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# irreducible polynomials over finite field

Theorem.β Over a finite field $F$, there exist irreducible polynomials of any degree.

Proof.β Let $n$ be a positive integer, $p$ be the characteristic of $F$, $\mathbb{F}_{p}$ be the prime subfield, and $p^{r}$ be the order of the field $F$.β Since $p^{r}\!-\!1$ is a divisor of $p^{{rn}}\!-\!1$, the zeros of the polynomial $X^{{p^{r}}}\!-\!X$ form inβ $G:=\mathbb{F}_{{p^{{rn}}}}$β a subfield isomorphic to $F$.β Thus, one can regard $F$ as a subfield of $G$.β Because

$[G\!:\!F]=\frac{[G\!:\!\mathbb{F}_{p}]}{[F\!:\!\mathbb{F}_{p}]}=\frac{rn}{r}=n,$ |

the minimal polynomial of a primitive element of the field extension $G/F$ is an irreducible polynomial of degree $n$ in the ring $F[X].$

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FiniteField

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## Mathematics Subject Classification

12E20*no label found*11T99

*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

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new question: A good question by Ron Castillo

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new question: A trascendental number. by Ron Castillo

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new question: young tableau and young projectors by zmth

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

Jun 13

new question: young tableau and young projectors by zmth