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

10.3. DEDEKIND DOMAINS 113
10.2.4. Corollary. Let (R, (p)) be a discrete valuation ring. Any nonzero element
in the fraction field K of R has a unique representation up n where u is a unit and
n ∈ Z.
10.2.5. Definition. Let K be a field. A surjective map v : K\{0} → Z satisfying
(1) v(xy) = v(x) + v(x).
(2) If x + y = 0, v(x + y) ≥ min(v(x), v(y).
is a valuation on K.
10.2.6. Proposition. Let v be a valuation on a field K. Then
R = {x|v(x) ≥ 0}
is the discrete valuation ring of v. A local parameter is any element p, v(p) = 1.
Proof. Clear from the definition.
10.2.7. Proposition. Let (R, (p)) be a discrete valuation ring with fraction field
K. The map
v : K\{0} → Z, up n ↦→ n
is a valuation and R is the discrete valuation ring of v.
Proof. Clear from the definitions.
10.2.8. Proposition. Let R, (p) be a discrete valuation ring. A finite module M
has decomposition
Where n1 ≥ · · · ≥ nk > 0.
M = R/(p n1 ) ⊕ · · · ⊕ R/(p nk ) ⊕ R n
Proof. This is a case of 10.1.7 and 10.1.9.
10.2.9. Exercise. (1) Let K be a field. Show that the subring K[[X 2 , X 3 ]] ⊂ K[[X]]
is not a discrete valuation ring.
10.3. Dedekind domains
10.3.1. Definition. A noetherian domain R, which is not a field, is a Dedekind
domain if all local rings RP at nonzero prime ideals are discrete valuation rings.
10.3.2. Proposition. Let R be a Dedekind domain.
(1) Any nonzero prime ideal is maximal.
(2) If U ⊂ R is multiplicative, then U −1 R is a field or a Dedekind domain.
Proof. (1) (0), P are the only prime ideals in a prime ideal P . (2) This is clear
from 5.2.11.
10.3.3. Proposition. Let R be a domain which is not a field. The following conditions
are equivalent.
(1) R is a Dedekind domain.
(2) Every nonzero proper ideal in R is a product of finitely many maximal ideals.
Proof. The primary ideals P (n) = P n . Conclusion by 9.5.4 and Chinese remainders
1.4.2.
10.3.4. Proposition. Let R be Dedekind domain.
(1) If R is a unique factorization domain then it is a principal ideal domain.