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

[billybob12345]

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.

 

[adriank]

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).

 

[Office_Shredder]

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 ⁴C_k (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

 

[billybob12345]

Thanks for the help. greatly appreciate it.