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Séminaire de Théorie des Nombres

Splitting Subspaces, Singer Cycles and Linear Recurrences

Sudhir Ghorpade

( Mumbai )

Salle de Conférences

11 mai 2012 à 14:00

\def\FF{{\mathbb F}} \def\PP{{\mathbb P}} \def\Fp{{\mathbb F}_p} \def\Fq{{\mathbb F}_q} \def\Fqm{{\mathbb F}_{q^m}} \def\Fqmn{{\mathbb F}_{q^{mn}}} \def\Fqtn{{\mathbb F}_{q^{2n}}} \def\mod{{\rm mod}} Finite fields have a remarkable property that finite dimensional vector spaces over them are naturally endowed with a compatible field structure. Indeed, we can simply ``move the dd'' so as to write \Fqd\FFqd\Fq^d \simeq \FF_{q^d}, where dd is any positive integer and, as usual, \Fq\Fq denotes the finite field with qq elements. This leads to some interesting notions where the field structure and the linear structure are intertwined. One such notion is that of a splitting subspace, which appears to go back at least to Niederreiter (1995) in connection with his work on pseudorandom number generation. Here is the definition. %{\bf Definition.} Fix positive integers m,nm,n and a prime power qq. Let α\Fqmn\alpha \in \Fqmn. % be a primitive element of \Fqmn\Fqmn in the sense that \Fqmn=\Fq(α)\Fqmn=\Fq(\alpha). An mm-dimensional \Fq\Fq-linear subspace WW of \Fqmn\Fqmn is said to be \emph{α\alpha-splitting} if \begin{displaymath} \Fqmn = W \oplus \alpha W \oplus \cdots \oplus \alpha^{n-1}W. \end{displaymath} Concerning splitting subspaces, Niederreiter asked the following \noindent {\bf Question.} Given %any α\Fqmn\alpha\in \Fqmn such that \Fqmn=\Fq(α)\Fqmn=\Fq(\alpha), what is the number of mm-dimensional α\alpha-splitting subspaces of \Fqmn\Fqmn? This question has been open for over 15 years. We will outline some recent progress as well as connections to topics such as Singer cycles (in general linear groups), linear recurrences, and primitive polynomials. %, and cryptography. En route, we will also notice an amusing connection with the Riemann zeta function and questions such as when are two polynomials in \Fq[X]\Fq[X] of a given positive degree relatively prime.