Is $\mathbb{Z}_2$ a projective $\mathbb{Z}$-Module?

Question

I suspect that it is not but I am not sure how to prove that something is not projective.

Answer 1

$\mathbb{Z}_2$ is not projective as a $\mathbb{Z}$ module.

Since every projective module is flat (projective modules are direct summands of free modules, free modules are flat, and direct summands of flat modules are flat), it suffices to show that $\mathbb{Z}_2$ is not flat as a $\mathbb{Z}$ module.

To see this, consider the map $$ \mathbb{Z} \xrightarrow{2} \mathbb{Z} $$ given by multiplication by 2. This is injective, but when tensored with $\mathbb{Z}_2$ it is no longer injective because the image is 0.

Answer 2

Another way to see it from the definition.

The universal property of projectives is that given a surjection $B \to C$ and a map $M \to C$, there must exist a lift $M \to B$ so that the composition $M \to B \to C$ is the same as given map (I don't know how to make commutative diagrams but I'll try to figure it out).

$\require{AMScd}$ \begin{CD} @. M \\ @. @VVV \\ B @>>>C\\ \end{CD}

And we want there to be a map $M \to B$ making the diagram commute.

In this case $\mathbb{Z}_2$ is $M$. Consider the surjection $\mathbb{Z}_4 \to \mathbb{Z}_2$ given by reducing elements mod $2$, and let the given map be the identity map. The only homomorphisms from $\mathbb{Z}_2 \to \mathbb{Z}_4$ are the $0$ homomorphism and the one given by doubling. The first one obviously would not make this diagram commute. On the other hand, the map given by doubling would have $1 \mapsto 2 \mapsto 0$, so the diagram fails to commute in both cases. So the universal property fails to hold.

Answer 3

Consider the exact sequence defined by the function $n:\mathbb{Z} \rightarrow \mathbb{Z}$; n(z)=nz, and the function $\pi: \mathbb{Z}\rightarrow \mathbb{Z} / n\mathbb{Z}$ ; $\pi(z)=z$(mod n). Notice the $\mathbb{Z}$-module homomorphisms from $\mathbb{Z} / n\mathbb{Z}$ to $\mathbb{Z}$ are trivial. On the other hand $\mathbb{Z}$-module homomorphisms from $\mathbb{Z} / n\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ since the number 1 can be sent to any element.

Why is any of this useful information? Because applying the $Hom_{\mathbb{Z}}(\mathbb{Z} / n\mathbb{Z},\star )$ functor to the original exact sequence gives $0$ maps to $0$ under $n'$ and 0 maps to $\mathbb{Z}/n\mathbb{Z}$ under $\pi'$, which can not be a surjective map, hence the sequence cannot be exact. A criteria for $\mathbb{Z}/n\mathbb{Z}$ to be a projective $\mathbb{Z}$ module is that the $Hom_{\mathbb{Z}}(\mathbb{Z} / n\mathbb{Z},\star )$ functor is exact.