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I read that the global dimension of $\mathbb Z/4\mathbb Z$ is not finite. I think that it's because that $4=2\cdot 2$ and $(2,2)\neq 1$, hence $\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z$ is not $\mathbb Z/4\mathbb Z$.

Is it the reason for this ? If it's not too complicated, I would really like to see an explanation for why the global dimension is not finite.

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up vote 1 down vote accepted

It's not too complicated (I got it from Lam, Lectures on Modules and Rings, Lemma 5.16, who claims it is an idea of Kaplansky):

Consider the exact sequence $0\to\mathbb Z/2\mathbb Z\stackrel{2\cdot }{\to} \mathbb Z/4\mathbb Z\to \mathbb Z/2\mathbb Z\to 0$. Then $\mathbb Z/2\mathbb Z$ is not projective, since it is not a direct summand of $\mathbb Z/4\mathbb Z$ as you remarked. Therefore this yields that $\mathbb Z/2\mathbb Z$ has infinite projective dimension.

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Is hard to believe that the first map is the multiplication by $2$ since this sends $\mathbb Z/2\mathbb Z$ to $0$, but excepting this everything is ok. –  user26857 Sep 26 '13 at 18:48
    
@YACP it is the map induced by multiplication by $2\colon\mathbb{Z}\to \mathbb{Z}$ and then modding out. –  Julian Kuelshammer Sep 27 '13 at 8:52
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Actually the first non-zero term of the sequence is $2\mathbb Z/4\mathbb Z$ and this is isomorphic to $\mathbb Z/2\mathbb Z$ by multiplication. Anyway, my remark was made in order to make it clear. –  user26857 Sep 27 '13 at 9:47
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