My lecture notes state the following with no further explanation:

If $R=\mathbb{Z}$, then every R-module admits a resolution of length 1. This implies that $Tor_i^{\mathbb{Z}}$ and $Ext_{\mathbb{Z}}^{i}$ vanishes as soon as $i>1$. This property is called '$\mathbb{Z}$ has cohomological dimension one'.

So I am guessing that that "admits a resolution of length 1" here is meant to mean that "admits both an injective and a projective resolution of length 1". Or is my guess wrong? Also, is this some triviality (I couldn't show it!) or is less obvious? If it's not too difficult, I'd be interested in a formal proof as well.


You only need projective resolutions; for Ext you resolve the first argument, not the second. If $A$ is an abelian group, first consider any surjection $f : \bigoplus_I \mathbb{Z} \to A$ from a free abelian group to $A$. The kernel of this surjection is another free abelian group $\bigoplus_J \mathbb{Z}$, because subgroups of free abelian groups are free abelian. So we have a resolution

$$0 \to \bigoplus_J \mathbb{Z} \to \bigoplus_I \mathbb{Z} \to A \to 0.$$

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    $\begingroup$ It's also true, both because injective abelian groups are exactly the divisible groups, so closed under quotient, and as a consequence of the ext statement, that short injective resolutions exist. $\endgroup$ – Kevin Carlson Jun 3 at 21:38

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