Factor x^{4}-2 into irreducible polynomials.

widdonod1t

widdonod1t

Answered question

2022-01-13

Factor x42 into irreducible polynomials.

Answer & Explanation

alexandrebaud43

alexandrebaud43

Beginner2022-01-14Added 36 answers

Step 1
You had a wrong sign in the factorization in R[x] (typo):
(x2+2)(x24)(x24)
is the factorization: the quadratic factor has no roots in R, so it's irreducible.
The polynomial is indeed irreducible in Q[x]: it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in R[x] and the only way you can combine the previous factorization in the required way would be (x2+2)(x22) and the factors are not in Q[x]. Or you can use Eisenstein's criterion, which applies to x42 for the ' 2.
Sorry, but the factorization in C[x] is wrong: it should be a product of linear factors and you get them from the factorization in R[x]
I'll do the case Z3[x]. The polynomial has no roots in Z3, but it could be a product of two (monic) quadratic factors:
x42=(x2+ax+b)(x2+cx+d)
Since bd=2=1, you can have b=1 and d=1 or b=2 and d=2. In the first case we have.
(x2+ax+1)(x2+cx+1)=x4+(a+c)x3
+(2+ac)X2+(a+c)x+1
and so you need a+c=0 and ac+2=0. Hence c=a and a2=2, which is impossible.
In the second case we have
(x2+ax+2)(x2+cx+2)=x4+(a+c)x3+(1+ac)x2+(2a+2c)x+1
Again a+c=0 and 1+ac=0. Since c=a, we obtain a2=1. Well, this is indeed possible: we have a=1 or a=2. Thus the factorization is
(x2+x+2)(x2+2x+2)

Jeffery Autrey

Jeffery Autrey

Beginner2022-01-15Added 35 answers

Step 1
The Z17 case remains. The units of Z17 form a cyclic group of order 16. The order of 2 is 8, as 24=1. Thus 2Z17, and your polynomial factorises as:
x42=(x2+2)(x22)
It cannot factorise further as if there were a linear factor, then Z17 would contain an element α with α4=2, so α16=1, which is impossible as for any element αZ17 we have α16=1.
To complete the question you should work out 2 in Z17. The easiest way to do this is check values of 2+17k, for k=0,1,2,3, till you get a perfect square.

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