Let R1、R2 be two integral domains and F1、F2 be their quotient field. Suppose F1 is isomorphic to F2, I want to know if R1 is isomorphic to R2 or |R1|=|R2|.

  1. If R1 and R2 are fields, then it is trivial.

  2. If only one of them is an integral domain, then it is not necessary for R1 isomorphic to R2. Ex: R1=Q R2=Z Their quotient fields are exact the same, but clearly they are not isomorphic.

  3. If neither of them is a field, then we also can't say R1 is isomorphic to R2. Ex: R1={\frac{a+b*(\sqrt -3)}{2}|a,b are integers and a+b is even} R2=Z[\sqrt -3]
  4. But I don't know whether |R1|=|R2|.

I think that's an interesting question to me because I find that the relationship between an integral domain and its quotient field is quite similar to a given set and the free group on it, and a basis and the free abelian group constructed by it. And all these have a certain "mapping property"(ex. free object on groups).In the free (abelian) group, we have this theorem:

Let X、Y be two sets, G1、G2 are two free groups on it, then |X|=|Y| if and only if G1 is isomorphic to G2.

So I think if the preceding 4 is true, we can get a similar conclusion about an integral domain and its quotient field.

Many thanks!


If $R$ is an integral domain and $F$ is its field of fractions, then $|R|=|F|$. Indeed, if $R$ is finite then it must be a field, so $R\cong F$ and $|R|=|F|$ trivially. If $R$ is infinite, then since every element of $F$ is a fraction of two elements of $R$, we have $|F|\leq |R|^2=|R|$. But of course $R$ naturally embeds in $F$ so $|R|\leq|F|$ as well, and so $|F|=|R|$.

So, if two integral domains have isomorphic fields of fractions (or even fields of fractions of the same cardinality), they must have the same cardinality.

  • $\begingroup$ OK,thank you very much! I have tried in this way, but I neglect the fact that | R | ²= | R |…… $\endgroup$ – Likemath Jan 17 '19 at 11:08

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