Pair of primitive normal elements of rational form over finite fields of characteristic 2

Himangshu Hazarika () and Dhiren Kumar Basnet ()
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Himangshu Hazarika: Tezpur University
Dhiren Kumar Basnet: Tezpur University

Indian Journal of Pure and Applied Mathematics, 2023, vol. 54, issue 1, 66-75

Abstract: Abstract An element $$\alpha \in {\mathbb {F}}_{q^m}$$ α ∈ F q m is primitive if it is a generator of $${\mathbb {F}}_{q^m}^*$$ F q m ∗ . An element $$\beta \in {\mathbb {F}}_{q^m}$$ β ∈ F q m is normal in $${\mathbb {F}}_{q^m}$$ F q m over $${\mathbb {F}}_q$$ F q if its Galois orbit is a spanning set for $${\mathbb {F}}_{q^m}$$ F q m as a vector space over $${\mathbb {F}}_q$$ F q . If an element is simultaneously primitive and normal, then the element is termed as primitive normal element. In this article, we establish a sufficient condition for the existence of a primitive normal element $$\alpha \in {\mathbb {F}}_{q^m}$$ α ∈ F q m , such that $$F(\alpha )$$ F ( α ) is also a primitive element of $${\mathbb {F}}_{q^m}$$ F q m over $${\mathbb {F}}_q$$ F q , where $$F(x)=\dfrac{ax^2+bx+c}{dx^2+ex+f} \in {\mathbb {F}}_{q^m}(x)$$ F ( x ) = a x 2 + b x + c d x 2 + e x + f ∈ F q m ( x ) (with $$a\ne 0, d \ne 0$$ a ≠ 0 , d ≠ 0 ) and $${\mathbb {F}}_{q^m}$$ F q m is a field of even characteristic. We conclude that every finite field $${\mathbb {F}}_{q^m}$$ F q m , contains such primitive normal element except for very few q and m.

Keywords: Finite field; Primitive element; Normal element; Character; 12E20; 11T23 (search for similar items in EconPapers)
Date: 2023
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DOI: 10.1007/s13226-022-00231-y

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