I discovered an elegant theorem. But it is easy to prove using some heavy altirary.
Theorem: Let be a polynomial with rational coefficients having a prime degree . If is irreducible (cannot be factored in terms of other non-constant polynomials having rational coefficients) then has exactly complex zeros.
Yes, exactly.
We can state this theorem in a different way (equivalent):
Let be a polynomial with integer coefficients and prime degree so that is irreducible (cannot be factord into non-constant polynomials with integer coefficients) then has exactly zeros.
Why?
In topsquark's example, that polynomial satisfies my conditions. Now, as promised, it has exactly 3 zeros (1 real 2 not). But, yes, we need that condition that they can be complex too.
Note, it should not be supprising because in field theory we usually deal in one big extension field, usually the algebraic closure. Which in this case is the closure of which are the algebraic numbers.