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Lacking imagination, for understanding purposes I would like to see an example of an integral domain (with unity) that is not a field but has a quotient field of finite characteristic. If convenient an examples with finite and infinite quotient fields are appreciated.

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What do you mean exactly by quotient field? Field of fractions? –  Mariano Suárez-Alvarez Jun 29 '11 at 18:48
Yep, field of fractions is another name for it. (I refer to Lang's Algebra.) –  Peter Patzt Jun 29 '11 at 18:50
A finite domain is a field, so a finite field of fractions means the original domain was a field. Quotient fields as in quotient rings would be easy though: the integers have Z/2Z as a finite quotient ring that is a field. –  Jack Schmidt Jun 29 '11 at 18:50
@Peter: please add that information to the question itself, so that it is self contained. –  Mariano Suárez-Alvarez Jun 29 '11 at 18:51
Any finite integral domain is a field (by counting). –  Mark Bennet Jun 29 '11 at 18:51

2 Answers 2

up vote 8 down vote accepted

The ring of polynomials $\mathbb{F}_p[T]$ has a fraction field of $\mathbb{F}_p(T)$, which is of characteristic $p$. In fact, any $\mathbb{F}_p$-algebra that is an integral domain will have a fraction field of characteristic $p$.

There will not be any examples of non-field integral domains whose fraction field is a finite field, because this would imply that the original integral domain was finite, and any finite integral domain is already a field.

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Assuming you mean field of fractions...

There is no commutative domain which has a finite field of fractions and which is not itself a field: a theorem of Wedderburn asserts that a finite commutative domain is a field.

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I thought that Wedderburn's theorem was that every finite division ring is a field (ie commutative). Every (commutative with 1) finite integral domain is a field is trivial by comparison. –  Mark Bennet Jun 29 '11 at 18:57
@Mark: Well, the theorem implies that, if you prefer :) –  Mariano Suárez-Alvarez Jun 29 '11 at 18:58

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