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I'm new to homological algebra. I was studying projective modules from Brown's book and was trying to work out some examples. I read here that $\mathbb{Q}$ is not a projective module over $\mathbb{Z}$, but I was wondering if $\mathbb{Q}G$ is a projective module over $\mathbb{Z}G$ if $G=\mathbb{Z}/n$ for any choice of integer $n\ge 1$. I do not know how to prove/disprove this. Could you please guide?

PS: I asked this question on Math Overflow a few hours back but unfortunately, it was severely downvoted for reasons of being "ill-formed" and "inappropriate for Math Overflow" rather than being improved or migrated here. I hope someone here could be kind to help me.

Jenna Fargo
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2 Answers2

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If $\mathbb{Q}G$ is projective over $\mathbb{Z}G$, it is a direct summand of a free $\mathbb{Z}G$-module $F$. But then $F$ is a free abelian group. Can $\mathbb{Q}$ be a subgroup of a free abelian group?

Matthé van der Lee
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To determine if $\mathbb{Q}G$ is projective over $\mathbb{Z}G$, we check if it’s a direct summand of a free $\mathbb{Z}G$-module. Take $y = \frac{1}{2} \cdot 1_G \in \mathbb{Q}G \setminus \mathbb{Z}G$. If $\mathbb{Q}G$ were projective, it must be projective as a $\mathbb{Z}$-module. But as a $\mathbb{Z}$-module, $\mathbb{Q}G \cong \bigoplus_{g \in G} \mathbb{Q}$, and $\mathbb{Q}$ is not projective over $\mathbb{Z}$ (e.g., $1$ and $\frac{1}{2}$ are linearly dependent, unlike in free modules). Since projectivity over $\mathbb{Z}G$ implies projectivity over $\mathbb{Z}$, and $\mathbb{Q}G$ is not projective over $\mathbb{Z}$, no such projectivity exists. Thus, $\mathbb{Q}G$ is not projective over $\mathbb{Z}G$.

Random Math Enthusiast
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