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The set of integers. Let R be an integral domain. There exists a field FR and an embedding f :RmapsFR such that every element of FR is of the form f(a)·f(b)-1 (a,b element of R, bnot equal0). Let F be a field, X an indeterminate over F, f a nonzero, nonconstant polynomial in F[X], a the leading coefficient of f. Then f can be written in the form f = af1*...ft*, where fi* is an irreducible, monic polynomial. Moreover, this decomposition is unique up to the order of the fi* Let x element of Rx, x not an element of UR and let x = pi1...pis = lambda1...lambdat be two expressions of x as a product of irreducible elements. Then s = t and by renumbering lambda1,...,lambdas, we can guarantee that pii and lambdai are associates (1 < i < s). Let f,g element of R[X] be primitive. Then fg is primitive. Let f element of 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 * element of R[X] is primitive. Let R be a unique factorization domain, a,b,pi element of R, pi irreducible. If pi|ab, then pi|a or pi|b. We say that R is a unique factorization domain if the following two conditions are satisfied: (1) If x element of Rx is not a unit of R, then x can be written as a product of irreducible elements of R.and (2) If x element of Rx is not a unit of R, and if x = pi1 ... pis = lambda1 ... lambdat are two expressions of x as a product of irreducible elements, then s = t and it is possible to renumber pi1 ... pis so that pi1 and lambda1 are associates (1 < i < s). A ring R having the property that every ideal is principal is called a principal ideal ring (PIR). If, in addition, R is an integral domain, then R is called a principal ideal domain (PID). Let R be a principal ideal domain. Then R is a unique factorization domain.

Let R be a ring x,y element of R. We say that x and y are associates if x = epsilony for some unit epsilon of R. Let R be a ring with identity 1, and let a element of R. If a has an inverse with respect to multiplication, then we say that a is a unit of R.

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

Corollary 3: Let F be a field, X1,...,Xn indeterminates over F. Then F[X1,...,Xn] 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,...,Xn] 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 element of Rx can be written in the form

(1)
x = epsilonproduct of primespiapi(x),

where api(x) is a nonnegative integer, epsilon is a unit of R, api(x) = 0 for all but a finite number of pi, and product of primes denotes the product over pi element of P for which api(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 pi element of P. Since factorization in R is unique, the decomposition (1) is unique, up to rearrangements of the pi element of P. (Here we use the fact that no two elements of P are associates.)

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

1. c|a1, ..., c|an.

2. If d element of R such that d|a1, ..., d|an, 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,...,an, 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,...,an element of R not all 0. Then a1,...,an have a g.c.d.

Proof: Let

ai = epsiloni product of primespiapi(ai),    epsiloni a unit of R (1 < i < n).

For each pi element of P, let api be the smallest of api(ai),..., api(an). Then since api(ai) = 0 for all but a finite number of pi, we see that api = 0 for all but a finite number of pi, and thus we may set

c = product of primespiapi.

Then c is a g.c.d. of a1,...,an. 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,pi element of R, pi irreducible, then pi|ab implies pi|a or pi|b. Let us observe that this is true in any unique factorization domain.

Proposition 5: Let R be a unique factorization domain, a,b,pi element of R, pi irreducible. If pi|ab, then pi|a or pi|b.

Proof: Since pi|ab, there exists k element of R such that ab = kpi. Without loss of generality, we may assume that anot equal0, not equal0, knot equal0. Since R is a UFD, we can write a and b as a product of irreducible elements. Since pi appears in a decomposition of kpi into a product of irreducible elements, the uniqueness of factorization implies that pi is an associate of either some irreducible element dividing a or some irreducible element dividing b. Thus, either pi|a or pi|b.

If 1 is a g.c.d of a1,...,an element of R, then we say that a1,...,an are relatively prime. Moreover, if c is a g.c.d. of a1,...,an element of R, then ai = kic (1 < i < n) for some ki element of R, and k1,...,kn are relatively prime.

Let us now turn to the study of polynomials f element of R[X]. Let f = a0 + a1X + ... + anXn, ai element of R, be a nonzero polynomial. If a0,...,an are relatively prime, then we say that f is primitive. Let c be a g.c.d. of a0,...,an and let ai = kic (1 < i < n), where ki element of R. then k0,...,kn are relatively prime so that

f * = k0 + k1X + ... + knXn

is primitive. Moreover,

f = cf *.

Thus we have proved

Lemma 6: Let f element of 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 * element of 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 element of Z[X].

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

4X5 + 8X3 + 4X2 + 12 = 4(X5 + 2X3 + X2 + 3),

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 element of R[X] be primitive. Then fg is primitive.

Proof: Let

f = a0 + a1X + ... + anXn,
g = b0 + a1X + ... + bnXm,
fg = c0 + c1X + ... + cm+nXm+n.

Let pi be an irreducible element of R. It suffices to prove that pi does not divide all of c0,...,cm+n. For then if e is a g.c.d. of c0,...,cm+n, then e is not divisible by any irreducible element of R and is thus a unit, so that c0,...,cm+n are relatively prime. Since f and g are primitive, pi does not divide all of a0,...,an and pi does not divide all of b0,...,bm. Let ai and bi be the first coefficients of f and g, respectively, which are not divisible by pi. We assert that ci+j is not divisible by pi. Indeed, assume that pi|ci+j. Then

gausss lemma

By the choice of ai and bj, we see that a0,...,ai-1,b0,...,bj-1 are all divisible by pi, so that terms of I and II are divisible by pi. Thus,

pi|aibj,

which implies that aibj = pik for some k element of R. But pi is irreducible in R, so that by the unique factorization in R and Proposition 5, pi must divide either ai or bj, which is a contradiction to the choice of ai and bj. Thus pidoes not divideci+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 element ofRx. If lambda is a g.c.d. of a and b, then a* = a/lambda, b* = b/lambda belong to R and x = a*/b*, with a* and b* relatively prime. Thus every nonzero x element of F can be written in the form x = a/b, a,b element of R, a and b relatively prime. Therefore, if f element of F[X] is nonzero, there exists a* element of F and f *element of 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 element of R[X] be primitive. Then f is irreducible in R[X] if and only if f is irreducible in F[X].

Proof: forward implication Assume that f is irreducible in R[X] and that f = gh, g,h element of F[X]. By the above discussion, we can write g = a*g*, h = b*h*, where g*, h* element of R[X], g* and h* are primitive, and a*, b* element of 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].

backwards implication Assume that f is irreducible in F[X], and assume that f = gh, g,h element ofR[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 element of R[X], fnot equal 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 element of 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 not equal a constant polynomial. By Lemma 6, we can write

(2)
f = a*f *, a* element of R,  f * element of R[X],   f * is primitive.

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

(3)
f * = f1...fn,  fi element of F[X],   fi irreducible in F[X].

Now we can write, fi = ai fi*, fi* element of R[X], fi* primitive, ai element of F. By Corollary 8, fi* is irreducible in R[X]. Moreover, by Theorem 7, f1*...fn* is primitive, so that by (3) and the fact that f * is primitive, we see that

f * = (a1...an)f1*...fn*

and a1...an is a unit epsilon of R. By (2),

f = (epsilona*)f1*...fn*.

Since R is a UFD, we can write epsilona* = pi1...pim, pii element of R, pii irreducible. But then as observed above, pii is irreducible in R[X] and

f = pi1...pim f1*...fn*

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 = pi1...pis = lambda1...lambdat

are two decompositions of f into a product of irreducible elements of R[X], then s = t and upon proper rearrangement of pi1,...,pis, we have pii and lambdai 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 element of 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].