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Polynomials
Polynomials with coefficients in a ring
The thing that our familiar polynomials all have in common, whether they’re in Z[x], Q[x], R[x],
or C[x] (i.e. their coefficients come from Z, Q, R, or C) is that Z, Q, R, or C are all commutative
rings. We can, of course, generalize this:
A polynomial p(x) with coefficients in a commutative ring R is an expression
in the form
p(x) = an xn + an−1 xn−1 + ... + a2 x2 + a1 x + a0
where n ∈ Z, n ≥ 0, ai ∈ R for i = 0...n.
The set of all polynomials with coefficients in R is denoted R[x].
Now, if you’re worried about just what other rings we might pull our coefficients from (matrices,
anyone?), examples outside of the usual Z, Q, R, or C will come from the Zm rings.
First, we’re going to tighten up the idea of a polynomial function; the rule will be that whatever
ring R the coefficients are in, p(x) will be a function from R to R. In other words, the allowable
inputs from x will come from the same ring.
Under this definition, if we view p(x) = x2 + 2x − 5 as a polynomial in Z[x], it doesn’t look like
this:
It looks like this:
Terminology
The degree of a polynomial p(x) = an xn + an−1 xn−1 + ... + a2 x2 + a1 x + a0 is n if an = 0 and
ai = 0 for all i > n.
an is the leading coefficient of p(x).
If an = 1 (or whatever the unity of the ring is, so these should be more properly called polynomials
with with coefficients in a commutative ring with unity...), we say that p(x) is monic.
Polynomials p(x) = C when C is a nonzero constant have degree 0. The polynomial p(x) = 0 has
no degree at all.
Examples:
• The polynomial p(x) = 4x5 + 2x2 + 1 can be viewed as a polynomial in Z[x], Q[x], R[x], or
C[x]. It has degree 5. It is not monic, since the leading coefficient is not 1.
• We could also view p(x) = 4x5 + 2x2 + 1 as a polynomial in Z5 [x] - all the coefficients are
elements of Z5 . It is still degree 5, and not monic.
• If we’re willing to bend the rules a little, though, we could also view p(x) = 4x5 + 2x2 + 1
as a polynomial in Z3 [x]...but we should immediately reduce its coefficients down modulo 3.
Then it becomes p(x) = 1x5 + 2x2 + 1. Now it’s degree 5 and monic.
• And, just for fun, what if we decided to view it a polynomial in Z4 [x]? Mod the coefficients
down, and it becomes p(x) = 0x5 + 2x2 + 1, so p(x) = 2x2 + 1. In Z4 [x], it’s a degree 2
polynomial, and not monic.
Addition and multiplication
Let
p(x) = an xn + an−1 xn−1 + ... + a1 x + a0
and
q(x) = bm xm + bm−1 xm−1 + ... + b1 x + b0
where the coefficients ai and bj are all elements of some ring R.
Let k = max{n, m}, let s = m + n, and let ai = 0 if i > n and bi = 0 if i > m.
The sum p(x) + q(x) is defined by
k
i
p(x) + q(x) =
i=0 (ai + bi )x
k
= (ak + bk )x + (ak−1 + bk−1 )xk−1 +
... + (a1 + b1 )x + (a0 + b0 )
The product p(x)q(x) is defined by
p(x)q(x) = cs xs + cs−1 xs−1 + ... + c1 x + c0
where the coefficients ct are computed from
ct =
i+j=t
ai bj
for t = 0, 1, ..., s
Arithmetic in Zm [x]
Example:
Consider p(x) = 2x2 + 3x + 4 as a polynomial in Z5 [x]. Find the value of p(4).
Example:
Let p(x) = 2x2 + 3x + 4 and q(x) = 5x + 6 be polynomials in Z7 [x].
• Find the sum p(x) + q(x).
• Find the product p(x)q(x).
Example:
If for some reason, we have coefficients outside the ring (say we start out with a polynomial in
Z[x] but decide to consider it in Z5 [x]), mod the coefficients into range. In particular, if negative
integers appear, replace them with the appropriate additive inverses for the coefficient ring.
What does the polynomial p(x) = 7x3 − 4x − 2 look like in Z5 [x]?
