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Polynomial RingsAs every student of high school algebra knows,
X + X2, 5 + X3, 17 + XY + Z2W, 3 + 2X
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
a0 + a1X + a2 + ...
where 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 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 + ...
![]() we will write
1 + 2X.
By this convention, the typical polynomial for which
a0 + a1X + ... + anXn.
The one exception to this convention will be the polynomial
0 + 0X + 0X2 + ...
which will be denoted simply by 0. The polynomial 0 will be called the zero polynomial. Let (1)
f = a0 + a1X + a2X2 +...,
(2)
g = b0 + b1X + b2X2 +...
be two polynomials. We say that f is equal to g, denoted
a0 = b0, a1 = b1, a2 = b2, ...
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
(3)
f + g = (a0 + b0) + (a1 + b1)X + (a2 + b2)X2 + ...
Thus, to add two polynomials, we merely add corresponding coefficients. Note that (3) actually defines a polynomial - that is
-f = (-a0) + (-a1)X + (-a2)X2 + ...
Multiplication of Polynomials Let f and g be defined by (1) and (2), respectively. We define the product
f · g = c0 + c1X + c2X2 + ...,
where c0 = a0 · b0 c1 = a0 · b1 + a1 · b0 c2 = a0 · b2 + a1 · b1 + a2 · b0 . . . cn = a0 · bn + a1 · bn-1 + ... + an-1 · b1 + an · b0
Let us see what this definition really says. Let us make the agreement that the polynomial Note that
ci = a0· bi + a1· bi-1 + ... + ai· b0.
The typical term in this sum is 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 (4)
h = c0 + c1X + ...
![]() ![]() Then
(f + g) · h = d0 + d1X + d2X2 + ...,
where (5)
di = (a0 + b0)ci + (a1 + b1)ci-1 +...+ (ai + bi)c0
(i = 0, 1, ...)
Moreover,
f · h = e0 + e1X + e2X2 + ...
g · h = f0 + f1X + f2X2 + ...
where (6)
ei = a0 · ci + a1 · ci-1 + ... + ai · c0
(7)
fi = b0 · ci + b1 · ci-1 + ... + bi · c0
Finally,
f · h + g · h = (e0 +f 0) + (e1 + f1)X + ....
However, by (5),(6),(7) and the distributive law in R, we see that
(f + g) · h = f · h + g · h.
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
1 = 1 + 0X + 0X2 + ....
(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
f = a0 + a1X + a2X2 +...,
then
f · 1 = c0 + c1X + c2X2 +...,
where c0 = a0 · 1 = a0 c1 = a1 · 1 + a0 · 0 = a1 c2 = a2 · 1 + a1 · 0 + a0 · 0 = a2 . . . cn = an · 1 + an-1 · 0 + ... + a0 · 0 = an.
Therefore, (2)Let f and g be given by
f = a0 + a1X + a2X2 +...,
g = b0 + b1X + b2X2 +...
and let
f · g = c0 + c1X + c2X2 + ...,
g · f = d0 + d1X + d2X2 + .....
But if R 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 + ... + amXm, am
![]()
g = b0 + b1X + ... + bnXn, bn
![]()
We may assume that f and g have this form since both are assumed to be nonzero. Then the coefficients of Xm+n in
Corollary 3: Let F be a field. Then Proof: F is a commutative integral domain with identity. Note, however, that Let R be a ring, X is indeterminate over R and f a polynomial in
f = a0 + a1X + ... + anXn, an
![]()
In this case, we say that f has degree n, and we write ![]() ![]() ![]() ![]() ![]() ![]()
for every integer n. Then let us set the degree of the zero polynomial equal to
Proposition 4: Let R be an integral domain, X and indeterminate over R, Proof: If either f or g is zero, then
f = a0 + a1X + ... + amXm, am
![]()
g = b0 + b1X + ... + bnXn, bn
![]()
Then
f · g = c0 + c1X + ... + cm+nXM+n,
where, in particular, Since we made the agreement that
Proposition 5: Let R be a ring, X an indeterminate over R, 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,
Theorem 6: Let F be a field, X an indeterminate over F, and let
(a) f = qg + r,
and
(b) deg(r) < deg(g).
Proof: If
f = an+1Xn+1 + anXn + ... + a0, ai
![]() ![]()
g = bmXm + ... + b0, bi
![]() ![]()
If Set
f ' = f - (an+1bm-1)Xn+1-m g.
Then a quick computation shows that
q = (an+1bm-1)Xn+1-m + q'
we see that 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
a0(X) + a1(X)Y + ... + an(X)Yn ai(X)
![]() or
b0 + b01X + b10Y + b20X2 + b11XY + b02Y2 + ....
Let us make the convention that |
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