
Unique Factorization in Polynomial Rings
We have seen that if F is a field and X is an indeterminate over F, then F[X] is a UFD. Thus, we can factor polynomials in one variable with coefficients in a field into a product of irreducible polynomials in an essentially unique way. Can the same be said for polynomials in several variables? In other words, if X_{1},...,X_{n} are indeterminates over F, is F[X_{1},...,X_{n}] a UFD? The only way which we have at our disposal to prove that F[X_{1},...,X_{n}] is a UFD is to verify that F[X_{1},...,X_{n}] is a PID. However, we have already seen that for n = 2, F[X_{1},...,X_{n}] is not a PID. Therefore, the machinery developed earlier is not sensitive enough to determine whether or not F[X_{1},...,X_{n}] is a UFD. We will prove that it is, but we will require several new ideas, which are due to Gauss, who first exploited them in the early part of the nineteenth century. Our main result will be
Theorem 1: Let R be a unique factorization domain and X an indeterminate over R. Then R[X] is a unique factorization domain.
Before we begin proving Theorem 1, let us mention two easy consequences of it.
Corollary 2: Let R be a unique factorization domain and let X_{1},...,X_{n} be indeterminates over R. Then R[X_{1},...,X_{n}] is a unique factorization domain.
Proof: Induction on n. The result is true for n = 1 by Theorem 1. Assume the result for n  1 (n > 1). Then R[X_{1},...,X_{n1}] is a UFD. Therefore, by Theorem 1, R[X_{1},...,X_{n}] = R[X_{1},...,X_{n1}][X_{n}] is a UFD.
Corollary 3: Let F be a field, X_{1},...,X_{n} indeterminates over F. Then F[X_{1},...,X_{n}] is a unique factorization domain.
Proof: F is a PID since its only ideals are (0) and F = (1). Therefore, F is a UFD by Theorem 1 in the section on arithmetic in a principal ideal domain. Therefore, F[X_{1},...,X_{n}] is a UFD by Corollary 2 above.
To prove Theorem 1, it will be necessary to establish some preliminary machinery. Throughout the rest of this section, let R be a UFD. Let P be a set of irreducible elements of R such that
(1) every irreducible element of R is an associate of some element of P and
(2) no two elements of P are associates.
Such a set can be constructed as follows: Let P^{*} denote the set of all irreducible elements of R. Define an equivalence relation ~ on P^{*} by setting x ~ y if and only if x and y are associates. Then P can be gotten by choosing one element from each equivalence class of P^{*} with respect to ~. Every element r R^{x} can be written in the form
(1)
where a_{}(x) is a nonnegative integer, is a unit of R, a_{}(x) = 0 for all but a finite number of , and denotes the product over P for which a_{}(x) > 0. The decomposition (1) follows from the fact that x can be written as a product of irreducible elements and every irreducible element of R is an associate of some P. Since factorization in R is unique, the decomposition (1) is unique, up to rearrangements of the P. (Here we use the fact that no two elements of P are associates.)
Let a_{1},...,a_{n} R. Then a greatest common divisor of a_{1},...,a_{n} is an element of R such that
1. ca_{1}, ..., ca_{n}.
2. If d R such that da_{1}, ..., da_{n}, then dc.
In the case n = 2, this definition coincides with the definition of g.c.d which we gave earlier. As before, we can prove that if d and d' are two g.c.d's of a_{1},...,a_{n}, then d and d' are associates. We also showed that if at least one of a and b is nonzero, then a and b have a g.c.d when R is a PID. We can generalize this statement.
Lemma 4: Let R be a UFD, a_{1},...,a_{n} R not all 0. Then a_{1},...,a_{n} have a g.c.d.
Proof: Let
a _{i} = _{i} ^{a(ai)}, _{i} a unit of R (1 < i < n).
For each P, let a_{} be the smallest of a_{}(a_{i}),..., a_{}(a_{n}). Then since a_{}(a_{i}) = 0 for all but a finite number of , we see that a_{} = 0 for all but a finite number of , and thus we may set
c = ^{a}.
