Proceedings of the Japan Academy, Series A, Mathematical Sciences

On an analog of Serre’s conjectures, Galois cohomology and defining equation of unipotent algebraic groups

Nguyêñ Quôć Thǎńg and Nguyêñ Duy Tân

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Abstract

In this note we establish the validity, in the case of unipotent group schemes over non-perfect fields, of an analog of Serre’s conjectures for algebraic groups, which relates properties of Galois (or flat) cohomology of unipotent group schemes to finite extensions of non-perfect fields. We also establish an interesting property of Russell’s defining equations of connected smooth one-dimensional unipotent groups over a field $k$.

Article information

Source
Proc. Japan Acad. Ser. A Math. Sci., Volume 83, Number 7 (2007), 93-98.

Dates
First available in Project Euclid: 18 January 2008

Permanent link to this document
https://projecteuclid.org/euclid.pja/1200672007

Digital Object Identifier
doi:10.3792/pjaa.83.93

Mathematical Reviews number (MathSciNet)
MR2361418

Zentralblatt MATH identifier
1190.11028

Subjects
Primary: 11E72: Galois cohomology of linear algebraic groups [See also 20G10]
Secondary: 12F05: Algebraic extensions 12F15: Inseparable extensions 20G10: Cohomology theory

Keywords
Serre’s conjectures unipotent groups Galois cohomology

Citation

Tân, Nguyêñ Duy; Thǎńg, Nguyêñ Quôć. On an analog of Serre’s conjectures, Galois cohomology and defining equation of unipotent algebraic groups. Proc. Japan Acad. Ser. A Math. Sci. 83 (2007), no. 7, 93--98. doi:10.3792/pjaa.83.93. https://projecteuclid.org/euclid.pja/1200672007


