projective dimension

**Veve** · May 22nd 2009, 07:54 AM

I have to find the projective dimension of Z/pZ as Z-module, Z/pZ-module and Z/(p^2)Z-module, p is prime.Could anyone help me?

**NonCommAlg** · May 22nd 2009, 09:09 AM

Originally Posted by Veve

I have to find the projective dimension of Z/pZ as Z-module, Z/pZ-module and Z/(p^2)Z-module, p is prime.Could anyone help me?

let $M=\mathbb{Z}/p\mathbb{Z}.$ clearly $M$ as $\mathbb{Z}/p\mathbb{Z}$ module is free and so projective. thus $\text{proj} \cdot \dim_{\mathbb{Z}/p\mathbb{Z}} M = 0.$

$M$ as $\mathbb{Z}$ module is torsion and so it's not projective. now consider the exact sequence $0 \to p\mathbb{Z} \to \mathbb{Z} \to M \to 0.$ as $\mathbb{Z}$ modules: $p\mathbb{Z} \cong \mathbb{Z}.$ hence $p\mathbb{Z}$ is projective. thus $\text{proj} \cdot \dim_{\mathbb{Z}} M = 1.$

finally looking at $M$ as $\mathbb{Z}/p^2 \mathbb{Z}$ module we have this exact sequence: $0 \to p \mathbb{Z}/p^2 \mathbb{Z} \to \mathbb{Z}/p^2 \mathbb{Z} \to M \to 0.$ but as $\mathbb{Z}/p^2 \mathbb{Z}$ module: $p \mathbb{Z}/p^2 \mathbb{Z} \cong M.$ therefore: $\text{proj} \cdot \dim_{\mathbb{Z}/p^2 \mathbb{Z}} M = \infty.$

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