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In the finite field GF$(25)$, each element is of the form $e = aw+b$ where $w^2 = 2$ in GF$(5)$ and $a$ and $b$ are ${0, 1, 2, 3}$ or ${4}$, elements in GF$(5)$.

For each element $e ,\exists m$ such that $e^m = 1$ in GF$(25)$. If this $m$ is the smallest integer such that $e^m = 1$, then $e$ is a primitive $m$th root of unity. If $m = 24$ is the smallest integer $m$ such that $e^m = 1$, then $e$ is called a primitive element.

Find all primitive elements $e = aw+b$ in the field GF$(25)$ where $w^2 = 2$

If $e$ is a primitive element in GF$(25)$, in other words, $e^{24}=1$, and for every integer $0 < k < m$, $e^k≠1$, then $e$ is a $24$th root of unity in GF$(25)$.

Does a minimal polynomial of $e$ with integer coefficients exist?

Thanks for help in advance.

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J. Linne
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1 Answers1

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A finite field of order $25$ can be obtained if we consider $\Bbb Z_5[x]/\langle x^2-2\rangle$ where $x^2-2$ is irreducible in $\Bbb Z_5$.

Any element of $\Bbb Z_5[x]/\langle x^2-2\rangle$ is of the form $a+bx$ where $x^2-2=0$.

In other words any element of $\Bbb Z_5[x]/\langle x^2-2\rangle$ is of the form $a+b\sqrt 2$.

Since for a finite field $F$,$F^*=F\setminus \{0\}$ is cyclic so there are $\phi(24)=8$ elements of order $24$ in $\Bbb Z_5[x]/\langle x^2-2\rangle$.

Note that the primitive elements of $\Bbb Z_5[x]/\langle x^2-2\rangle=\Bbb Z_5(\sqrt 2)$ are

$1+\sqrt 2,1+2\sqrt 2,1+3\sqrt 2,1+4\sqrt 2,3+\sqrt 2,3+2\sqrt 2,3+3\sqrt 2,3+4\sqrt 2.$

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