Abstract Algebra Question Polynomials? 10 points for informative answer.

As you have found the factorization for p(x) in C,
p(x)=(x-α)(x-(α)) ̅(x-β)(x-(β)) ̅, where α=(1+i)/√2,β=(-1+i)/√2.
In R[x],
p(x)
=[x^2-(α+(α)) ̅x+α(α)) ̅] [x^2-(β+(β)) ̅x+β(β)) ̅]
=(x^2-√2 x+1) (x^2+√2 x+1).

In Q[x], if p(x) is irreducible in Z[x], then so is in Q[x].
It is obvious that p(x) is irreducible in Z[x]. So p(x) is irreducible in Q[x].
Or, if you want to use the Eisenstein criterion, let us consider p(x-1).
It is expanded as follows:
x^4-4x^3+6x^2-4x+2. Then prime 2 divides all the coefficients of x^i (0≤i≤3)
and 2^2 does not divide the constant term 2.
By the Eisenstein criterion, p(x-1) is irreducible over Q.
So p(x) is also irreducible over Q.

In the end, let us consider Z3[x] case.
Case 1: p(x) has a linear factor
If p(x) has a linear factor x-e, then we must have f(e)≡0 (mod 3).
However, p(0)≡0,p(1)≡2,p(-1)≡3 (mod 3),
which shows that p(x) has no linear factors.
Case 2: p(x) is decomposed into two quadratic factors in Z3[x]
Suppose p(x)=(x^2+ax+b)(x^2+cx+d) in Z3[x].
The RHS is expanded as
x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd.
Equating with x^4+1, we have
a≡2,b≡2,c≡1,d≡2 (mod 3).
So, p(x)=(x^2+2x+2)(x^2+x+2) in Z3[x].

There are no proper ideals in a field.
Let I be any nonzero ideal in a field F.
and let x be in I. Then x*x^-1 is in I
so 1 is in I and I = F.
Let's factor p(x) in R first.
Write x^4 + 1 = x^4 + 2x^2 + 1 - 2x^2 =
(x^2 + √2x + 1)(x^2-√2x + 1).
To find the factors in C, set each
of the quadratic factors above to 0 and solve.
To see p is irreducible over Q use the factor
theorem. Any possible rational root must be
1 or -1. But neither is a solution of p(x) = 0.
That leaves the problem of finding the factors in Z_3.
I'll let you show that the factors must look like
(x^2+ax+b)(x^2-ax+b).
Now equate coefficients on both sides
to find a and b(mod 3).