Let be a prime and let be the th cyclotomic polynomial. (Remember that is irreducible over .) Let be a prime and let denote any fixed primitive th root of unity.
- Show that if then .
- Suppose and let denote the order of in . (That is, is minimal such that mod .) Show that is minimal such that . Conclude that the minimal polynomial of over has degree .
- Show that for any integer not divisible by . Conclude that, in , is the product of distinct irreducible polynomials of degree .
- As an example, find the degrees of the irreducible factors of over .
- Mod , we have . So as desired.
- Note that precisely when is a root of . Since mod , we have that divides . By this previous exercise, divides . Since is a th root of unity, . Conversely, suppose . Now , so divides by Lagrange’s theorem in the group . Thus mod , and so divides . So is minimal such that .
Next we claim that . The inclusion is clear since . for some , since finite fields are essentially unique. By the preceding argument, , so using this previous exercise.
- It is clear that . Conversely, since does not divide a$, by Bezout’s identity there exist such that . Now . So provided does not divides .
Now consider as a polynomial over . Let be the distinct primitive th roots of unity. By part (b), has degree over for each , so the minimal polynomial of over has degree . So the irreducible factors of have degree , and there are such factors. Moreover, since is separable, its factors are distinct.
- If mod 7, then by part (a). Evidently 2 and 4 have order 3 mod 7, so if mod 7 then factors into two irreducible cubics. Similarly, 3 and 5 have order 6 mod 7, so that if mod 7 then is irreducible. Since 6 has order 2 mod 7, if mod 7, then factors into three irreducible quadratics.