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Let $R$ be a finite commutative local ring with identity. If $M$ is a finite $R$-module it is necessarily projective?

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Do you mean $R$ is finite-dimensional? Over local rings projective is equivalent to free. – Andrew Aug 8 '12 at 21:57
No, $R$ is a finite ring. Yes, over local rings projective is equivalent to free by a Theorem of Kaplansky. – zacarias Aug 8 '12 at 22:00
up vote 5 down vote accepted

Let $p$ be a prime number. Let $R = \mathbb{Z}/p^2\mathbb{Z}$. Let $M = R/pR$. Since the number of elements of $M$ is $p$, $M$ cannot be free. Hence $M$ cannot be projective.

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As Andrew pointed out, every finitely-generated projective module over a local ring is free (in fact, the hypothesis of being finitely-generated can be dropped - this is a theorem of Kaplansky). Hence, it remains to show that there exists a non-free module. But we have the following characterisation:

A commutative ring $R$ is a field if and only if every module is free.

For a concrete example: if $R$ is a commutative ring which is not a field, let $I$ be a non-zero proper ideal. Then $R/I$ is module, which is not free (its annihilator is $I$).

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Direct counter-example: Let $R=\mathbb F_p[t]/(t^2),M=\mathbb F_p = R/(t).$ Then $M$ is finite but not free, as $M$ is torsion with $Ann(M)=(t).$

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