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Let $R$ be an integral domain with field of fractions $K$, and let $I$ be a fractional ideal. If $I$ is projective then every $R$-linear maps $I\to R$ is multiplication by an element of $K$.

The three definitions of a projective module I have been playing with are: \begin{align*} (i)\; &\text{Every exact sequence } 0\to M\to N\to I\to 0\text{ splits}\\ (ii)\; &\text{The existence of a lifting homomorphism}\\ (iii)\;&I \text{ is the direct summand of a free module}. \end{align*} I am not sure what modules to use for $M$ and $N$, nor which definition should be used.

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Why can't $I\cap R=\emptyset$? –  Ward Dec 12 '12 at 20:34
    
@Ward There exists $\alpha\in R$ with $\alpha I\subset R$. Since $I$ is an $R$-module, we thus also have that $\alpha I\subset I$. Hence $\alpha I\subset I\cap R \Rightarrow I\cap R \neq\emptyset$. –  John Martin Dec 12 '12 at 21:44
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up vote 2 down vote accepted

The projectivity of $I$ is not needed. One can suppose $I\ne 0$. First for all $b/c\in I$, we have $b\in I$ and $ \phi(b)=\phi(c(b/c))=c\phi(b/c)$, so $$\phi(b/c)=\phi(b)/c.$$ Now fix $a_0\in I\cap R$ non-zero. Let $x=b/c\in I$. Then $$ a_0\phi(x)=\phi(a_0x)=\phi(a_0b/c)=b\phi(a_0/c)=b\phi(a_0)/c. $$ So $$\phi(x)=(\phi(a_0)a_0^{-1})x, \quad \forall x\in I.$$

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