How to show a module is not projective

Let $R$ be the ring $K[X]/(X^r)$ for some field $K$ and $r>1$, and consider the module $XR$. I can show that this module is not projective by observing that the ring is local, so all projective modules are direct sums of copies of $R$, and $XR$ is not one of these because it doesn't have the right dimension.

In particular, this means that $XR$ cannot satisfy the lifting property, but I haven't been able to find an example of a map that doesn't lift - can anybody think of a good example in this case?

Looking around previous questions suggests that if the module I want to show isn't projective is a quotient of the ring then the identity generally won't lift, but there doesn't seem to be a neat strategy for dealing with ideals.

-
@MartinBrandenburg, in some cases. If a ring is non-trivially a direct product, for example, there are projectives which are annihilated by some of the factors. Here $R$ is local, so this does not happen, though (because projectives are just free) –  Mariano Suárez-Alvarez Mar 21 '12 at 16:58

It's a good idea to map from a projective module onto the module you want to show is not projective. What about the homomorphism $$R\to RX,\;r\mapsto rX?$$
Since your ring is local, its projective modules are free. Since your ring has dimension $r$ over $K$, all its finitely generated projective modules have $K$-dimension divisible by $r$, yet your module $XR$ has dimension $r-1$. It follows that $XR$ is not projective.
Ouch. ${}{}{}{}$ –  Mariano Suárez-Alvarez Mar 21 '12 at 17:11