Polynomial Rings

As every student of high school algebra knows,

are all examples of polynomials. In these examples, the coefficients of the polynomial all belong to the field of real numbers. In this section we will generalize the notion of a polynomial to allow coefficients in an arbitrary ring. In doing so, we will show that the set of all polynomials in X having coefficients in the ring R is itself a ring, with respect to suitably defined addition and multiplication of polynomials. These rings of polynomials provide us with yet another interesting class of rings. Moreover, rings of polynomials are important in themselves, since they will provide us with a technical tool by means of which we can study the properties of arbitrary rings.

Let R be a ring and let X be a symbol. (X is an indeterminate.) Then a polynomial in X with coefficients in R is a formal sum

where a_i R for all i; and a_i = 0 for i sufficiently large. The set of all polynomials in X with coefficients in R will be denoted by R[X]. Note that every element of R is contained in R[X]. The elements of R are call constant polynomials.

A word of caution. The element X is not an unknown or variable element of R. It is not an element of R in any sense and the reader should avoid thinking of it as an element of R. It is a definite fixed element of R[X].

Let us adopt a notational convention for writing polynomials. In specifying a polynomial, we will omit all terms which appear with a zero coefficient. Thus, instead of

1 + 2X + 0X² + 0X³ + ... Z[X],

we will write

By this convention, the typical polynomial for which a_i = 0 for i > n will be denoted

The one exception to this convention will be the polynomial

which will be denoted simply by 0. The polynomial 0 will be called the zero polynomial.

Let

be two polynomials. We say that f is equal to g, denoted f = g, if

Thus, two polynomials are equal if their coefficients are equal.

Addition of Polynomials

Let f and g be given by (1) and (2), respectively. Then the sum f + g is defined by

Thus, to add two polynomials, we merely add corresponding coefficients. Note that (3) actually defines a polynomial - that is a_i + b_i = 0 for all sufficiently large i. Suppose that a_i = 0 for i > N and that b_i = 0 for i > M. Then a_i + b_i = 0 for all i > max(M,N), where max(M,N) denotes the larger of M and N. It is trivial to see that with respect to the operation +, R[X] becomes an abelian group. The identity element is 0 and the inverse of f under addition is

Multiplication of Polynomials

Let f and g be defined by (1) and (2), respectively. We define the product f · g by

where

c₀ = a₀ · b₀

c₁ = a₀ · b₁ + a₁ · b₀

c₂ = a₀ · b₂ + a₁ · b₁ + a₂ · b₀

.

.

.

c_n = a₀ · b_n + a₁ · b_n-1 + ... + a_n-1 · b₁ + a_n · b₀

Let us see what this definition really says. Let us make the agreement that the polynomial a₀X⁰ is just the polynomial a₀ + 0X + 0X2 + .... Then the definition of multiplication essentially amount to the following: To form f · g, take the product of each term of f by each term of g using the rule a_iXⁱ · b_jX^j = a_i · b_jX^i+j (i,j > 0), and collect all terms containing the same power of X. Thus, our definition of multiplication coincides with the familiar definition of multiplication of polynomials with coefficients in R.

Note that f · g is a polynomial in X with coefficients in R - that is, c_i = 0 for i sufficiently large. Indeed, let M and N be so large that a_i = 0 for all i > M and b_i = 0 for i > N. Assume that i > M + N. Then

The typical term in this sum is a_j· b_i-j (0 < j < i). Since i > M +N, we must have either j > M or i - j > N. [Because if both of j < M, i - j < N hold, then we have i = j + (i - j) < M + N.] Therefore, for every i > M + N, we must have either a_j = 0 or b_i-j = 0 (0 < j < i), so that every term in the expression for c_i is zero. Therefore, c_i = 0 for i > M + N, and f · g is a polynomial.

It is not obvious that multiplication of polynomials is associative or that the distributive laws are satisfied. By way of illustration, let us verify the right distributive law. Let f and g be given by (1) and (2), respectively, and let

h = c₀ + c₁X + ... R[X], c_i R.

Then

where

Moreover,

where

Finally,

However, by (5),(6),(7) and the distributive law in R, we see that e_i + f_i = d_i, so that

This proves the right distributive law in R[X]. Similarly, it is possible to prove the left distributive law and the associativity of multiplication. These can be proven easily by an interested reader. Thus we have

Theorem 1: R[X] is a ring.

The ring R[X] is called the ring of polynomials in X over R. The next result is typical in the theory of polynomial rings. It asserts that certain properties of the ring R carry over to the ring R[X].

Theorem 2: Let R be a ring, and X an indeterminate over R. Then

(1) If R is a ring with identity 1, then R[X] is a ring with identity

(2) If R is commutative, then R[X] is commutative.

(3)If R is an integral domain, then R[X] is an integral domain.

Proof: (1) Let f R[X] be given by

then

where

c₀ = a₀ · 1 = a₀

c₁ = a₁ · 1 + a₀ · 0 = a₁

c₂ = a₂ · 1 + a₁ · 0 + a₀ · 0 = a₂

.

.

.

c_n = a_n · 1 + a_n-1 · 0 + ... + a₀ · 0 = a_n.

Therefore, f · 1 = f. Similarly, 1 · f = f. Therefore, R[X] is a ring with the identity 1.

(2)Let f and g be given by

and let

But if R is commutative, a_i · b_j = b_j · a_i for all i, j and therefore c_n = d_n for all n. Thus, f · g = g · f. Since f and g are arbitrary elements of R[X], we see that R[X] is commutative.

