1

Let $R$ be a ring (not necessarily commutative and not necessarily with unit). Recall the definition of $R$-$\bmod$ as an abelian group $A$ on which $R$ acts on the left respecting the following axioms:

  1. $r \cdot (a+b) = r\cdot a + r\cdot b$
  2. $(r+s)\cdot(a) = r\cdot a + r\cdot b$
  3. $(r\cdot s)\cdot a = r\cdot(s\cdot a)$

for each possible choice of $r, s \in R$ and $a \in A$. Let's suppose to have $S$ subring of $R$. In this case $A$ can be seen as a $S$-$mod$ by restricting the previous action.

And now comes the question.

If $A$ is projective as $R$-$\bmod$, will it be projective again if seen as $S$-$\bmod$ using the previous identification?

This question arises spontaneously since I am studying some basic topics from Homological Algebra and properties like that could be helpful to solve some exercises and to have a better vision of the topic.

I tried to solve it by myself using explicitly the definition of projective object as a module that satisfies the requested commutative diagram, but withoun making any relevant progress. On the one hand I am pretty sure that the answer is no, but on the other hand I hope should be yes because it seems to be useful.

Any hint is appreciated.

Michael Hardy
  • 1
Biagio
  • 413
  • 1
    Well, what about some easy examples, like $\mathbb{Z} \subset \mathbb{Q}$? – Zhen Lin Jun 13 '15 at 17:15
  • @ZhenLin seems to be a good point. May ask you for more details regarding the reason for which $\mathbb{Q}$ is not projective as a $\mathbb{Z}$-module, please? [Just to be sure to have understood: you are suggesting to pick $\mathbb{Q}$ as a $\mathbb{Q}$ module - clearly projective because free - and then think about it as a $\mathbb{Z}$-module. Here, it should be no more projective, right?] – Biagio Jun 13 '15 at 17:37
  • 2
    @Biagio: There are a couple of ways to see that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module. See for example http://math.stackexchange.com/questions/506256/prove-that-operatornamehom-bbbz-bbbq-bbbz-0-and-show-that-b?rq=1. – Kopper Jun 13 '15 at 17:42
  • As another comment, $\mathbb{Q}$ is a flat $\mathbb{Z}$-module, which is maybe a helpful exercise – Kopper Jun 13 '15 at 17:44
  • Yes, that's clear [regarding the projectivity]. Speaking about the flatness, yes, I see the point - it is helpful to know. Dear users "Kopper" and "Zhen Lin", thank you very much for your help and clever hints. Cheers – Biagio Jun 13 '15 at 17:48

0 Answers0