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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
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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? –  mt_ 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

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up vote 2 down vote accepted

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

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