Abstract Algebra Question Polynomials? 10 points for informative answer.

Hello Experts,

If I have p(x) = x^4 + 1 and I need to break it in C,R,Q and Z_3:

I have already found for C it's just to find 4 complex roots. But I am stuck in R,Q and Z_3.

What I am trying to do is to find prime ideal of one of the fields and use Eisenstein criterion but what are the ideals of fields except 0?
I know there is a Gauss Lemma too but again I can't proceed.

I will appreciate any help. Thanks in advance.

For Z₃, note that none of x = 0, 1, 2 (mod 3) are roots.

Since this is quartic, there remains the possibility that
x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d) for some a, b, c, d in Z₃.
==> x^4 + 1 = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.

Comparing cubic terms, we need a + c = 0 (mod 3) ==> c = -a (mod 3)
==> x^4 + 1 = x^4 + (b + d - a^2)x^2 + a(d - b)x + bd.

Comparing the constant terms, we have bd = 1 (mod 3)
==> b = d = 1 or 2 (mod 3).

Letting b = d yields
x^4 + 1 = x^4 + (2b - a^2)x^2 + 1.

If b = d = 1 (mod 3), then 2 - a^2 = 0 (mod 3).
However, the squares mod 3 are 0, 1; so this is not a valid possibility.

If b = d = 2 (mod 3), then 1 - a^2 = 0 (mod 3) ==> a = -1, 1 mod 3.

Letting a = 1, b = d = 2, and c = -1 = 2 (mod 3) yields
x^4 + 1 = (x^2 + x + 2)(x^2 + 2x + 2) (mod 3).
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For Q, note that letting x = y+1 yields
p(y+1) = (y+1)^4 + 1 = y^4 + 4y^3 + 6y^2 + 4y + 2.
(Note that p(x) is irreducible <==> p(y+1) is irreducible.)

However, p(y+1) is irreducible over Q by applying Eisenstein's Irreducibility Criterion
using the prime p = 2.

Hence, p(x) is irreducible over Q.
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Finally for R, note that this has no linear factors, because its roots are not real.

However,
x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2
............= (x^2 + 1)^2 - (x√2)^2
............= (x^2 + x√2 + 1)(x^2 - x√2 + 1), and we are done.