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  • \item[1)]$k$ has no extensions of degree $p$;
  • \item[2)]$k$ has no extensions of degree divisible by $p$;
  • \item[3)]$k$ has no normal extensions of degree $p$;
  • \item[4)]$k$ has no normal extensions of degree divisible by $p$;
  • \item[5)]$k$ has no Galois extensions of degree $p$;
  • \item[6)]$k$ has no Galois extensions of degree divisible by $p$;
  • \item[7)]Every separable $p$-polynomial in one variable is universal;
  • \item[8)]Every $p$-polynomial in one variable is universal.
  • \item[9)]$\mathrm{H}^{1}(k,G) =0$ for any smooth unipotent k-group G. Then we have the following diagram of relations \endenumerate \beginequation* \includegraphicsPJA8307A-01fig01.eps \endequation* All other related implications or non-implications between the statements above follow from these diagrams. \endthm \beginrmk* 1) One might add the 10-th condition, saying that $\mathrm{H}^{1}_{fppf}(k,G) =0$ for all unipotent $k$-group schemes $G$, which is equivalent to conditions 2) and 8), but it is a bit difficult to draw the square above of relations with ten vertices. 2) From above (Theorem 3 and Corollary 7) we see that the two conditions in Theorem 6 are not the same. 3) We give an example, which shows that the condition “$\mathrm{H}^{1}(k,G)=0$ for all smooth unipotent $k$-groups $G$” and the condition “$\mathrm{H}^{1}(k,G)=0$ for all connected and smooth unipotent $k$-groups $G$” are not the same. Indeed, take any field $k$ of characteristic $p>0$ such that certain separable $p$-polynomial $f(T)$ in one variable with coefficients in $k$ (e.g. the Artin - Schreier map $\wp $) is not surjective as a map $k^{+} \to k^{+}$. Take $a \in k^{+} \setminus f(k^{+})$. We claim that for the perfect closure $K:= k^{-p^\infty}$ of $k$, we have $K^{+} \ne f(K^{+})$. If not, $K^{+}=f(K^{+})$, and we have $a \in f(K^{+})$, so $a=f(x)$, $x \in K$. By assumption, we have $k \ne k(x)$. Since $x$ is a root of the separable polynomial $f(T)-a$, $k(x)/k$ is a separable extension. But $k(x) \subset K$ and $K/k$ is purely inseparable, hence so is $k(x)/k$. Thus $k=k(x)$, which is impossible. Therefore $f(K^{+}) \ne K^{+}$. Let $G:= \mathit{Ker} (f)$. Then $G$ is a finite (smooth) étale unipotent $k$-group scheme with $\mathrm{H}^{1}(K,G) = K^{+}/f(K^{+}) \ne 0$, while $\mathrm{H}^{1}(K,H)=0$ for all connected smooth unipotent $k$-groups $H$, since $K $ is perfect. \endrmk* \sectionEquations defining commutative unipotent groups of exponent $p$. We first recall some notations and results in Part 1 of [KMT?]. Let $k$ be a non-perfect field of characteristic $p>0$. It is known that the endomorphism ring $R:= \mathit{End}_{k-gr}(\mathbf{G}_{a})$ can be identified with the noncommutative polynomial $k$-algebra with one indeterminate $F$ subjected to the relation $F\lambda=\lambda^{p}F$, for all $\lambda \in k$. A pair $(n,\alpha)$ with $n\in \mathbf{N}$ and $\alpha=\sum a_{i}F^{i}\in k[F]$ is called admissible if either (i) $n=0$ or (ii) $a_{0}\not=0$ and $a_{i}\not\in k^{p}$ for some $i>0$. For a commutative affine $k$-group scheme $G$, denote $ M(G):=\mathrm{Hom}_{k-gr}(G,\mathbf{G}_{a})$, which is a left $R$-module, hence also a left $k[F]$-module in a natural way. On the other hand, any given left $k[F]$-module $M$ can be considered as a (commutative) $p$-Lie algebra with zero multiplication and $p$-power given by $m^{[p]} = Fm,$ for all $m \in M$. The universal envelopping $k$-algebra $U(M)$ of $M$ has a natural Hopf algebra structure, and the affine $k$-group scheme corresponding to $U(M)$ is denoted by $D(M)$. It has been shown that (cf. [DG; Chap. IV, Sec. 3, no. 6.2, p. 520]) there is an anti-equivalence between the category of commutative $k$-group schemes with the category of left $k[F]$-modules, via $G \mapsto M(G)$; $M \mapsto D(M)$, where the algebraic $k$-group schemes correspond to finitely generated modules. Let $M(n,\alpha)$ be the left $k[F]$-module on a set of 2 generators $x,y$ defined by the relation $F^{n} y=\alpha x$, $\alpha \in k[F]$. Then, there is a natural bijective correspondence $G\mapsto M(G)$; $M\mapsto D(M)$ between the unipotent groups $G$ of dimension 1 and left $k[F]$-modules $M=M(n,\alpha)$, where $(n,\alpha)$ are admissible pairs. More precisely, if $G$ is defined by the equation $y^{p^{n}}=a_{0}x+a_{1}x^{p}+\cdots+a_{r}x^{p^{r}}$, where $a_{0}\not=0$ and $a_{i}\notin k^{p}$ for some $i$, then to $G$ one assigns $M(G):=M(n,\alpha)$, where $\alpha=a_{0}+a_{1}F+\cdots+a_{r}F^{r}$. For $\alpha=\sum a_{i}F^{i}\in k[F]$, let $\alpha^{(\nu)}=\sum a_{i}^{p^\nu}F^{i}$. \beginprop Let $k$ be a non-perfect field of characteristic $p>0$, and let $G_{1} , G_{2}$ be unipotent smooth $k$-groups of dimension 1, defined by $\{(x,y)\in \mathbf{G}_{a}^{2} | y^{p^{m}}= x+a_{1}x+\cdots+a_{r}x^{p^{r}}, \exists i, a_{i}\not\in k^{p}\}$, $\{(x,y)\in \mathbf{G}_{a}^{2} | y^{p^{n}}= x+b_{1}x+\cdots+b_{s}x^{p^{s}}, \exists j, b_{j}\not\in k^{p}\}$ respectively. If the groups $\mathrm{Hom}_{k-gr}(G_{1}, G_{2})$ and $\mathrm{Hom}_{k-gr}(G_{2}, G_{1})$ are both nontrivial then $m=n$ and the following two sets of indices coincide: \beginequation* \i\mid a_i\not\in k^p equiv\i\mid b_i\not\in k^p. \endequation* \endprop \beginproof[Proof (Sketch)] We first prove the second statement of Proposition 10 in the particular case when $n= m = 1$. \beginlem Let $k$ be a non-perfect field of characteristic $p>0$, and let $G_{1} , G_{2}$ be unipotent smooth $k$-groups of dimension 1, defined by $\{(x,y)\in \mathbf{G}_{a}^{2}| y^{p}= x+a_{1}x+\cdots+a_{r}x^{p^{r}}, \exists i, a_{i}\not\in k^{p}\},$ $\{(x,y)\in \mathbf{G}_{a}^{2}| y^{p}= x+b_{1}x+\cdots+b_{s}x^{p^{s}}, \exists j, b_{j}\not\in k^{p}\}$ respectively. Assume that $\mathrm{Hom}_{k-gr}(G_{1}, G_{2})$ and $\mathrm{Hom}_{k-gr}(G_{2}, G_{1})$ are nontrivial. Then we have \beginequation* \i\mid a_i\not\in k^p equiv\i\mid b_i\not\in k^p. \endequation* \endlem Now, we proceed to prove the proposition. Let \beginequation* \alpha=1+a_1F+\cdots+a_rF^r, \beta=1+b_1F+\cdots+b_sF^s. \endequation* Then we have $G_{1}\simeq D(M(m,\alpha)), G_{2}\simeq D(M(n,\beta))$, where $(m,\alpha), (n,\beta)$ are admissible pairs and by assumption, $\mathrm{Hom}_{k[F]}(M(m,\alpha),M(n,\beta))$ and $\mathrm{Hom}_{k[F]}(M(n,\beta), M(m,\alpha))$ are nontrivial. Assume that $n>m$. Then by [KMT?], Theorem 5.3.1, we have $\mathrm{Hom}_{k[F]}(M(m,\alpha),M(n,\beta))=0$. So $n\leq m$. Similarly, $m\leq n$ and we get $n=m$. Let $G_{1}^\prime=G_{1}^{(p^{n-1})}$ and $G_{2}^\prime=G_{2}^{(p^{n-1})}$. Then from [KMT?], Proposition 5.4.1 it follows that $\mathrm{Hom}_{k-gr}(G^\prime_{1}, G^\prime_{2})$ and $\mathrm{Hom}_{k-gr}(G^\prime_{2}, G^\prime_{1})$ are nontrivial. Let $r^\prime=\max \{i\mid a_{i}\not\in k^{p}\},s^\prime= \max\{j: b_{j}\not\in k^{p}\}.$ Thus we have $G_{1}^\prime\simeq D(M(1,\alpha^\prime)), G_{2}^\prime\simeq D(M(1,\beta^\prime))$, where $\alpha=1+a_{1}F+\cdots+a_{r^\prime}F^{r^\prime}$, $\beta=1+b_{1}F+\cdots+b_{s^\prime}F^{s^\prime}$. By Lemma 11, we get $\{i\mid a_{i}\not\in k^{p}\}\equiv\{i\mid b_{i}\not\in k^{p}\}$ and Proposition 10 follows \endproof \begincor Let $k$ be a non-perfect field of characteristic $p>0$, and let $G$ be an unipotent $k$-group of dimension 1. If there are k-isomorphisms of G with the following k-groups $\{(x,y)\in \mathbf{G}_{a}^{2} | y^{p^{m}} =x+a_{1}x^{p}+\cdots+a_{r}x^{p^{r}}, \exists i, a_{i}\not\in k^{p}$\} and $\{(x,y)\in \mathbf{G}_{a}^{2} | y^{p^{n}} = x+b_{1}x^{p}+\cdots+b_{s}x^{p^{s}}, \exists j, b_{j}\not\in k^{p}\}$. then $m=n$ and $\{i\mid a_{i}\not\in k^{p}\}\equiv\{i\mid b_{i}\not\in k^{p}\}$. \endcor \section*Acknowledgements. We would like to thank Prof. S. Mori and the referee for pointing out some innacuracies in the preliminary version of the paper and very helpful suggestions. We thank Prof. Lê D. T., Mathematics Section, the Abdus Salam I. C. T. P. and S. I. D. A. and the second author also thanks Max Plank Institut für Mathematik, Germany, for the hospitality and support, which help us to realize this work. This paper contains some results presented in the talk by the second author given in a Colloqium talk in Eichstaett University and in a Esnault - Viehweg seminar talk in Essen University. Thanks are due also to Prof. J. Rohlfs, S. Schroerer and E. Viehweg for remarks, questions, and discussion related to results presented here. \beginthebibliography99
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