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§2. Factorial Properties of Polynomial Rings 59

Observe that the units R in R v is the group of all x with v(x) = 0 and the
∗
v
nonunits form the unique maximal ideal M v which equals the set of all x in F with
v(x)> 0. The residue class field k(v) is the quotient ring R v /M v . If the valuation
is trivial, that is, if v(F ) = 0, then R v = F and M v = 0. Otherwise R v = F and
∗
∗
M v = 0. If the value group is v(F ) = mZ, then there exists an element t in R v
with v(t) = m. Then it follows that M v = R v t and every x in F canberepresented
∗
r
as x = t u,where u is in R and r is the integer where v(x) = rm. Note that v is
∗
v
m · ord t .
∗
∗
∗
If k is a subfield of F with v(k ) = 0, then k ⊂ R and the restriction of the
v
residue class morphism R v → k(v) to k → k(v) is a monomorphism of fields. In
this case we say that the valuation v is trivial on k.
§2. Factorial Properties of Polynomial Rings
Valuations are useful for discussing factorization properties of polynomial rings. Let
F be a field with a valuation v and define v on F[x 1 ,... , x n ] by the relation
+

i 1
v + x ... x i n .
n = min v a i 1 ...i n
a i,...i n 1
(2.1) Proposition (Gauss’s Lemma). Let F be a field with a valuation v. For f, g
in F[x 1 ,... , x n ] we have
+
+
+
v ( fg) = v ( f ) + v (g).
Proof. Consider the case of one variable x.Since v (cf ) = v(c) + v ( f ), we can
+
+
+ + m
assume that v ( f ) = v (g) = 0sothat f, g are in R v [x]. If f (x) = a m x +· · ·+a 0
n
and g(x) = b n x + ··· + b 0 , then there exists indices i and j where a i and b j are
units and a p and b q are not units for p < i and q < j. From this we see that the
coefficient of x i+ j in f (x)g(x) is a unit since v(a) = v(a + b) if v(a)<v(b). Thus
+
we have v ( fg) = 0, and this proves the formula in the case of one variable.
For n variables choose d > deg( fg). Then substitute x i = x d i−1 for all i. This
maps
n−1
d
f (x 1 ,... , x n ) → f (x) = f x, x ,... , x d
∗
such that v ( f ) = v ( f ).Inthisway the n variable case is reduced to the 1
+
+
∗
variable result which proves the proposition.
(2.2) Theorem. If R is a factorial ring, then R[x 1 ,... , x n ] is a factorial ring.
Proof. Since R[x 1 ,... , x n ] = R[x 1 ,... , x n−1 ][x n ], it suffices, by induction on n,
to prove the theorem for n = 1and x 1 = x.If F is the field of fractions of R, then
F[x] is a factorial ring since F[x] is a principal ring. Thus every f in R[x] ⊂ F[x]is
a product f = f 1 ... f r , where the f i are irreducibles in F[x]. For each irreducible
p in R,wehave