$K$ is not projective as $R$ module where $K$ is quotient field of a PID $R$.

Question

Let $R$ be a P.I.D. (not a field). Let $K$ be the quotient field of $R.$ How can I show $K$ is not projective as $R$-module?

I can only prove that $K$ is not free as $R$ module. Since $K$= $S^{-1}R$ for $S=R-\{0\}.$ But I cannot show that $K$ is not direct summand of a free module. Help me. Thanks.

Answer 1

Let $R$ be any noetherian integral domain, but not a field, (You dont have to assume PID, in that case free and projective are the same anyway) and $K$ its fraction field.

It is very easy to show that $\operatorname{Hom}_R(K,\bigoplus\limits_{i \in I} R)=0$ for any index set $I$. In particular $K$ cannot even be a submodule of a free $R$-module, let alone a direct summand.

If you want to show that the Hom-set is zero, note that $K$ is divisible, while a free $R$-module has no non-trivial divisible submodules.