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.
Non-Free Finitely Generated Injective Modules over a Local Ring
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commutative-algebra
ring-theory
modules
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1Let $k$ be a field. Then $(k, (0))$ is a local ring, and $k$ is a finitely-generated injective (and projective and free!) $k$-module. – 2012-12-10
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0Did you try Wikipedia? http://en.wikipedia.org/wiki/Injective_module Several examples there. – 2012-12-10
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1Oops! I forgot to mention that the module has to be non-free. – 2012-12-10
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0The 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? – 2012-12-10
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0@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*? – 2012-12-10
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0@HaskellCurry you're welcome. I don't know if it's in Lam's book I'm afraid. – 2012-12-10
1 Answers
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An example would be $R = k[x,y]/(xy,x^2,y^2), \mathfrak{m}=(x,y)$, with $I$ the unique indecomposable injective.