Alphabeta Math
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Rational points do not detect the group-scheme structure of alpha_p and mu_p

Statement refuted

For group schemes of finite type over an algebraically closed field k, the abstract group of k-rational points determines their group-scheme structure. Even a fixed underlying k-scheme together with that abstract group determines the group law.

Facts & Assumptions

[F2]

The additive and multiplicative group schemes have the displayed structure morphisms over arbitrary algebras. (The group schemes Ga, Gm, and GLn)

[F3]

Affine scheme morphisms correspond to algebra maps, and product coordinate rings are tensor products. (Affine schemes are contravariantly equivalent to commutative rings, Affine fibre products are spectra of tensor products)

[F4]

We assume the Axiom of Choice, inherited through the closed-subgroup criterion in [F1] and its affine quotient supplier. The coefficient comparisons are finite algebraic calculations. (The Axiom of Choice)

Counterexample

Let k be algebraically closed of characteristic p>0. Set αp=Spec⁡k[x]/(xp),μp=Spec⁡k[t]/(tp−1). Then αp(k)={0} and μp(k)={1} are isomorphic singleton groups. Their underlying k-schemes are isomorphic by t=1+x. Nevertheless they are not isomorphic as k-group schemes: αp has additive comultiplication x↦x⊗1+1⊗x, while in the coordinate x=t−1 the multiplicative law of μp has x↦x⊗1+1⊗x+x⊗x. The proof below excludes every group-scheme isomorphism, not merely the displayed scheme isomorphism.

Given: AC and an algebraically closed field k of characteristic p>0 and the two displayed schemes.

1.1F1F2F3F4givenalgebra

For every commutative k-algebra R, αp(R)={a∈R:ap=0} is an additive subgroup of R: (a+b)p=ap+bp and (−a)p=(−1)pap. Also μp(R)={u∈R×:up=1} is a multiplicative subgroup of R×. The closed immersions into Ga and Gm therefore give the induced group laws by [F1]–[F2]. AC is carried through the criterion in [F1], as recorded in [F4]. Each coordinate algebra has dimension p over k, so is finite type. The formula t=1+x identifies their underlying rings because (1+x)p−1=xp. In a field ap=0 forces a=0, and tp=1 forces (t−1)p=0, hence t=1. Thus both rational-point groups are singleton while both schemes retain a nonzero nilpotent coordinate.

2.1F1F2F3step 1.1algebra

Every group-scheme homomorphism f:αp→Gm corresponds by [F3] to a unit g(x)=∑j=0p−1cjxj in k[x]/(xp) satisfying g(0)=1 and g(x+y)=g(x)g(y) in k[x,y]/(xp,yp). Compare coefficients of xr−1y for 1≤r<p: the left side has coefficient rcr, the right side cr−1c1. Since c0=1 and 1,…,p−1 are invertible in k, induction gives cr=c1r/r!. Now compare the coefficient of xp−1y: the left side is zero because every term of g(x+y) has total degree less than p, while the right side is cp−1c1=c1p/(p−1)!. Thus c1=0 and all cr=0 for r>0. This also covers p=2. Hence every such homomorphism is the trivial one, g=1.

3.1F1F2F3step 1.1step 2.1algebra∎

If αp≅μp as group schemes, compose that isomorphism with the closed subgroup inclusion μp↪Gm. The result would be nontrivial: on coordinate rings the inclusion pulls t back to its nonconstant class in k[t]/(tp−1), and an isomorphism cannot send t−1≠0 to zero. This contradicts step 2.1. Thus the two group schemes are not isomorphic despite their isomorphic underlying schemes and rational-point groups. Nilpotent test algebras distinguish their laws; for example their common coordinate x=t−1 has the additional product term x⊗x for μp written in the counterexample.

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