# oct26 - f has no root so the only possibility is that f...

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Math 5125 Wednesday, October 26 October 26, Ungraded Homework Exercise 9.4.1 on page 311 Determine whether the following polynomials are irreducible in the rings indicated. For those that are reducible, determine their factorization into irre- ducibles. The notation F p denotes the ﬁnite ﬁeld Z / p Z , p a prime. (a) x 2 + x + 1 in F 2 [ x ] (b) x 3 + x + 1 in F 3 [ x ] (c) x 4 + 1 in F 5 [ x ] (d) x 4 + 10 x 2 + 1 in Z [ x ] In each case, let f ( x ) indicate the relevant polynomial. (a) Since f ( 0 ) = f ( 1 ) = 1 6 = 0, we see that f has no roots in F 2 . This means that f has no degree 1 factor. If f is not irreducible, then f must be the product of two degree 1 polynomials, and we have just shown that this is not the case. Therefore f is irreducible. (b) Here f ( 1 ) = 0, so x + 2 is a factor and we have f ( x ) = ( x + 2 )( x 2 + x + 2 ) . Since x 2 + x + 2 has no root in F 3 , we see that x 2 + x + 2 is irreducible and therefore ( x + 2 )( x 2 + x + 2 ) is the factorization of f into irreducibles. (c) Since f ( 0 ) = 1 and f ( 1 ) = f ( 2 ) = f ( 3 ) = f ( 4 ) = 1, we see that
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Unformatted text preview: f has no root, so the only possibility is that f factors into two degree 2 irreducible factors. By inspection (or apply the root theorem to y 2 + 1) f = ( x 2 + 2 )( x 2 + 3 ) , and x 2 + 2 and x 2 + 3 are irreducible. (d) If f has a degree 1 factor, then it must be of the form x + a where a = ± 1, which is not the case. Therefore either f is irreducible or it is a product of two irreducible degree 2 factors. If it is the latter, then we must have x 4 + 10 x 2 + 1 = ( x 2 + ax + d )( x 2 + bx + d ) where a , b ∈ Z and d = ± 1. The coefﬁcient of x 3 shows that a + b = 0, and then by con-sidering the coefﬁcient of x 2 , we ﬁnd that 2 d-a 2 = 10, which is not possible. Therefore x 4 + 10 x 2 + 1 is irreducible over Z ....
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