## Thursday, January 29, 2009

### Ring of Polynomials

If you are not comfortable with the idea of rings, start here.

Definition 1: Polynomial in one indeterminate with coefficients in a ring A

P : N → A such that P = { n ∈ N : Pn ≠ 0 }

where N is the set of natural numbers (that is, 0, 1, 2, ... )

As a convention, this mapping is usually represented in the form:

anxn + an-1xn-1 + ... + a1x + a0

where n ∈ N.

Example 1:

5x2 + 3x + 2 is a polynomial.

5,3,2 are all integers which is a ring (see here for details on integers)

{ 2 → 5, 1 → 3, 0 → 2 }

(P + Q)n = Pn + Qn

Example 2:

(5x2 + 3x + 2) + (3x2 + x + 5) = (5 + 3)x2 + (3+1)x + (2+5) = 8x2 + 4x + 7

Definition 3: Polynomial Multiplication

(PQ)n = ∑(i+j=n) Pi*Qj

Example 3:

(5x2 + 3x + 2)(3x2 + x + 5) = (5*3)x(2+2) + (5*1 + 3*3)x(2+1) + (5*5 + 3*1 + 2*3)x(2+0) + (3*5 + 2*1)x(1+0) + (2*5)x0

Definition 4: A[X]

The set of all polynomials with coefficients in A

Example 4:

5x2 + 3x + 2 ∈ Z[X] where Z is the set of all integers.

Lemma 1:

If A is a ring, then A[X] is a ring. If A is a commutative ring, then A[X] is a commutative ring.

Proof:

(1) I will show that A[X] has all the properties of a ring.

Pn + Qn = (P + Q)n = (Q + P)n = Qn + Pn

(Pn + Qn) + Rn = (P + Q)n + Rn = ([P + Q] + R)n = (P + [Q + R])n

= Pn + (Q + R)n = Pn + (Qn + Rn)

This is the 0 polynomial. Pn + 0 = (P + 0)n = Pn

Pn + -Pn = (P + -P)n = 0

(6) Associative Property for Multiplication

∑(i+j+k=n) (Pi*Qj)*Rk = [(P*Q)*R]n = [P*(Q*R)]n = ∑ (i+j+k=n)Pi*(Qj*Rk)

(7) Distributive Property for Multiplication

∑(i+j=n) Pi(Qj + Rj) = ∑(i+j=n)Pi(Q + R)j = [P(Q+R)]n
= (PQ + PR)
n = ∑(i+j=n)(PiQj) + ∑(i+j=n)(PiRj)

We can use the same argument to prove:

∑(i+j=n) (Qj + Rj)Pi = ∑(i+j=n)(QjPi) + ∑(i+j=n)(RjPi)

(8) Commutative Property for Multiplication is true if A has the Commutative Property for Multiplication

Assume that A is commutative.

∑(i+j=n)PiQj = (PQ)n = (QP)n = ∑(i+j=n)QjPi

QED

References
• Jean-Pierre Tignol, , World Scientific, 2001