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Let $R$ be a commutative ring with unit.

Is there an example of a direct sum of $R$-modules $$R^{(I)} \cong K \oplus H$$ where $R^{(I)}$ is free but $K$ is not free ?

Clearly $R$ can't be a PID.

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    $\begingroup$ Actually you are asking for a projective module $K$ which is not free. $\endgroup$
    – user26857
    Jul 16, 2014 at 13:59

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Let, for example $R = \mathbb Z/(6)$. Then, as $\def\Z{\mathbb Z}\Z/(6)$-modules, $\Z/(6) \cong \Z/(2) \oplus \Z/(3)$. But a free $\Z/(6)$-module cannot have two or three elements only.

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    $\begingroup$ An example where $K$ is not free but $H$ is would be even better. ;-) $\endgroup$
    – egreg
    Jul 16, 2014 at 15:02
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The tangent bundle of the sphere is not free (hairy ball theorem), but it becomes free after being summed with the normal bundle of the sphere (which is free) - the sum of the two is just the restriction of the tangent bundle of $\mathbf R^3$ to the sphere.

This gives an example over the ring of $C^\infty$ functions on the sphere, and where the modules are the modules of $C^\infty$ sections of the corresponding bundles.

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