Commutative algebra - Department of Mathematical Sciences - old ...

8.4. POWER SERIES RINGS 97
8.3.11. Exercise. (1) Show that if R[X] is noetherian, then R is noetherian.
(2) Show that the subring Z[2X, 2X 2 , . . . ] ⊂ Z[X] is not noetherian. Conclude that the
extension is not finite.
8.4. Power series rings
8.4.1. Proposition. Let R be a noetherian ring. Then the power series ring R[[X]]
is noetherian.
Proof. Let P ⊂ R[[X]] be a prime ideal. Then P +(X)/(X) = (a1, . . . , an) ⊂ R
is a finite ideal. If X ∈ P then P = (a1, . . . , an, X) is finite. Suppose X /∈ P
and choose fi = ai + terms of positive degree ∈ P . If g ∈ P then Xg1 =
g − b11f1 + · · · + bn1fn ∈ P ∩ (X) for some bi1 ∈ R. Since P is prime, g1 ∈ P .
Now Xg2 = g1 − b12f1 + · · · + bn2fn ∈ P ∩ (X) and so on. Put hi = bikX k ,
then g = hifi. P is finite and R[[X]] is noetherian by 8.2.12.
8.4.2. Proposition. Let R be a principal ideal domain. Then the power series ring
R[[X]] is a unique factorization domain.
Proof. Let P ⊂ R[[X]] be a minimal nonzero prime ideal. Then P + (X)/(X) =
(a) ⊂ R Suppose P = (X) and choose f = a + terms of positive degree ∈ P .
If g ∈ P then Xg1 = g − b1f ∈ P ∩ (X) for some b1 ∈ R. Since P is prime,
g1 ∈ P . Now Xg2 = g1 − b2f ∈ P ∩ (X) and so on. Put h = bkX k , then
g = hf. P = (f) is principal. The conditions of 1.5.3 are satisfied since R[[X]] is
noetherian 8.4.1.
8.4.3. Proposition. Let I ⊂ R be an ideal in a noetherian ring. Then there is a
canonical isomorphism
R[[X]]/IR[[X]] R/I[[X]]
Proof. The projection R[[X]] → R[[X]]/IR[[X]] factors over R/I[[X]] since I is
finitely generated. Then there is an inverse to the homomorphism 1.9.8.
8.4.4. Corollary. If P ⊂ R is a prime ideal in a noetherian ring, then P R[[X]] ⊂
R[[X]] is a prime ideal.
8.4.5. Proposition. Let R be a noetherian ring.
(1) The inclusion R[X] ⊂ R[[X]] is a flat homomorphism.
(2) The inclusion R ⊂ R[[X]] is a faithfully flat homomorphism.
Proof. (1) Let I ⊂ R[X] be an ideal and let ai ⊗ fi ∈ K = Ker(I ⊗R[X] R[[X]] → R[[X]]. Then aifi = 0. Write fi = gi + Xnhi and get ai ⊗ fi =
Xn (ai ⊗ hi). It follows that K ⊂
n Xn (I ⊗R[X] R[[X]]). I ⊗R[X] R[[X]]
is a finite R[[X]]-module and 1 + aX is a unit, so conclusion by 8.3.8. (2) The
homomorphism is flat by (1). For any maximal ideal P ⊂ R the homomorphism
RP → RP [[X]] is local and flat.
8.4.6. Exercise. (1) Show that if R[[X]] is noetherian, then R is noetherian.