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 X₁,...,X_n are indeterminates over F, is F[X₁,...,X_n] a UFD? The only way which we have at our disposal to prove that F[X₁,...,X_n] is a UFD is to verify that F[X₁,...,X_n] is a PID. However, we have already seen that for n = 2, F[X₁,...,X_n] is not a PID. Therefore, the machinery developed earlier is not sensitive enough to determine whether or not F[X₁,...,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 X₁,...,X_n be indeterminates over R. Then R[X₁,...,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[X₁,...,X_n-1] is a UFD. Therefore, by Theorem 1, R[X₁,...,X_n] = R[X₁,...,X_n-1][X_n] is a UFD.

Corollary 3: Let F be a field, X₁,...,X_n indeterminates over F. Then F[X₁,...,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[X₁,...,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

x = ^a(x),

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 a₁,...,a_n R. Then a greatest common divisor of a₁,...,a_n is an element of R such that

1. c|a₁, ..., c|a_n.

2. If d R such that d|a₁, ..., d|a_n, then d|c.

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 a₁,...,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, a₁,...,a_n R not all 0. Then a₁,...,a_n have a g.c.d.

Proof: Let

a_i = _i ^a(a_i),    _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 a₁,...,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 a₁,...,a_n R, then we say that a₁,...,a_n are relatively prime. Moreover, if c is a g.c.d. of a₁,...,a_n R, then a_i = k_ic (1 < i < n) for some k_i R, and k₁,...,k_n are relatively prime.

Let us now turn to the study of polynomials f R[X]. Let f = a₀ + a₁X + ... + a_nXⁿ, a_i R, be a nonzero polynomial. If a₀,...,a_n are relatively prime, then we say that f is primitive. Let c be a g.c.d. of a₀,...,a_n and let a_i = k_ic (1 < i < n), where k_i R. then k₀,...,k_n are relatively prime so that

is primitive. Moreover,

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

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

4X⁵ + 8X³ + 4X² + 12 Z[X].

Then a g.c.d. of 4,8,4, and 12 is 4,

and X⁵ + 2X³ + X² + 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

Let be an irreducible element of R. It suffices to prove that does not divide all of c₀,...,c_m+n. For then if e is a g.c.d. of c₀,...,c_m+n, then e is not divisible by any irreducible element of R and is thus a unit, so that c₀,...,c_m+n are relatively prime. Since f and g are primitive, does not divide all of a₀,...,a_n and does not divide all of b₀,...,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 a₀,...,a_i-1,b₀,...,b_j-1 are all divisible by , so that terms of I and II are divisible by . Thus,

|a_ib_j,

which implies that a_ib_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

f = a^*f ^*, a^* R,  f ^* R[X],   f ^* is primitive.

Since F[X] is a UFD, we can write

f ^* = f₁...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, f₁^*...f_n^* is primitive, so that by (3) and the fact that f ^* is primitive, we see that

and a₁...a_n is a unit of R. By (2),

f = (a^*)f₁^*...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 f₁^*...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].