Here is a classic little lemma on polynomials:

**Lemma.** *Let be a commutative ring with nilradical . Then .*

That is, a polynomial is invertible if and only if is invertible and are nilpotent. This is the formulation you find in Atiyahâ€“MacDonald, exercise 1.2(i). Note that when is a domain (or more generally a reduced ring), this just says that (via the constant polynomials).

*Proof.* First suppose with and . Multiplying by we may assume ; this does not change the nilpotence assumption on . Then is nilpotent since it is a sum of nilpotent elements in the commutative ring . Thus we get for some , and

shows that . Conversely, let , and suppose for some polynomial . Without loss of generality we may assume . If is a domain, then the leading term of is , so we must have and and are constant. In the general case, we conclude that for any ring homomorphism to a domain . In particular, taking for any prime ideal shows and . Taking the intersection over all prime ideals gives