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 + a₁x^n-1 + ... + a_(n-1)x + a_n

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 : P_n ≠ 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 + ... + a_nXⁿ

In this case, each a_i 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 a_i.

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

Definition 2: Addition of Polynomials

(P + Q)_n = P_n + Q_n

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) P_i*Q_j

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 = P_i + Q_i for all i.

Q + P = Q_i + P_i for all i.

Since A is a ring, P_i + Q_i = Q_i + P_i

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

(P + Q) + R = (P_i + Q_i) + R_i for all i

P + (Q + R) = P_i + (Q_i + R_i) for all i

(P_i + Q_i) + R_i = P_i + (Q_i + R_i) since A is a ring.

(c) A[X] has an additive identity

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

Then for any polynomial P, it is clear that P + Q = P_i + 0 = P_i 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: Q_i = -P_i

Then

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

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

PQ = ∑ (i+j=n) P_i*Q_j for all n

(PQ)R = ∑ (i+j+k=n) (P_i*Q_j)R_k for all n =

= ∑ (i+j+k=n) P_i*(Q_jR_k) for all n

QR = (j+k=n) Q_j*R_k for all n

P(QR) = ∑ (i+j+k=n) P_i*(Q_jR_k) for all n

(f) A[X] has a distributive rule

Q + R = Q_j + Rj for all j

P(Q+R) = ∑ (i+j=n) P_i*(Q_j + R_j) =

= ∑ (i+j=n) P_iQ_j + P_iR_j

PQ = ∑ (i+j=n) P_i*Q_j for all n

PR = ∑ (i+j=n) P_i*R_j for all n.

PQ + PR = ∑ (i+j=n) P_iQ_j + ∑(i + j=n) P_iR_j for all n =

= ∑(i+j=n) P_iQ_j + P_iR_j

(Q+R)P = ∑ (i+j=n) (Q_j + R_j)*P_i =

= ∑ (i+j=n) P_iQ_j + P_iR_j

(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) P_i*Q_j for all n =

= ∑ (i+j=n) Q_j*P_i 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) P_i*Q_j for all n

(c) QP = ∑ (i+j=n) Q_j*P_i for all n

(d) So, therefore for all n, ∑ (i+j=n) P_i*Q_j = ∑ (i+j=n) Q_j*P_i

QED

References

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