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 : P_n ≠ 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:

a_nxⁿ + a_n-1x^n-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 }

Definition 2: Polynomial Addition

(P + Q)_n = P_n + Q_n

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) P_i*Q_j

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.

(2) Commutative Property for Addition:

P_n + Q_n = (P + Q)_n = (Q + P)_n = Q_n + P_n

(3) Associative Property for Addition

(P_n + Q_n) + R_n = (P + Q)_n + R_n = ([P + Q] + R)_n = (P + [Q + R])_n

= P_n + (Q + R)_n = P_n + (Q_n + R_n)

(4) Additive Identity

This is the 0 polynomial. P_n + 0 = (P + 0)_n = P_n

(5) Additive Inverse

P_n + -P_n = (P + -P)_n = 0

(6) Associative Property for Multiplication

∑(i+j+k=n) (P_i*Q_j)*R_k = [(P*Q)*R]_n = [P*(Q*R)]_n = ∑ (i+j+k=n)P_i*(Q_j*R_k)

(7) Distributive Property for Multiplication

∑(i+j=n) P_i(Q_j + R_j) = ∑(i+j=n)P_i(Q + R)_j = [P(Q+R)]_n
= (PQ + PR)_n = ∑(i+j=n)(P_iQ_j) + ∑(i+j=n)(P_iR_j)

We can use the same argument to prove:

∑(i+j=n) (Q_j + R_j)P_i = ∑(i+j=n)(Q_jP_i) + ∑(i+j=n)(R_jP_i)

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

Assume that A is commutative.

∑(i+j=n)P_iQ_j = (PQ)_n = (QP)_n = ∑(i+j=n)Q_jP_i

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001