Then c is a g.c.d. of a_{1},...,a_{n}. We leave the details of this verification to the interested reader.
In an earlier section, we showed that if R is a principal ideal domain, a,b, R, irreducible, then ab implies a or b. Let us observe that this is true in any unique factorization domain.
Proposition 5: Let R be a unique factorization domain, a,b, R, irreducible. If ab, then a or b.
Proof: Since ab, there exists k R such that ab = k. Without loss of generality, we may assume that a0, 0, k0. Since R is a UFD, we can write a and b as a product of irreducible elements. Since appears in a decomposition of k into a product of irreducible elements, the uniqueness of factorization implies that is an associate of either some irreducible element dividing a or some irreducible element dividing b. Thus, either a or b.
If 1 is a g.c.d of a_{1},...,a_{n} R, then we say that a_{1},...,a_{n} are relatively prime. Moreover, if c is a g.c.d. of a_{1},...,a_{n} R, then a_{i} = k_{i}c (1 < i < n) for some k_{i} R, and k_{1},...,k_{n} are relatively prime.
Let us now turn to the study of polynomials f R[X]. Let f = a_{0} + a_{1}X + ... + a_{n}X^{n}, a_{i} R, be a nonzero polynomial. If a_{0},...,a_{n} are relatively prime, then we say that f is primitive. Let c be a g.c.d. of a_{0},...,a_{n} and let a_{i} = k_{i}c (1 < i < n), where k_{i} R. then k_{0},...,k_{n} are relatively prime so that
f^{ *} = k_{0} + k_{1}X + ... + k_{n}X^{n}
is primitive. Moreover,
f = cf^{ *}.
Thus we have proved
Lemma 6: Let f R[X] be nonzero and let c be a g.c.d of the coefficients of f. Then f can be written in the form
f = cf^{ *},
where f^{ *} R[X] is primitive.
The element c of Lemma 6 is called the content of f. It is uniquely determined up to multiplication by units of R.
For example, consider the polynomial
4X ^{5} + 8X ^{3} + 4X ^{2} + 12 Z[X].
Then a g.c.d. of 4,8,4, and 12 is 4,
4X^{5} + 8X^{3} + 4X^{2} + 12 = 4(X^{5} + 2X^{3} + X^{2} + 3),
and X^{5} + 2X^{3} + X^{2} + 3 is primitive.
The key to the proof of Theorem 1 is the following result , which is referred to as Gauss's lemma.
Theorem 7: Let f,g R[X] be primitive. Then fg is primitive.
Proof: Let
f = a_{0} + a_{1}X + ... + a_{n}X^{n},
g = b_{0} + a_{1}X + ... + b_{n}X^{m},
fg = c_{0} + c_{1}X + ... + c_{m+n}X^{m+n}.
Let be an irreducible element of R. It suffices to prove that does not divide all of c_{0},...,c_{m+n}. For then if e is a g.c.d. of c_{0},...,c_{m+n}, then e is not divisible by any irreducible element of R and is thus a unit, so that c_{0},...,c_{m+n} are relatively prime. Since f and g are primitive, does not divide all of a_{0},...,a_{n} and does not divide all of b_{0},...,b_{m}. Let a_{i} and b_{i} be the first coefficients of f and g, respectively, which are not divisible by . We assert that c_{i+j} is not divisible by . Indeed, assume that c_{i+j}. Then
By the choice of a_{i} and b_{j}, we see that a_{0},...,a_{i1},b_{0},...,b_{j1} are all divisible by , so that terms of I and II are divisible by . Thus,
a _{i}b _{j},
which implies that a_{i}b_{j} = k for some k R. But is irreducible in R, so that by the unique factorization in R and Proposition 5, must divide either a_{i} or b_{j}, which is a contradiction to the choice of a_{i} and b_{j}. Thus c_{i+j} as asserted.
