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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.

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@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

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up vote 5 down vote accepted

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 $$R\to RX,\;r\mapsto rX?$$

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Thank you! I had foolishly thought earlier that I could lift the identity over this, so I'm glad you made me reconsider it! –  Matt Pressland Mar 21 '12 at 16:56

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.

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Sure, this is what I say in the first paragraph. I was looking particularly for a non-lifting map partly because I was wondering if there was one that could be easily written down, and partly because that might apply better to other examples. Thinking about it, maybe my title is misleading, but I can't think of a better concise description of the question. –  Matt Pressland Mar 21 '12 at 17:04
    
Ouch. ${}{}{}{}$ –  Mariano Suárez-Alvarez Mar 21 '12 at 17:11
    
Sorry! It's probably worth having this answer here though in case somebody finds this question later. –  Matt Pressland Mar 21 '12 at 17:15

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