Assume that $1\in R$
I have a question which is as follows:
Let $R$ be a commutative integral domain and $I$ a non-zero ideal of $R$. Let $F$ be the field of fractions of $R$. Suppose that $I\oplus R^{(m)}\cong R^{(n)}$ as $R$-modules, where $R^{(k)}$ denotes $R\oplus R\oplus \cdots \oplus R$ $k$-times.
Show that the above isomorphism extends to give $F^{(m+1)}=F^{(n)}$ and hence that $n=m+1$.
Now I think that the idea of this question is to show that the field of fractions of a non-trivial ideal $I$ is equal to the field of fractions $R$, which I can do as follows (I think):
I we let $F_I$ denote the field of fractions of $I$ and $F_R$ denote the field of fractions of $R$ and take the definition that the field of fractions is the localisation at all non-zero divisors or equivalently it the field of equivalence classes of fractions then:
if we take $\frac{a}{b}\in F_I$ and $\frac{x}{y}\in F_R$ we have $\frac{a}{b}\frac{x}{y}=\frac{ax}{by}$ but as $I$ is an ideal we have that $ax,by\in I$ hence $\frac{ax}{by}\in F_I$ and hence $F_I\lhd F_R$ and so $F_R=F_I$
Extending The Isomorphism
If we have that $f:I\oplus R^{(m)}\rightarrow R^{(n)}$ is an isomorphism then we may extend this to be an isomorphism of $f':F_{I\oplus R^{(m)}}\rightarrow F_{R^{(n)}}$ by $f'\left ( \frac{r_1}{r_2} \right )= \frac{f(r_1)}{f(r_2)}$ for $r_1,r_2\in I\oplus R^{(m)}$
Showing $m+1=n$
Now the field of fractions of a direct sum is the sum of the field of fractions so that if $F_{I\oplus R^{(m)}}\cong F_{R^{(n)}}$ then we have that $F_I\oplus F^{(m)}\cong F^{(n)}$.
Then as from above we have $F_I=F_R$ this gives that:
$F_R^{(m+1)}\cong F^{(n)}$
But then as these are vector spaces the dimension must be the same so that we have that $m+1=n$
Is what I have done above correct?
Thanks for any help.
(Already answered in comments) I also want to know why I can't just say that if we have:
$I\oplus R^{(m)}\cong R^{(n)}$ then we may assume w.l.o.g that $n\geq m$ and quotient by $R^{(m)}$ to get that:
$I\cong R^{(n-m)}$ and hence $I\cong R$ and so the result follows trivially?
Let $S\subset R$ be the multiplicative set $S=R\setminus \{0\}$ so that all $R$-modules and in particular all ideals $I\subset R$ have a fraction vector space $S^{-1}I$ over the field $F=S^{-1}R$.
Your equality of $R$-modules $I\oplus R^{(m)}\cong R^{(n)}$ implies after applying $S^{-1}$ the equality of $F$-vector spaces $S^{-1}I\oplus S^{-1}R^{(m)}\cong S^{-1}R^{(n)}$ and thus $F\oplus F^{(m)}\cong F^{(n)}$, which of course yields the required equality $1+m=n$.
Remark
That $F=S^{-1}I$ if $I$ contains a non-zero element $0\neq i \in I$ follows from the formula $\frac ab=(ib)^{-1} (ia) \; (a,b\neq0\in R)$