Field of Fractions: Proof it is a Field

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Discussion Overview

The discussion revolves around the proof that the field of fractions is a field, particularly in the context of integral domains and the implications of finiteness. Participants explore the definitions and properties of fields and integral domains, as well as the relevance of the finiteness condition in the proof.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Brendan questions how the lack of finiteness in the domain affects the proof that the field of fractions is a field.
  • Some participants suggest that the finiteness of the domain is irrelevant to the proof, indicating that the proof should be similar to that of the rational numbers, which are infinite.
  • One participant clarifies that the definition of a field does not inherently include a finiteness condition and suggests demonstrating that an arbitrary non-zero element in the field of fractions has an inverse.
  • Another participant references a known result that states if an integral domain is finite, then it is a field, and outlines a proof involving the cancellation property.
  • Further elaboration is provided on the implications of injective functions in finite versus infinite sets, highlighting that injectivity does not guarantee surjectivity in infinite domains, which is relevant to the discussion of multiplicative inverses.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the finiteness condition in proving that the field of fractions is a field. While some argue it is irrelevant, others emphasize the importance of understanding the implications of finiteness in integral domains.

Contextual Notes

There are unresolved assumptions regarding the properties of injective functions in finite and infinite domains, and the discussion does not reach a consensus on the implications of these properties for the proof in question.

beetle2
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Hi guys,

I know that for integral domains with finte elements that if we show that each element has a multiplicative inverse then it is a field.

I need to show that the field of fractions is a field.

As the domain is not finite how does that effect the proof of being a field?


regards
Brendan
 
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Whether the domain is finite or not will be irrelevant unless you have a pretty wacky proof.

Your proof should probably be inspired by how you would prove that the rational numbers are a field, which is obviously infinite
 
The definition of a field is an integral domain in which each non-zero element has an inverse. I don't see where you get the finiteness condition from.
So just take an arbitrary non-zero element in the field of fractions and show that it has an inverse.
 
Is this what you meant by the first question? "If an integral domain is finite then it is a field." This is fairly well-known. To prove it: Suppose x is a nonzero element of a finite integral domain; then find positive integers k > l such that xk = xl. Then prove that xxk-l-1 = 1.
 
If adriank is correct, and you mean prove A. "If an integral domain is finite then it is a field." and B. "Why does this fail if the integral domain is not finite?" then consider the following:

First, if you have an integral domain D any a (not 0) inside D then the function
mult_a: D --> D given by
mult_a(x) = a*x
is injective -exactly- because D is an integral domain:
mult_a(x) = mult_a(y) => a*x = a*y => x=y by cancellation property of integral domains.

But if is D finite, mult_a is also surjective, so for some x, mult_a(x) = 1. So what does this mean about x?

Second, if D is not finite then you can't conclude anything: the function f: Z --> Z (Z = integers) given by f(x) = 2*x is injective but certainly not surjective, and Z is an integral domain.

Anyways, that an injective function f: A --> A is surjective if A is finite is a very important fact in mathematics. That this can fail if A is infinite is equally important. You can use this last fact to construct a (nontrivial) group G so that G x G is isomorphic to G, for example.


Skolem
 

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