Affine Algebraic Geometry: Proceedings of the Conference by Kayo Masuda, Hideo Kojima, Takashi Kishimoto

By Kayo Masuda, Hideo Kojima, Takashi Kishimoto

The current quantity grew out of a global convention on affine algebraic geometry held in Osaka, Japan in the course of 3-6 March 2011 and is devoted to Professor Masayoshi Miyanishi at the get together of his seventieth birthday. It comprises sixteen refereed articles within the parts of affine algebraic geometry, commutative algebra and similar fields, that have been the operating fields of Professor Miyanishi for nearly 50 years. Readers should be capable of finding contemporary traits in those components too. the subjects include either algebraic and analytic, in addition to either affine and projective, difficulties. the entire effects handled during this quantity are new and unique which as a result will supply clean examine difficulties to discover. This quantity is acceptable for graduate scholars and researchers in those components.

Readership: Graduate scholars and researchers in affine algebraic geometry.

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8 of the Abhyankar-Moh-Suzuki theorem. 4 below. 3. There is an isomorphism Aut(Xd,e ) Nd,e /Gd,e . Proof. 1]. f This yields an exact sequence (see [5, Thm. 1]) 1 → Gd,e → Nd,e → Aut(Xd,e ) → 1 , as claimed. 4. If e2 ≡ 1 mod d then (23) Nd,e + − Nd,e ∗T Nd,e , while for e2 ≡ 1 mod d (24) Nd,e + Nd,e ∗N + Nd,e . 14 below. 4 due to the fact that the subgroup Gd,e is normal in every group that participates in (23) and (24). 3 below. April 1, 2013 10:34 28 Lai Fun - 8643 - Affine Algebraic Geometry - Proceedings 9in x 6in affine-master I.

Consider further an irreducible acyclic curve Ca, b given in A2 by equation ya − xb = 0, where a, b ≥ 1 and gcd(a, b) = 1. In the following proposition we describe the stabilizer Stab(Ca, b ) for a singular such curve. Consider a one-parameter subgroup Ta,b of the torus T, (15) Ta,b = {γa,b (t) | t ∈ Gm } ⊆ Stab(Ca, b ), where γa,b (t) : (x, y) −→ (ta x, tb y) . 12. If min{a, b} > 1 then Stab(Ca, b ) = Ta,b . Proof. Letting C = Ca, b and Γ = Stab(C), we consider the pointwise stabilizer Γ0 = {γ ∈ Γ | γ|C = idC } ⊆ Γ .

Thus we may assume that G ⊆ Aff(A2k ). It remains to show that G has a fixed point in A2 . Observe that G ⊆ Aff(A2k ) admits a representation in GL(3, k) by matrices of the form ⎛ ⎞ ∗∗∗ ⎝∗ ∗ ∗⎠. Since G is reductive and char (k) = 0, G is geometrically reduc001 tive. Therefore the G-invariant plane L0 = {x3 = 0} in A3k with coordinates (x1 , x2 , x3 ) has a G-invariant complement, say, R, which meets the parallel G-invariant plane L1 = {x3 = 1} in a fixed vector. This yields a fixed point of G in A2k G L1 .

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