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.
A ring and a quotient field with finite characteristic
2
$\begingroup$
abstract-algebra
-
1What do you mean exactly by quotient field? Field of fractions? – 2011-06-29
-
1Yep, field of fractions is another name for it. (I refer to Lang's Algebra.) – 2011-06-29
-
2A 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
-
0@Peter: please add that information to the question itself, so that it is self contained. – 2011-06-29
-
0Any finite integral domain is a field (by counting). – 2011-06-29
-
0Thank 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
-
0@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
2 Answers
8
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.
3
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.
-
3I 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
-
0@Mark: Well, the theorem *implies* that, if you prefer :) – 2011-06-29