Exercise 8.6.3

Show that the polynomial f = x 4 4 x 2 + x + 1 of Example 8.6.7 is irreducible over and has four real roots.

Answers

Proof. By Gauss Lemma (Theorem A.3.2), it is sufficient to prove that f has no non trivial factorization in [ x ] . Since the reduction modulo 2 of f is f ¯ = x 4 + x + 1 is of the same degree, it is sufficient to prove that f ¯ is irreducible in 𝔽 2 [ x ] (a non trivial factorization in [ x ] would give a factorisation in 𝔽 2 [ x ] by projection). This is the case if f ¯ has no root in 𝔽 4 𝔽 2 (an irreducible factor g ¯ of f ¯ of degree 2 would give a root of g in 𝔽 2 [ x ] g 𝔽 4 ).

As any element α 𝔽 4 satisfies α 4 = α = α , f ¯ ( α ) = α 4 + α + 1 = 1 0 , so f ¯ is irreducible over 𝔽 2 . Therefore

x 4 4 x 2 + x + 1 is irreducible over

f ( 3 ) = 43 > 0 , f ( 2 ) = 1 < 0 , f ( 0 ) = 1 > 0 , f ( 1 ) = 1 < 0 , f ( 2 ) = 3 > 0 . As f is continuous, the Intermediate Value Theorem shows the existence of the roots x 1 ] 3 , 2 [ , x 2 ] 2 , 0 [ , x 3 ] 0 , 1 [ , x 4 ] 1 , 3 [ . As deg ( f ) = 4 , f has no other root, so all the roots of f are real. □

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2022-07-19 00:00
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