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

  • 4
    $\begingroup$ Can it be a submodule of a free $Z$-module? $\endgroup$ – Mariano Suárez-Álvarez Mar 23 '17 at 4:46
  • $\begingroup$ Presumably you mean $\mathbb Z/2\mathbb Z$ and not the $2$-adic integers? $\endgroup$ – Thomas Andrews Mar 23 '17 at 13:21

$\mathbb{Z}_2$ 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.

  • 2
    $\begingroup$ It is very ikely that flatness is something that the OP is not very familiar with, in view of the fact that he s strugling with projectiveness! $\endgroup$ – Mariano Suárez-Álvarez Mar 23 '17 at 5:11

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.

  • $\begingroup$ I can't seem to figure out how to get a diagonal arrow. If anyone knows better than me, feel free to edit so I can see how the code is done. Thank you in advance. $\endgroup$ – Alfred Yerger Mar 23 '17 at 5:06
  • 2
    $\begingroup$ It is easier if you take $B$ to be $\mathbb Z$, as in that case there is only the zero morphism. $\endgroup$ – Mariano Suárez-Álvarez Mar 23 '17 at 5:12
  • 1
    $\begingroup$ As far as I know, AMScd doesn't have diagonal arrows. More flexible packages like xymatrix are unfortunately not supported by MathJax (as of now.) $\endgroup$ – BW. Mar 23 '17 at 5:13
  • $\begingroup$ I appreciate both comments. I guess we'll have to live regarding the diagram. I think it's clear enough what I mean at this point. I also agree that if $B$ is $\mathbb{Z}$ life gets easier. I just went with the first thing that popped into my head. $\endgroup$ – Alfred Yerger Mar 23 '17 at 5:15

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.

  • $\begingroup$ It is also useful to notice the same exact sequence is a counter example to Z being an injective Z-module. The key property to notice is the exact sequence does not split, for the reasons I answered. $\endgroup$ – Morph Mar 23 '17 at 7:58
  • $\begingroup$ A generalization to this question is any finite abelian group is not a projective Z-module, since every projective module is a submodule of a free module hence can not have elements of finite order. This can be extended to any abelian group with torsion. However Z is a free Z-module hence projective Z-module. By the classification of modules over PID's you can prove a finitely generated Z-module is projective iff it is free. $\endgroup$ – Morph Mar 23 '17 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.