Non-Free Finitely Generated Injective Modules over a Local Ring

I was wondering if someone could be so kind as to provide an example of a local ring $(R,\frak{m})$ and a non-free finitely generated injective module over $R$. Thank you very much! I tried searching everywhere online, but my attempts have been a total failure so far.

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Let $k$ be a field. Then $(k, (0))$ is a local ring, and $k$ is a finitely-generated injective (and projective and free!) $k$-module. – Zhen Lin Dec 10 '12 at 18:26
Did you try Wikipedia? en.wikipedia.org/wiki/Injective_module Several examples there. – Fredrik Meyer Dec 10 '12 at 18:27
Oops! I forgot to mention that the module has to be non-free. – Haskell Curry Dec 10 '12 at 18:29
The simplest non-self-injective local algebra I can think of is $R = k[x,y]/(xy,x^2,y^2), \mathfrak{m}=(x,y)$. There will be a unique indecomposable injective. Have you tried it? – m_t_ Dec 10 '12 at 19:21
@mt_: Thanks! Actually, you could make that an official answer. By the way, does this example appear in Lam's book Lectures on Modules and Rings? – Haskell Curry Dec 10 '12 at 20:04

An example would be $R = k[x,y]/(xy,x^2,y^2), \mathfrak{m}=(x,y)$, with $I$ the unique indecomposable injective.