(3) By definition of an integral domain it suffices to show that if R is an integral domain, then the product of two nonzero polynomials with coefficients in R is nonzero. Let

f = a₀ + a₁X + ... + a_mX^m,   a_m 0,

g = b₀ + b₁X + ... + b_nXⁿ,   b_n 0.

We may assume that f and g have this form since both are assumed to be nonzero. Then the coefficients of X^m+n in f · g is given by a_m · b_n. However, since a_m 0, b_n 0, and R is an integral domain, we have a_m · b_n 0. Thus, f · g has a nonzero coefficient and hence nonzero.

Corollary 3: Let F be a field. Then F[X] is a commutative integral domain.

Proof: F is a commutative integral domain with identity.

Note, however, that F[X] is not a field since, for example, X has no multiplicative inverse.

Let R be a ring, X is indeterminate over R and f a polynomial in R[X]. If f is not the zero polynomial, then

f = a₀ + a₁X + ... + a_nXⁿ,   a_n 0.

In this case, we say that f has degree n, and we write deg(f) = n. Defining the degree of the zero polynomial, 0, presents something of a problem. Strictly as a mater of convenience, let us introduce a symbol - called minus infinity. We will perform arithmetic with - according to the following rules:

- + n = -,    - + (-) = -,   - < n

for every integer n. Then let us set the degree of the zero polynomial equal to -. Thus the zero polynomial has smaller degree than any nonzero polynomial. The reason for this strange choice of degree for 0 is explained by the following result:

Proposition 4: Let R be an integral domain, X and indeterminate over R, f,g R[X]. Then deg(fg) = deg(f) + deg(g).

Proof: If either f or g is zero, then f · g = 0 and both sides of the above equation are -. Therefore, it suffices to assume that f and g are both nonzero. Assume that deg(f) = m, deg(g) = n. Then f and g have the form

f = a₀ + a₁X + ... + a_mX^m,   a_m 0,

g = b₀ + b₁X + ... + b_nXⁿ,   b_n 0.

Then f · g is of the form

where, in particular, c_m+n = a_m· b_n. But since R is an integral domain and a_m and b_n are nonzero, we see that c_m+n 0. Therefore, deg(f+g) = m+n.

Since we made the agreement that - < n for all integers n, we may state the following analogue of Proposition 4.

Proposition 5: Let R be a ring, X an indeterminate over R, f,g R[X]. Then deg(f + g) < max(deg(f),deg(g)), where max(deg(f),deg(g)) denotes the larger of deg(f) and deg(g). If deg(f) deg(g), then the inequality may be replaced by equality.

We have thus far considered the addition, subtraction, and multiplication of polynomials. Let us now begin to consider the problem of one polynomial by another. Let us recall what happens in the case of polynomials with real coefficients. Let f and g be polynomials with real coefficients, g 0. In high school algebra we considered the problem of "dividing" f by g. We learned a procedure called "long division", whereby f/g could be written as a quotient q plus a remainder of the form r/g, where q and r are polynomials. For example, if f = X⁵ + 2X³ + X + 1 and g = 2X³ + 2, we have q = 1/2X² + 1, r = -X² + X - 1. The reader is urged to carry out this computation in order to recall the procedure. By inspection of the long division process, we can see that deg(r) < deg(g). Thus we can express the process of long division as follows: There exist polynomials q and r such that f = qg + r, and deg(r) < deg(g). It makes sense to ask whether such a process of long division can be carried out in a polynomial ring. Note that in the above example, the coefficients lie in Z, but nevertheless, the coefficients of q and r involve rational numbers. This suggests that it would be wise to confine our investigations to polynomial rings over a field. With this restriction, the process of long division can be carried over to polynomial rings. The resulting process in F[X] is called the division algorithm in F[X].

Theorem 6: Let F be a field, X an indeterminate over F, and let f, g F[X], g 0. Then there exist polynomial q and r belonging to F[X] such that

and

Proof: If f = 0, then we may set q = r = 0. Then deg(g) < 0, since g 0, so that deg(r) < deg(g). Thus we may assume that deg(f) > 0. Let us proceed by induction of the degree of f. We leave it to the interested reader to verify the theorem if deg(f) = 0, in which case f is a nonzero constant. Let us assume the theorem for polynomials of degree < n, and let us assume that f has degree n+1. Suppose that

f = a_n+1Xⁿ⁺¹ + a_nXⁿ + ... + a₀,   a_i F,   a_n+1 0,

g = b_mX^m + ... + b₀,   b_i F,   b_m 0,

If m > n+1, then we may set f = r, 0 = q. Therefore, we may assume that n+1 > m.

Set

Then a quick computation shows that deg(f ') is at most n. Therefore, by induction there exist polynomials q' and r such that f ' = q'g +r, deg(r) < deg(g). But then, if we set

we see that f = qg +r, deg(r) < deg(g). Thus the induction is complete.

Remark: The polynomials q and r satisfying both (a) and (b) above are unique.

Thus far, we have considered polynomial rings in only one indeterminate X. However, if R[X] is a polynomial ring and Y is an indeterminate over R[X], then we may form the polynomial ring R[X][Y]. A typical element of R[X][Y] is of the form

a₀(X) + a₁(X)Y + ... + a_n(X)Yⁿ   a_i(X) R[X],

or

Let us make the convention that XY = YX in R[X][Y]. Then, in particular, we see that R[X][Y] = R[Y][X]. We will denote R[X][Y] simply by R[X,Y]. Then by what we have said, R[X,Y] = R[Y,X]. This ring is said to be a polynomial in two variables (indeterminates) over R. Similarly, we can construct rings R[X₁,...,X_n] in n variables over R. Such rings are of critical importance to geometrical investigations, since geometrical objects (curves, surfaces, etc.) are described by equations in several variables.