Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.