Let be a commutative ring with . Let be a monic polynomial of degree (that is, the leading coefficient of is 1), and let bars denote the natural projection .
- Show that every element of is of the form for some polynomial of degree less than .
- Prove that if are distinct and have degree less than , then .
- Suppose for some of degree less than . Prove that is a zero divisor in .
- If for some nilpotent element , prove that is nilpotent in .
- Let be a prime, let , and let for some . Prove that is nilpotent in .
We begin with a lemma.
Lemma: Let be a commutative ring with . If is monic, then for all , . Proof: The inequality always holds. Write and , where and have degree less than and , respectively. Then ; since , the degree of is the sum of the degrees of and .
- Let , and write , where the degree of is less than . We proceed by induction on the degree of .
For the base case, if the degree of is less than , then we may use itself to represent . For the inductive step, suppose that for some , if is a polynomial of degree at most then there exists of degree at most such that . Let have degree . We may write , where the degree of is at most . Note that since , . Now . Note that the degree of is at most , so that the degree of is at most . Thus by the induction hypothesis, we have for some polynomial of degree less than .
- Suppose have degree less than and that . Suppose . Then for some . Note, however, that has degree less than , while has degree at least . Thus we have a contradiction.
- Suppose , where both and have degree less than . Then in , but by the previous part, neither nor is zero. Thus is a zero divisor in ..
- Suppose in . Now , so that is nilpotent in .
- Using Fermat’s Little Theorem, we have . Thus is nilpotent in .