Is (x^4)+1 Reducible over Z5 and Z?

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In summary, the conversation discusses the reducibility of the polynomial [(x^4)+1] over Z5 and Z. The speaker has attempted to solve it by comparing coefficients and using Eisenstein's Criterion, but is still unable to find a solution. They seek further assistance and express their gratitude for the help.
  • #1
billybob12345
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I am trying to figure out if the polynomial [(x^4)+1] is reducible over Z5 and also over Z.

For Z5, i tried:
f(0) = 1
F(1) = 2 = f(2) = f(3) = f(4)
Since neither are zero, i tried
f(x) = (ax^2 + bx + c)(dx^2 + ex + f)

I compared the coefficients but am unable to solve it.

For Z, i have no idea how to do it.

Please help! Thanks.
 
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  • #2
Say a = d = 1, so suppose x4 + 1 = (x2 + bx + c)(x2 + ex + f). Comparing coefficients gives:

(x3) 0 = b + e
(x2) 0 = c + f + be
(x1) 0 = bf + ce
(x0) 1 = cf.

Fiddling around should give either a contradiction somewhere, or some factorization. Now 0 = b + e, so b = -e. Thus the x2 part gives 0 = c + f - b2, so b2 = c + f.

Working in Z, cf = 1, so c = f = ±1 and c + f = ±2. But this is a contradiction, since b2 = c + f and c + f is not a perfect square. Many irreducibility proofs follow a similar pattern.

In Z5, since cf = 1, c + f is 0, 2, or 3 (by looking at each possible pair of c, f). But c + f = b2, so c + f = 0 (since 2 and 3 are not perfect squares), so since cf = 1 as well, you have c, f = 2, 3 in some order. Also, b = -e = 0 since c + f = 0 = b2. You can check that x4 = (x2 + 2)(x2 + 3).
 
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  • #3
I'm not sure what theorems you have for Z, but you can use Eisenstein's Criterion for this (which works more generally also) by noting:
(1)f(x) is irreducible if and only if f(t+a) is irreducible for any constant a (so you're changing variables to t by the transformation x=t+a)... note that f(x) = g(x)h(x) if and only if f(t+a) = g(t+a)h(t+a)

(2) Substitute t+1 for x in the original polynomial

(3) note that 4Ck (4 choose k) is divisible by 2 for all k not 0 or 4. Hence the coefficients in the new polynomial satisfy Eisenstein's Criterion
 
  • #4
Thanks for the help. greatly appreciate it.
 

What is an irreducible polynomial over Z?

An irreducible polynomial over Z is a polynomial with integer coefficients that cannot be factored into polynomials with smaller degrees and integer coefficients.

What is the difference between reducible and irreducible polynomials over Z?

A reducible polynomial over Z can be factored into polynomials with smaller degrees and integer coefficients, while an irreducible polynomial over Z cannot.

How can I determine if a polynomial over Z is irreducible?

There are a few methods for determining if a polynomial over Z is irreducible. One method is to check if the polynomial has any integer roots. If it does not, then it may be irreducible. Another method is to use the Eisenstein's criterion, which states that if a prime number divides all coefficients except the leading coefficient of the polynomial, then the polynomial is irreducible.

Can an irreducible polynomial over Z have complex roots?

No, an irreducible polynomial over Z can only have real roots. This is because the coefficients of the polynomial are integers, and complex roots would require the coefficients to be complex numbers as well.

Why are irreducible polynomials over Z important in mathematics?

Irreducible polynomials over Z are important because they provide a way to construct fields, which are mathematical structures that are crucial in many branches of mathematics. Additionally, they are used in various areas of number theory, algebra, and cryptography.

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