Would there be an easy way to generate polynomials with number fields the same as the cyclotomic polynomial of degree 4 or more generally n1?
Yes  given the right software [e.g. PariGP], and understanding what it does.
Your polynomial,
f=x^4+7*x^3x^27*x+1 is irreducible with [according to PariGP] Galois group D4. I also used Pari to find a "reduced" polynomial defining the same field,
g = x^4  2*x^3  6*x^2 + 7*x + 11 with D = 5^3 * 41
The polynomial f (mod 3) has a repeated quadratic factor because of the "extraneous" factor 3^4 in the discriminant. The polynomial g (mod 3) splits into 2 distinct quadratic factors.
The splitting field contains the quadratic fields defined by t^2  t  1, t^2  t  10, and t^2  t  51 [the last being a defining polynomial for Q(sqrt(D))]
Both f and g split into quadratic factors in the field defined by t^2  t  1; g factors as
(x^2  x + (t  3)) * (x^2  x + (t  4)) where t^2  t  1 = 0.
This means that f and g (mod p) will factor into two quadratic factors whenever 5 is a quadratic residue (mod p), i.e. p == 1 or 1 (mod 5).
However, whether either quadratic factor splits further (mod p) is a more complicated proposition; it cannot be described by rational integer congruences (this is because D4 is not an Abelian group). For example, one of the quadratic factors (mod 11) splits into linear factors; both quadratic factors (mod 31) remain irreducible; both quadratic factors (mod 131) split into linear factors.
If p is congruent to 2 or 3 (mod 5), g (mod p) either remains irreducible, or splits into irreducible quadratic factors. These cases are not describable with rational integer congruences.
I decline to generate more defining polynomials for the field of 7th roots of unity.
Last fiddled with by Dr Sardonicus on 20170216 at 16:01
