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 Atiyah–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