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

64 4. FRACTION CONSTRUCTIONS
4.6. The polynomial ring is factorial
4.6.1. Definition. Let R be a unique factorization domain and let f = anX n +
· · · + a0 be a polynomial over R. Then the content of polynomial f, c(f), is the
greatest common divisor of the coefficients a0, . . . , an.
4.6.2. Proposition (Gauss’ lemma). Let R be a unique factorization domain. For
polynomials f, g ∈ R[X]
c(fg) = c(f)c(g)
Proof. Assume by cancellation that c(f), c(g) are units in R. For any irreducible
p ∈ R the projections of f, g in R/(p)[X] are nonzero. Since R has unique factorization
the ideal (p) is a prime ideal. It follows that the projection of the product
fg in R/(p)[X] is also nonzero and therefore p is not a common divisor of the
coefficients of the product fg.
4.6.3. Proposition. Let R be a unique factorization domain. Then the ring of
polynomials R[X] is a unique factorization domain.
Proof. Let K be the fraction field of R, then the polynomial ring K[X] is a principal
ideal domain. Let f ∈ R[X] and use unique factorization in K[X] to get
0 = a ∈ R and p1, . . . , pn ∈ R[X], irreducible in K[X], such that
af = p1 . . . pn
Assume by 4.6.2 that a = 1 and c(p1), . . . , c(pn) are units in R. Apply 4.6.2 and
1.6.5 to see that p1, . . . , pn are irreducible in R[X]. An irreducible p ∈ R generates
a prime ideal (p) ⊂ R[X]. A non constant irreducible p ∈ R[X] generates a prime
ideal (p) ⊂ K[X] and therefore also a prime ideal (p) ⊂ R[X]. So conditions
1.5.3 are satisfied.
4.6.4. Theorem. Let K be a field. Then the polynomial ring K[X1, . . . , Xn] is a
unique factorization domain.
Proof. Follows by induction from 4.6.3.
4.6.5. Exercise. (1) Let f ∈ Z[X] be monic and assume f = gh where g, h ∈ Q[X]
are monic. Show that g, h ∈ Z[X].
(2) Let f ∈ Z[X] be monic and irreducible in Z/(n)[X]. Show that f is irreducible
Q[X].
(3) Let K be a field. Show that the polynomial ring K[X1, X2, . . . ] in countable many
variables is a unique factorization domain