = NB H A. So A/NA embeds into B/NB, and this ring extension satisfies the hypotheses of the lemma.

We identify A and 5 with their natural images in B. Note that B is a commutative affine domain. Define a fc-algebra R by (Ai+mB \ mB B \ A2 + mB J ' Then R is a prime Pi-algebra over k. Since S = k -f m, the commutative algebras Ai + mB = Ai + S + m(A ® 5) are affine and Noetherian, so that also R is affine and Noetherian. Since Ai + mB = (Ai ® k) + (A ® m) is a direct sum of k-vector spaces, it follows that the center of R is C = ((Ai O A2) -f mB) • I2. The action of G on 5 extends to a rational action on B by fixing A pointwise.

Where x is an indeterminate. The minimal prime ideals of RG are q\ = k[x] and q2 = &, embedded into RG in the obvious way. Hence GK(RG/qi) — GK(k) = 0, but GK(RG /q2) = GK(k[x]) = 1. So equidimensionality does not hold. Moreover, RG (and a fortiori R) are not homogeneous as right i^-modules: the GK-dimension of q\ and q2 as ideals of RG are 1 and 0, respectively. ) We have PnRG=(xkW 2)=**Meo. Hence $(P) — {pi,P2} where px — qx — k[x] ® 0 and p2 — xk[x] © fc. Clearly, p\ is a minimal prime ideal of RG, but p2 is not.

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