Let F denote the quotient field of R. Every nonzero element x of F can be written in the form x = a/b, a,b R^{x}. If is a g.c.d. of a and b, then a^{*} = a/, b^{*} = b/ belong to R and x = a^{*}/b^{*}, with a^{*} and b^{*} relatively prime. Thus every nonzero x F can be written in the form x = a/b, a,b R, a and b relatively prime. Therefore, if f F[X] is nonzero, there exists a^{*} F and f^{ *} R[X] such that f = a^{*}f^{ *}. Moreover, it is clear by choosing a^{*} appropriately, we may assume that f^{ *} is primitive, by Lemma 6. Note that F[X] is a UFD by Proposition 13 of the section on principal ideal arithmetic.
Corollary 8: Let f R[X] be primitive. Then f is irreducible in R[X] if and only if f is irreducible in F[X].
Proof: Assume that f is irreducible in R[X] and that f = gh, g,h F[X]. By the above discussion, we can write g = a^{*}g^{*}, h = b^{*}h^{*}, where g^{*}, h^{*} R[X], g^{*} and h^{*} are primitive, and a^{*}, b^{*} F. Then f = (a^{*}b^{*})(g^{*}h^{*}). By Theorem 7, g^{*}h^{*} is primitive. Therefore, since f is primitive a^{*}b^{*} is a unit of R. But then since f = (a^{*}b^{*})g^{*}h^{*} is irreducible, either g^{*} or h^{*} is a unit of R[X]. Thus, in particular either g^{*} or h^{*} is a constant polynomial so that either g or h is a unit of F[X]. Thus, f is irreducible in F[X].
Assume that f is irreducible in F[X], and assume that f = gh, g,h R[X]. Then by the irreducibility of f in F[X], we see that on of g, h must be a unit of F[X], that is, a constant polynomial. Say g is a constant polynomial. Then g is a unit of R, since f = gh and f is primitive. Thus f is irreducible in R[X].
Proof of Theorem 1: Let f R[X], f a unit of R[X]. Let us first show that f can be written as a product of irreducible elements of R[X]. If f is a constant polynomial, then f R and f can be written as a product of irreducible elements of R, since R is a UFD, and an irreducible element of R is an irreducible element of R[X], Thus we may assume that f a constant polynomial. By Lemma 6, we can write
(2)
f = a ^{*}f^{ *}, a ^{*} R, f^{ *} R[X], f^{ *} is primitive.
Since F[X] is a UFD, we can write
(3)
f^{ *} = f_{1}... f_{n}, f_{i} F[X], f_{i} irreducible in F[X].
Now we can write, f_{i} = a_{i} f_{i}^{*}, f_{i}^{*} R[X], f_{i}^{*} primitive, a_{i} F. By Corollary 8, f_{i}^{*} is irreducible in R[X]. Moreover, by Theorem 7, f_{1}^{*}...f_{n}^{*} is primitive, so that by (3) and the fact that f^{ *} is primitive, we see that
f^{ *} = (a_{1}...a_{n})f_{1}^{*}...f_{n}^{*}
and a_{1}...a_{n} is a unit of R. By (2),
f = ( a ^{*}) f_{1}^{*}... f_{n}^{*}.
Since R is a UFD, we can write a^{*} = _{1}..._{m}, _{i} R, _{i} irreducible. But then as observed above, _{i} is irreducible in R[X] and
f = _{1}... _{m} f_{1}^{*}... f_{n}^{*}
is a decomposition of f into the product of irreducible elements of R[X].
In order to complete the proof of Theorem 1, we must show that if
f = _{1}... _{s} = _{1}... _{t}
are two decompositions of f into a product of irreducible elements of R[X], then s = t and upon proper rearrangement of _{1},...,_{s}, we have _{i} and _{i} are associates (1 < i < s). This is proved using Proposition 5 plus the same reasoning used in the proof of Theorem 12 of the section on arithmetic in principal ideal domains.
From the proof of Theorem 1, we can use the following useful result.
Corollary 9: Let R be a unique factorization domain and let f R[X] be a primitive polynomial. Further, let F denote the quotient field of R. Then the factorization of f into a product of irreducible factors in F[X] can already be carried out in R[X].
