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Prove that there exists an irreducible polynomial of degree 10 over the the field of 25 elements.

I know that the multiplicative group of non-zero elements of any finite field is cyclic. So how can I proceed my work? Thanks!

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    $\begingroup$ Conceptually? We know that there exists a field with $\;5^{20}\;$ elements, namely $\;\Bbb F_{5^{20}}\;$ , and since $\;2\mid 20\;$ we get that $\;\Bbb F_{5^2}\;$ is a subfield of it, and in fact $\;\dim_{\Bbb F_{5^2}}\Bbb F_{5^{20}}=10\;$. But also $\;\Bbb F_{5^{20}}\cong\Bbb F_{5^2}[x]/\langle f(x)\rangle\;$ , with $\;f(x)\;$ irreducible of degree $\;10\;$ ... $\endgroup$ – DonAntonio Apr 10 '16 at 9:42
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    $\begingroup$ @Joanpemo: Proving the existence of this field doesn't require proving there exists an irreducible polynomial of degree $10$? $\endgroup$ – Bernard Apr 10 '16 at 10:31
  • $\begingroup$ @Bernard but the existence of the field $\mathbb{F}_{p^n}$ can be shown alternatively as being the splitting field of $f(t)=t^{p^{n}}-t$ over $\mathbb{F}_p$, right? That would avoid the circular argument. $\endgroup$ – user281593 Apr 10 '16 at 11:03
  • $\begingroup$ @Bernard In fact it requires the existence of an irreducible pol. of degree $\;20\;$ in $\;\Bbb F_5[x]\;$ . It can also be done with the help of some little stuff from Fields extensions and Galois theory. $\endgroup$ – DonAntonio Apr 10 '16 at 11:04
  • $\begingroup$ AH...exactly what @user314159 just wrote a few second ago! $\endgroup$ – DonAntonio Apr 10 '16 at 11:04

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