View Single Post 2017-02-16, 01:18 #5 Dr Sardonicus   Feb 2017 Nowhere 2×1,901 Posts The property I am getting at is the n-1 degree polynomial has factors either 0 or 1 (mod n) when the polynomial is evaluated for any "x" value. What is important here is the number field, not the particular polynomial used to define it. By rescaling the variable, you can also sneak in finitely many primes which violate the property. For example, x^2 + 5*x + 25 has D = -3 * 5^2 so defines the same quadratic field as x^2 + x + 1, but assumes values divisible by 5 (in fact by 25) whenever x is an integer divisible by 5. Similarly, x^2 + 7*x + 49 has the requisite property, but discriminant D = -3 * 7^2, which has a prime factor other than 3. Similarly, the quartic polynomial x^4 + 5*x + 5 has D = 5^3 * 11^2, but defines the same field as the cyclotomic polynomial x^4 + x^3 + x^2 + x + 1, and has the requisite property above. [In this case, I chose the "extraneous" factor 11 to be congruent to 1 (mod 5).] So it is prudent to allow for finitely many exceptional primes in the above property [also for finitely many "extraneous" square factors in the discriminant]. This allows rescaling to produce a monic polynomial [lead coefficient 1] defining the same field as any given (nonzero, nonlinear, irreducible) polynomial in Z[x]. Calling the polynomial f(x), and assuming it to be irreducible, monic and in Z[x], and using p instead of n (since you want n to be a prime), one interpretation of the above property is as follows: With at most finitely many exceptions, if q is a prime not congruent either to 1 or 0 (mod p), then the reduction f(x) (mod q) has no linear factors. For if q is a prime dividing f(r), r a rational integer, then (x - Mod(r, q)) is a linear factor of the reduction of f(x) (mod q). Alternatively (with possibly finitely many exceptions), if f(x) (mod q) has a linear factor, then q is congruent to 1 (mod p). Now the cyclotomic field K of p-th roots of unity is characterized by the fact that the primes q which split into p-1 distinct prime ideal factors in K/Q are precisely those which are congruent to 1 (mod p). This property extends to the cyclotomic field of m-th roots of unity, for any m > 2. The degree over Q is phi(m), which is less than m - 1 when m is composite. Fortuitously, cyclotomic polynomials have discriminant equal to the discriminant of the number field they define -- no extraneous prime factors. Number fields whose (field) discriminants only have one (finite) prime factor are unusal. BTW, Stickelberger's criterion says that the discriminant of any monic polynomial in Z[x] is congruent either to 1 or 0 (mod 4). Thus, polynomial discriminants equal to an odd power of any prime congruent to 3 (mod 4) are impossible. If p is a prime congruent to 3 (mod 4), the discriminant of the cyclotomic polynomial for the primitive p-th roots of unity is D = -p^(p - 2).  