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
Proof: For where in we define the degree of to be ; thus is the index of the highest nonzero coefficient of If is an integral domain we leave it as an exercise to prove that
But then, for is impossible to have That is, is an integral domain.
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
Proof: Suppose that the primitive element in is irreducible in but is reducible in Thus
where are in and are of positive degree. Now
where and where Also
where and are primitive in Thus
By Lemma 3.11.3, is primitive, whence the content of the right-hand side is . Since is primitive, the content of the left-hand side is ; but then [ where is a unit]; the implication of this is that [ where is a unit], and we have obtained a nontrivial factorization of in contrary to hypothesis. (Note: this factorization is nontrivial since each of are of the same degree as so cannot be units in (see Problem 4).) We leave the converse half of the lemma as an exercise.
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
Proof: When we consider as an element in by Lemma 3.9.5, we can factor it as
where are irreducible polynomials in Each where and ; moreover, where and where is primitive in Thus each
where and where is primitive. Since is irreducible in must also be irreducible in hence by Lemma 3.11.4 it is irreducible in
Using the primitivity of and of we can read off the content of the left-hand side as and that of the right-hand side as Thus
We have factored in as a product of irreducible elements.
Can we factor it in another way? If
where the are irreducible in by the primitivity of each must be primitive, hence irreducible in by Lemma 3.11.4. But by Lemma 3.9.5 we know unique factorization in ; the net result of this is that the and the are equal (up to associates) in some order [This means that where is a unit from We have to show that is a unit from Note that where therefore But then where is a unit from by ], hence has a unique factorization as a product of irreducibles 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
Proof: Let be an arbitrary element in We can write in a unique way as where is in and where in is primitive. By Lemma 3.11.5 we can decompose in a unique way as the product of irreducible elements of What about ? Suppose that
in ; then
Therefore, each must be of degree 0, that is, it must be an element of In other words, the only factorizations of as an element of are those it had as an element of In particular, an irreducible element in is still irreducible in Since is a unique factorization domain, has a unique factorization as a product of irreducible elements of hence of
Putting together the unique factorization of in the form where is primitive and where with the unique factorizability of and of we have proved the theorem.
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.