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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?2011-06-29
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    Yep, field of fractions is another name for it. (I refer to Lang's Algebra.)2011-06-29
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    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.2011-06-29
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    @Peter: please add that information to the question itself, so that it is self contained.2011-06-29
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    Any finite integral domain is a field (by counting).2011-06-29
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    Thank you Jack. I do remember that a finite integral domain is a field. So the infinite example of Zev is explaining most of this kind of structure then.2011-06-29
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    @Peter: yup. The domains can be a little weirder, but are basically like that. One could have $\mathbb{F}_p[T^2,T^3]$ as a different example, or $\mathbb{F}_p[X,Y]$, but I think "polynomial ring" is a healthy idea of what sort of rings you are looking for.2011-06-29

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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.2011-06-29
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    @Mark: Well, the theorem *implies* that, if you prefer :)2011-06-29