Tuesday, December 04, 2007

Polynomials defined

A polynomial expression is a mathematical construct that if often used and rarely defined. The most common form is a polynomial in one variable which can be represented as:
a0xn + a1xn-1 + ... + a(n-1)x + an

In today's blog, I will present a more formal definition and then show the very fundamental result that a polynomial is a ring. If you are not familiar with rings, then please start here.

The content in today's blog is taken from Jean Tignol's excellent Galois' Theory of Algebraic Equations.

Definition 1: Polynomial in one indeterminant with coefficients in ring A

P: N → A such that { n ∈ N : Pn ≠ 0 } is finite. The set of all polynomials in one indeterminant with coefficients in ring A can be denoted as A[X].

Example 1:

Let us consider the polynomial:

a0 + a1X + ... + anXn

In this case, each ai is a coefficient in ring A.

We can see that for each i, that i ∈ N and that there exists an integer n ∈ N such that:

if i ≤ n, then i ∈ N maps to ai.

if i is greater than n, then i ∈ N maps to 0.

(P + Q)n = Pn + Qn

In other words, each of the maps corresponding to the same natural number are added together to form a new map.

Definition 3: Multiplication of Polynomials

(PQ)n = ∑ (i+j=n) Pi*Qj

In other words, each of the maps corresponding to the multiplication of all maps where the index of one plus the index of the other = n.

Lemma 1: If A is a ring, then A[X] is a ring which is commutative if and only if A is commutative

Proof:

(1) A[X] is a ring since:

(a) A[X] has a commutative operation of addition

P + Q = Pi + Qi for all i.

Q + P = Qi + Pi for all i.

Since A is a ring, Pi + Qi = Qi + Pi

(b) A[X] has an associative rule for addition

(P + Q) + R = (Pi + Qi) + Ri for all i

P + (Q + R) = Pi + (Qi + Ri) for all i

(Pi + Qi) + Ri = Pi + (Qi + Ri) since A is a ring.

(c) A[X] has an additive identity

Let Q be a polynomial such that all Qi map to 0.

Then for any polynomial P, it is clear that P + Q = Pi + 0 = Pi for all i.

(d) A[X] has an additive inverse

For any polynomial P:

Let Q be a polynomial derived from P such that for all i: Qi = -Pi

Then

P + Q = Pi + -Pi = 0 for all i.

(e) A[X] has an associative rule for multiplication

PQ = ∑ (i+j=n) Pi*Qj for all n

(PQ)R = ∑ (i+j+k=n) (Pi*Qj)Rk for all n =

= ∑ (i+j+k=n) Pi*(QjRk) for all n

QR = (j+k=n) Qj*Rk for all n

P(QR) = ∑ (i+j+k=n) Pi*(QjRk) for all n

(f) A[X] has a distributive rule

Q + R = Qj + Rj for all j

P(Q+R) = ∑ (i+j=n) Pi*(Qj + Rj) =

= ∑ (i+j=n) PiQj + PiRj

PQ = ∑ (i+j=n) Pi*Qj for all n

PR = ∑ (i+j=n) Pi*Rj for all n.

PQ + PR = ∑ (i+j=n) PiQj + ∑(i + j=n) PiRj for all n =

= ∑(i+j=n) PiQj + PiRj

(Q+R)P = ∑ (i+j=n) (Qj + Rj)*Pi =

= ∑ (i+j=n) PiQj + PiRj

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

(a) Assume that A is a commutative ring

(b) PQ = ∑ (i+j=n) Pi*Qj for all n =

= ∑ (i+j=n) Qj*Pi for all n = QP

(3) If A[X] is a commutative ring, then A is a commutative ring

(a) Assume that A[X] is a commutative ring

(b) QP = PQ = ∑ (i+j=n) Pi*Qj for all n

(c) QP = ∑ (i+j=n) Qj*Pi for all n

(d) So, therefore for all n, ∑ (i+j=n) Pi*Qj = ∑ (i+j=n) Qj*Pi

QED

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