Polynomial Rings Over Commutative Rings

In defining the polynomial ring in one variable over a field no essential use was made of the fact that was a field; all that was used was that was a commutative ring. The field nature of only made itself felt in proving that was a Euclidean ring.

Thus we can imitate what we did with fields for more general rings. While some properties may be lost, such as ``Euclideanism'', we shall see that enough remain to lead us to interesting results. The subject could have been developed in this generality from the outset, and we could have obtained the particular results about by specializing the ring to be a field. However, we felt that it would be healthier to go from the concrete to the abstract rather than from the abstract to the concrete. The price we pay for this repetition, but even that serves a purpose, namely, that of consolidating the ideas. Because of the experience gained in treating polynomials over fields, we can afford to be a little sketchier in the proofs here.

Let be a commutative ring with unit element. By the polynomial ring in over we shall mean the set of formal symbols



where are in and where equality, addition, and multiplication are defined exactly as they were in Section 3.9. As in that section, is a commutative ring with unit element.

We now define the ring of polynomials in the -variables over as follows: Let the polynomial ring in over is called the ring of polynomials in over Its elements are of the form



where equality and addition are defined coefficientwise and where multiplication is defined by use of the distributive law and the rule of exponents



Of particular importance is the case in which is a field; here we obtain the ring of polynomials in -variables over a field.

Of interest to us will be the influence of the structure of on that of The first result in this direction is

LEMMA 3.11.1: If is an integral domain, then so is


Making successive use of the lemma immediately yields the

COROLLARY: If is an integral domain, then so is

In particular, when is a field, must be an integral domain. As such, we can construct its field of quotients; we call this the field of rational functions in over and denote it by This field plays a vital role in algebraic geometry. For us it shall be of utmost importance in our discussion, in Chapter 5, of Galois theory.

However, we want deeper interrelations between the structures of and of than that expressed in Lemma 3.11.1. Our development now turns in that direction.

Exactly in the same way as we did for Euclidean rings, we can speak about divisibility, units, etc., in arbitrary integral domains, with unit element. Two elements in are said to be associates if where is a unit in An element which is not a unit in will be called irreducible (or a prime element) if, whenever with both in then one of or must be a unit in An irreducible element is thus an element which cannot be factored in a ``nontrivial'' way.

DEFINITION: An integral domain, with unit element is a unique factorization domain if

(a) Any nonzero element in is either a unit or can be written as the product of a finite number of irreducible elements of

(b) The decomposition in part (a) is unique up to the order and associates of the irreducible elements.

Theorem 3.7.2 asserts that a Euclidean ring is a unique factorization domain. The converse, however, is false; for example, the ring where is a field, is not even a principal ideal ring (hence is certainly not Euclidean), but as we shall soon see it is a unique factorization domain.

In general commutative rings we may speak about the greatest common divisors of elements; the main difficulty is that these, in general, might not exist. However, in unique factorization domains their existence is assured. This fact is not difficult to prove and we leave it as an exercise; equally easy are the other parts of

LEMMA 3.11.2: If R is a unique factorization domain and if are in then and have a greatest common divisor in Moreover, if and are relatively prime (i.e., ), whenever then

COROLLARY: If is an irreducible element and then or

We now wish to transfer the appropriate version of the Gauss' lemma ( Theorem 3.10.1), which we proved for polynomials with integer coefficients, to the ring where is a unique factorization domain.

Given the polynomial



in then the content of is defined to be the greatest common divisor of It is unique within units of We shall denote the content of by A polynomial in is said to be primitive if its content is 1 (that is, is a unit in ) . Given any polynomial we can write where and where is primitive. (Prove!) Except for multiplication by units of this decomposition of as an element of by a primitive polynomial in is unique. (Prove!)

The proof of Lemma 3.10.1 goes over completely to our present situation; the only change that must be made in the proof is to replace the prime number by an irreducible element of Thus we have

LEMMA 3.11.3: If is a unique factorization domain, then the product of two primitive polynomials in is again a primitive polynomial in

Given in we can write where and where and are primitive. Thus



By Lemma 3.11.3, is primitive. Hence the content of is that is, it is We have proved the

COROLLARY: If is a unique factorization domain and if are in then (up to units).

By a simple induction, the corollary extends to the product of a finite number of polynomials to read



Let R be a unique factorization domain. Being an integral domain, by Theorem 3.6.1, it has a field of quotients We can consider to be a subring of Given any polynomial then where and where (Prove!) It is natural to ask for the relation, in terms of reducibility and irreducibility, of a polynomial in considered as a polynomial in the larger ring

LEMMA 3.11.4: If in is both primitive and irreducible as an element of then it is irreducible as an element of Conversely, if the primitive in is irreducible as an element of it is also irreducible as an element of


LEMMA 3.11.5: If is a unique factorization domain and if is a primitive polynomial in then it can be factored in a unique way as the product of irreducible elements in


We now have all the necessary information to prove the principal theorem of this section.

THEOREM 3.11.1: If is a unique factorization domain, then so is


Given as a unique factorization domain, then is also a unique factorization domain. Thus is also a unique factorization domain. Continuing in this pattern we obtain

COROLLARY 1: If is a unique factorization domain then so is

A special case of Corollary 1 but of independent interest and importance is

COROLLARY 2: If is a field then is a unique factorization domain.