### Theorem 6.6. Let R be a UFD. Then R[X] is a UFD. Proof. Suppose

```Theorem 6.6. Let R be a UFD. Then R[X] is a UFD.
Proof. Suppose that f is a nonzero, nonunit element of R[X]. Suppose first that
deg f = 0, or in other words, f ∈ R; then the only way we can factor f is as
a product of elements of R, and since R is a UFD, such a factorisation of f is
essentially unique.
So assume deg f > 0. Then f = αg where α is the content of f , and hence
(by Proposition 6.1) g is primitive in R[X]. We now consider g as an element of
Q[X] where Q is the field of fractions of R. We know Q[X] is a Euclidean domain,
and hence Q[X] is a UFD; thus we can write g = g1 · · · gk where g1 , . . . , gk are
irreducible elements of Q[X] (and hence deg gi > 0 for i = 1, . . . , k). Then by
Proposition 6.3, for each i = 1, . . . , k we have gi = βi gi∗ where βi ∈ Q and gi∗ is
primitive in R[X]. By Theorem 6.4, each gi∗ is irreducible in R[X]. Also, by Gauss’
Lemma, g1∗ · · · gk∗ is primitive in R[X]. Thus we have
g = βg1∗ · · · gk∗ , where β = β1 · · · βk .
By Proposition 6.2, we have that β is a unit in R. Thus we have f = αβg1∗ · · · gk∗ ;
if αβ is a unit then αβg1∗ is irreducible, and otherwise we can factor αβ as αβ =
γ1 · · · γt where γ1 , . . . , γt are irreducible elements of R (and hence irreducible elements of R[X]). This gives us a factorisation of f as a product of irreducible
elements of R[X].
Now suppose still that deg f > 0, and that we also have f = δ1 · · · δs h1 · · · hm
where δ1 , . . . , δs are irreducible elements of R, and h1 , . . . , hm are irreducible elements of R[X] with deg hj > 0 for j = 1, . . . , m. Thus δ1 · · · δs is the content of f ,
as is γ1 · · · γt ; since R is a UFD, we have s = t and, by reordering the δi , we have
that γi and δi are associates for i = 1, . . . , s. Hence
f = γ1 · · · γs g1∗ · · · gk∗ = wγ1 · · · γs h1 · · · hm
where w is a unit. Since R[X] is an integral domain and γ1 · · · γ2 6= 0, we have
g1∗ · · · gk∗ = wh1 · · · hm .
We have g1∗ , . . . , gk∗ , wh1 , h2 , . . . , hm irreducible (and thus primitive) in R[X], and
hence by Gauss’ Lemma, they are irreducible in Q[X] where Q is the field of fractions of R. Since Q is a field, we know Q[X] is a Euclidean domain and hence
a UFD; thus k = m, and reordering the hi , we have that for each i = 1, . . . , k,
gi∗ = ui hi where ui is a unit in Q[X]. Thus ui is a nonzero element of Q. Proposition 6.4 imples that ui a unit in R, and hence gi∗ and hi are associates in R[X] for
i = 1, . . . , k.
This shows that in R[X], factorisation into products of irreducibles is essentially
unique; hence R[X] is a UFD. 1
```