Let be a field. By the ring of polynomials in the indeterminate, written
as we mean the set of all symbols
where can be any nonnegative integer and where the coefficients are all in In order to make a ring out of we must be able to recognize when two elements in it are equal, we must be able to add and multiply elements of so that the axioms defining a ring hold true for This will be our initial goal.
We could avoid the phrase ``the set of all symbols'' used above by introducing an appropriate apparatus of sequences but it seems more desirable to follow a path which is somewhat familiar to most readers.
DEFINITION: If and are in then if and only if for every integer
Thus two polynomials are declared to be equal if and only if their corresponding coefficients are equal.
DEFINITION: If and are both in then
where for each
In other words, add two polynomials by adding their coefficients and collecting terms. To add and we consider as and add, according to the recipe given in the definition, to obtain as their sum
The most complicated item, and the only one left for us to define for is the multiplication.
DEFINITION: If and then
This definition says nothing more than: multiply the two polynomials by multiplying out the symbols formally, use the relation and collect terms.
Therefore according to our definition,
If you multiply these together high-school style you will see that you get the same answer. Our definition of product is the one the reader has always known.
Without further ado we assert that is a ring with these operations, its multiplication is commutative, and it has a unit element. We leave the verification of the ring axioms to the reader.
DEFINITION: If and then the degree of written as is
That is, the degree of is the largest integer for which the th coefficient of is not 0. We do not define the degree of the zero polynomial. We say a polynomial is a constant if its degree is 0. The degree function defined on the nonzero elements of will provide us with the function needed in order that be a Euclidean ring.
LEMMA 3.9.1: If are two nonzero elements of then
Proof: Suppose that and and that and Therefore and By definition,
We claim that and for That can be seen at a glance by its definition. What about for ? is the sum of terms of the form ; since then either or But then one of or is so that ; since is the sum of a bunch of zeros it itself is 0, and our claim has been established. Thus the highest nonzero coefficient of is whence
COROLLARY: If are nonzero elements in then
and since this result is immediate from the lemma.
COROLLARY: is an integral domain.
Proof: We leave the proof of this corollary to the reader.
Since is an integral domain, in light of Theorem 3.6.1 we can construct for it its field of quotients. This field merely consists of all quotients of polynomials and is called the field of rational functions in over The function defined for all in satisfies
1. is a nonnegative integer.
2. for all in
In order for to be a Euclidean ring with the degree function acting as the -function of a Euclidean ring we still need that given there exist in such that
where either or This is provided us by
LEMMA 3.9.2 (THE DIVISION ALGORITHM): Given two polynomials and in then there exist two polynomials and in such that
Proof: The proof is actually nothing more than the ``long-division'' process we all used in school to divide one polynomial by another.
If the degree of is smaller than that of there is nothing to prove, for merely put and we certainly have that
So we may assume that and where and
Thus so by induction on the degree of we may assume that
where or But then
from which, by transposing, we arrive at
If we put
we do indeed have that where and where or This proves the lemma.
This last lemma fills the gap needed to exhibit as a Euclidean ring and we now have the right to say
THEOREM 3.9.1: is a Euclidean ring.
All the results of Section 3.7 now carry over and we list these, for our particular case, as the following lemmas. It could be very instructive for the reader to try to prove these directly, adapting the arguments used in Section 3.7 for our particular ring and its Euclidean function, the degree.
LEMMA 3.9.3: is a principal ideal ring.
LEMMA 3.9.4: Given two polynomials in they have a greatest common divisor which can be realized as
What corresponds to a prime element?
DEFINITION: A polynomial in is said to be irreducible over if whenever
with then one of or has degree 0 (i.e., is a constant).
Irreducibility depends on the field; for instance the polynomial is irreducible over the real field but not over the complex field, for there
LEMMA 3.9.5: Any polynomial in can be written in a unique manner as a product of irreducible polynomials in
LEMMA 3.9.6: The ideal in is a maximal ideal if and only is irreducible over
In Chapter 5 we shall return to take a much closer look at this field but for now we should like to compute an example.
Let be the field of rational numbers and consider the polynomial
in As is easily verified, it is irreducible over whence is a field. What do its elements look like? Let the ideal in generated by
Any element in is a coset of the form of the ideal with in Now, given any polynomial by the division algorithm,
where or Thus where are in ; consequently
since is in Hence by the addition and multiplication in
If we put then every element in is of the form with in What about ? Since
(since is the zero element of ) we see that Also, if
whence is in How can this be, since every element in has degree at least 3? Only if
that is, only if Thus every element in has a unique representation as where By Lemma 3.9.6, is a field. It would be instructive to see this directly; all that it entails is proving that if then it has an inverse of the form Hence we must solve for in the relation
where not all of are Multiplying the relation out and using we obtain
We can try to solve these three equations in the three unknowns When we do so we find that a solution exists if and only if
Therefore the problem of proving directly that is a field boils down to proving that the only solution in rational numbers of
is the solution We now proceed to show this. If a solution exists in rationals, by clearing of denominators we can show that a solution exists where are integers. Thus we may assume that are integers satisfying (1). We now assert that we may assume that have no common divisor other than 1, for if and where is their greatest common divisor, then substituting in (1) we obtain
The problem has thus been reduced to proving that (1) has no solutions in integers which are relatively prime. But then (1) implies that is even, so that is even; substituting in (1) gives us
Thus and so, is even; Substituting in (1) we obtain
Thus and so is even! But then have 2 as a common factor! This contradicts that they are relatively prime, and we have proved that the equation
has no rational solution other than Therefore we can solve for and is seen, directly, to be a field.