Example:
The previous convention is useful for doing polynomial subtraction: p(x) − q(x) is the sum of p(x)
and the additive inverse of q(x).
Let p(x) = 2x2 + 3x + 4 and q(x) = 5x + 6 be polynomials in Z7 [x]. Find p(x) − q(x).
Algebraic propeties of R[x]
Proposition:
If R is a commutative ring, then R[x] is a commutative ring.
• Closure of addition and multiplication.
Since the coefficients for each term are are computed by ai +bi (addition), and ct = i+j=t ai bj
(multiplication), the results of those operations have coefficients in whatever ring they started
out in (since the rings are closed under those operations).
• Addition is commutative.
p(x) + q(x) =
k
i=0
i
(ai + bi )x =
k
i=0
(bi + ai )xi = q(x) + p(x)
• Addition is associative.
p(x) + [q(x) + r(x)] =
k
i=0
[ai + (bi + ci )]xi =
k
i=0
[(ai + bi ) + ci ]xi = [p(x) + q(x)] + r(x)
• The additive identity is the zero polynomial: p(x) = 0. For q(x) of degree n
p(x) + q(x) =
n
i=0
(0 + bi )xi =
n
i=0
bi xi = q(x)
• Additive inverses exist. The additive inverse of p(x) =
by
n
i=0
ai xi , denoted −p(x), is defined
n
i=0
(−ai )xi
Here, −ai is the additive inverse of ai in whatever ring the ai are coming from. Then
p(x) + [−p(x)] =
n
i=0
i
[ai + (−ai )]x =
n
0xi = 0
i=0
• Multiplication distributes over addition.
Follows from the computation of coefficients:
ct =
i+j=t
ai (bj + cj ) =
(ai bj + ai cj ) =
i+j=t
i+j=t
ai bj +
i+j=t
ai cj
• Multiplication is associative.
Follows from the computation of coefficients, but it’s a bit messy because of the double sum:
For the product p(x)q(x), coefficients are
cients
r+s=t
i+j=r ai bj cs =
i+j=t
r+s=t
ai bj , and product [p(x)q(x)]r(x) has coeffi
i+j=r
ai bj cs =
i+j+s=t ai bj cs
For the product q(x)r(x), coefficients are r+s=j br cs , and product p(x)[q(x)r(x)] has coefficients
i+j=t ai
r+s=j br cs =
i+j=t
r+s=j ai br cs =
i+r+s=t ai br cs
• Multiplication is commutative.
The coefficients in the computation of p(x)q(x) are given by ct =
i+j=t bi aj ... which are the coefficients in the computation of q(x)p(x).
i+j=t
ai bj = ct =
Proposition:
If R is a commutative ring with unity, then R[x] is a commutative ring with unity.
The constant polynomial p(x) = 1 is the multiplicative identity; under the definition of multiplication, 1q(x) = q(x).
Note that all our key examples, Z[x], Q[x], R[x], C[x], and the Zm [x] rings fit this description.
Proposition:
If R is an integral domain, then R[x] is an integral domain.
Recall the definining feature of integral domain is “has no zero divisors.” If we have a product
p(x)q(x) = 0, then either p(x) or q(x) (or both) must be the zero polynomial. If p(x) = 0 and
q(x) = 0, then there is no way for the product of the leading coefficients to be zero if those
coefficients come from an integral domain; i.e. if p(x) is degree n (so an = 0) and q(x) is degree
m (so bm = 0), the first term in the product is an bm xm+n . Since an and bn are nonzero elements
of an integral domain, their product cannot be zero, and so p(x)q(x) cannot be the zero polynomial.
Finally:
If R is a field, then R[x] is... not a field.
A ring of polynomials can’t be a field...because at this point, we require multiplicative inverses.
And no polynomials other than degree zero polynomials have multplicative inverses. If p(x) = C
1 is its multiplicative inverse. However, if p(x) = xn for
(a constant polynomial), then p−1 (x) = C
n > 0, the only quantity for which p(x)q(x) = 1 is q(x) = 1 ... and 1 is not a polynomial.
p(x)
p(x)
There is no polynomial q(x) such that p(x)q(x) = 1 under the rules of polynomial multiplication,
and the multiplicative inverse property is not satsfied.