# Can the field of fractions of an integral domain R be free as an R-module?

We know any integral domain R when extended to a quotient field F, then F is free as an F-module on the set {1}. Can this field be free as an R-module?

• If $R$ is already field, then $R\cong F$, so it will be true in that case. – Jared Apr 16 '14 at 18:44
• You seem to be confusing the quotient field (as in, quotient of a ring by a maximal ideal) and the field of fractions. – tomasz Apr 16 '14 at 20:30
• @tomasz I think that sometimes the term quotient field is used to designate what I'd call a fraction field, but really it shouldn't. – blue Apr 16 '14 at 20:47
• i mean the field of fractions. – user2902293 Apr 17 '14 at 7:42

If $F$ is free of rank one over $R$ then $F=R\cdot1=R$ and $R$ was a field to begin with.

Otherwise suppose it is free of rank greater than one, so it has at least two summands which are copies of $R$ which must be generated by some fractions, which means $R\frac{a}{b}\cap R\frac{c}{d}=0$ must hold for some values $a,b,c,d\in R$... is that possible?

Any free $R$ module is isomorphic to $F=R\oplus R\oplus R \oplus ...$

Notice that if it has more than one component it has a proper ideal $I=R\oplus O\oplus 0\oplus...$ which is not possible for a field so we must have $F=R$. Thus, $R$ also must be a field.

If you mean only $(F,+)\cong (R\oplus R...R,+)$ and do not use natural multiplication on $R\oplus R..$ then it is possible.

Example: Let $R=Z_p$ where $p$ is prime and Let $M=Z_p\oplus Z_p$ then $M$ is an free $R$ module with rank $2$ then $(F,+)\cong M$ where $F=Z_p(i)\cong Z_p[x]/(x^2+1)$. (which means that by putting appropriate $*$ on $M$, we can get $(M,*,+)$ as field )

• We're talking about the direct sum as a module not a ring so how do you speak of ideals of it? – blue Apr 16 '14 at 20:04
• @seaturtles: thanks for your warning I edited the answer. – mesel Apr 16 '14 at 20:27
• I think OP meant to ask about the field of fractions of $R$. – tomasz Apr 16 '14 at 20:33