# [Modules] Show that Q/Z is injective and not projective

Consider the set of all rational numbers Q as a Z-module, Z as a submodule of Q and M:=Q/Z the quotient. Show that M is an injective module, but not a projective module.

## Answer

First we show that M is injective: take $f: M_1 \to M$ and $i: M_1 \to M_2$ an arbitrary injection. We want to show that $f$ extends to $M_2$. Suppose we can extend to $f_H$ on some submodule $M_1 \subset H \subset M_2$, then taking $m \in M_2$ either $Span(m) \cap H = 0$ or $nm \in H$ for some $n \in \mathbb{N}$. In the former case we can extend arbitrarily, in the latter case then we can enforce $f(nm) = nf(m)$, so we can define $f(m) = \frac{1}{n} f(nm)$ using the fact that $\mathbb{Q}/\mathbb{Z}$ is divisible.

Since we see that we can extend at each stage, we use the axiom of choice. Specifically Zorn's lemma. Let $(H, \phi)$ be the set of pairs such that $\phi$ is an extension of $f$ to $H$, and $H$ is as above. Define an order such that $(H_1, \phi_1) < (H_2, \phi_2)$ if $H_1 \subset H_2$ and $\phi_2|_{H_1} = \phi_1$. Then if $(H_i, \phi_i), i \in I$ is an ascending chain we can always take the union $(\cup_i H_i, \cup \phi_i)$, so every chain has an upper bound. Thus there is some maximal pair $(H_\infty, \phi_\infty)$. Suppose $H_\infty$ was not $M_2$, then we could take some element $m \in M_2 - H_{\infty}$ and use the above process to extend, thus $H_\infty = M_2$ and injectivity is proved. In fact we only used that $\mathbb{Q}/\mathbb{Z}$ was a divisible Abelian group, so we have shown a fortiori that any divisible Abelian group is injective in the category of Abelian groups.

To see that $\mathbb{Q}/\mathbb{Z}$ is not projective, apply $Hom(\mathbb{Q}/\mathbb{Z}, -)$ to the exact sequence $$0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0,$$ then we get the sequence $$0 \to 0 \to 0 \to \text{Hom}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z}).$$ Since the map from $0 \to Hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ is obviously not surjective, $\mathbb{Q}/\mathbb{Z}$ cannot be projective. Once again, the only thing we have actually used is the following: no torsion abelian group maps to $\mathbb{Q}$ or $\mathbb{Z}$, and every torsion abelian group has at least one nonzero map to $\mathbb{Q}/\mathbb{Z}$ (one can see this from, for instance, another Zorn's lemma argument or the aforementioned injectivity of $\mathbb{Q}/\mathbb{Z}$). So we've actually proved that: no torsion abelian group can possibly be projective.

- answered
- 680 views
- $5.00

### Related Questions

- Tensor Product
- Let $R$ be an integral domain and $M$ a finitely generated $R$-module. Show that $rank(M/Tor(M))$=$rank(M)$
- Question on a subgroup of permutations
- Homomorphism
- Mean Value Theorem
- Suppose that $(ab)^3 = a^3 b^3$ for all $a, b \in G$. Prove that G must be an abelian goup [Group Theory].
- Study of Affine Group
- [Modules] Show that $h_3$ is injective given comutative diagram