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Home » Math Homework Help » Algebra Homework Help » Pigeon Hole Principle
Pigeon Hole Principle
If n objects are distributed over m places and if n > m, then some places receives at least two objects.

Theorem: Prove that finite integral domain is a field.

Proof: We know that an Integral domain is a commutative ring R without zero division, i.e. if a, b R then ab = 0 either a = 0 or b = 0.

A field, on the other hand, is a commutative ring with unit element in which every non-zero element has a multiplicative inverse in the ring.

Let D be a finite integral domain. In order to prove that D is a field we must show that

1. There exists an element 1 D such that a . 1 = a for every a D i.e.

2.For every element a ≠ 0, a D, there exists an element b D such that ab = 1, i.e. the multiplicative inverse of every elements in D exists.

Let O(D) = n. let x1, x2, x3, …. , xn be all the elements of D and suppose that a ≠ 0, a D.

The elements x1a, x2a, x3a, … , xna D (closure property in D).

We claim that these elements are all distinct.

Let if possible xia = xja for i ≠ j.

xja – xja = 0 for i ≠ j.

(xi – xj)a = 0 for i ≠ j.

Since D is an integral domain and a ≠ 0, so

(xi – xj)a = 0 xi – xj = 0

xi = xj contradicting i ≠ j.

Thus x1a, x2a, x3a, …. , xna are n distinct elements belonging to D, which has exactly n elements. By the pigeon-hole principle, all these elements must account for all the elements of D. Thus every element y D must be of the form xia for some xi. In particular a D, so

a = xi0a for some xi0 D.

∴ a = x10a = ax10. (∵ D is a commutative ring)

Now we shall show that x10 acts as a unit element for every element of D. Let y be an arbitrary element of D.

∴ y D y = x1a for some xi D and so.

yxi0 = (xia)xi0 = xi(axi0)            (associative law holds in D).

= xia = y     (∵ ax10 = a).

Thus x10 is a unit element for D and we write it as 1. Now 1 D, so it must be multiple of a there exists a, b D such that

1 = ab b is the inverse of a ∀ a ≠ 0 D.

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