# Confusion in properties related to $M$ being $R$ module and being $\Bbb{Z}$ module

I am self-studying some basic modules contents from some introductory books about commutative algebra. When studying the properties of some special modules and special functors, like projective/injective/flat module, tensor and Hom functors, some kind of subtle things seem quite confused to me.

For simplicity let’s take $$R$$ as a commutative ring. Then I know that a $$R$$ module is also a $$\Bbb{Z}$$ module, via a "underlying/forget" functor $$U$$.

The first question about my confusion: This functor is faithful but not full, is it right？

Secondly, if $$f\colon A \to B$$ is an injective/surjective $$R$$ homomorphism, then it is also an injective/surjective $$\Bbb{Z}$$ homomorphism, because the underlying/forgetful map is the same, right?

Third, if $$M$$ is projective/injective/flat $$R$$ module, it doesn’t mean $$M$$ is also projective/injective/flat as $$\Bbb{Z}$$ module? Is it right?

I knew that the tensor product doesn’t need to be coincided, for asking a question a few days ago: Examples for $A \otimes_R B \ne A \otimes_\Bbb{Z} B$

Any examples or explanations or references are appreciated. Thanks a lot!

• 1) yes, 2) yes, 3) yes, consider $M=R=\mathbb Z/2$. Commented Aug 10, 2022 at 6:00
• @KentaS Thanks! This example seems using the fact that the category of $\Bbb{Z}$ modules has more objects then the category of $\Bbb{Z}/2$ modules. Commented Aug 10, 2022 at 14:33
• I do not believe so, $M=R=\mathbb Z\times\mathbb Z/2$ works equally as well. Commented Aug 10, 2022 at 15:04
• @KentaS $\Bbb{Z} \times \Bbb{Z}/2$ is an excellent example. I think I maybe get it now. Commented Aug 10, 2022 at 16:25

With the excellent examples given by Kenta S, I think I can clear the confusion myself now. I try to write down the details to see if I understand it correctly.

Take $$R = \Bbb{Z} \times \Bbb{Z}/2$$, the product of rings $$\Bbb{Z}$$ and $$\Bbb{Z}/2$$, is still a ring.(Rings of this kind have never come to my mind before). Then $$\Bbb{Z}$$ and $$\Bbb{Z}/2$$ are both $$R$$ modules, with multiplication defined by $$(a, b) \cdot n = a\cdot n$$ for both cases.

My first question is solved by the map defined on the hom sets:

$$U: \mathrm{Hom}_{R}(R, \Bbb{Z}/2) \to \mathrm{Hom}_{\Bbb{Z}}(R, \Bbb{Z}/2) \\ f \mapsto \text{forgetting multiplication on } f$$

being injective but not surjective. In other words, there is an abelian group homomorphism from $$R$$ to $$\Bbb{Z}/2$$ that cannot be realized as an $$R$$ module homomorphism.

For such an abelian group homomorphism $$f$$, take $$f(0, \bar{1}) = f(1, \bar{0}) = \bar{1}$$. Then $$f$$ is defined by $$f(a, \bar{b}) = (a+b)\bar{1}$$. Such an $$f$$ cannot be a $$R$$ module homomorphism for otherwise:

$$\bar{0} = f(0, \bar{0}) = f((1, \bar{0})\cdot (0, \bar{1})) = (1, \bar{0})\cdot\bar{1} = \bar{1}$$ being a contradiction.

(Btw during this question I found another question the solve my first confusion too: Is a group homomorphism a module homomorphism?)

And for the third question, I think it should be checking that

$$\Bbb{Z} \xrightarrow[]{\pi} \Bbb{Z}/2 \to 0$$

is exact while

$$\mathrm{Hom}_{\Bbb{Z}}(R, \Bbb{Z}) \xrightarrow[]{\pi_*} \mathrm{Hom}_{\Bbb{Z}}(R, \Bbb{Z}/2) \to 0$$ is not exact. This is shown by that the previous definition of $$f$$, taking $$f(0, \bar{1}) = f(1, \bar{0}) = \bar{1}$$, $$f(a, \bar{b}) = (a+b)\bar{1}$$, belongs to $$\mathrm{Hom}_{\Bbb{Z}}(R, \Bbb{Z}/2)$$, but not in the image of $$\pi_{*}$$. Otherwise suppose $$f = \pi_{*}(g)$$ for some $$g \in \mathrm{Hom}_{\Bbb{Z}}(R, \Bbb{Z})$$, i.e. $$f = \pi \circ g$$. Then $$2g(0, \bar{1}) = g(0, \bar{0}) = 0$$ giving that $$g(0, \bar{1}) = 0$$ and $$\bar{1} = f(0, \bar{1}) = \pi(g(0, \bar{1})) = \pi(0) = \bar{0}$$ makes a contradiction.

Hence we know that $$R$$ cannot be projective as $$\Bbb{Z}$$ module, but it's projective as $$R$$ module for being free with rank $$1$$. And it's easy to check that

$$\mathrm{Hom}_{R}(R, \Bbb{Z}) \xrightarrow[]{\pi_*} \mathrm{Hom}_{R}(R, \Bbb{Z}/2) \to 0$$

is exact. I think it's an example for illustrating that for the same object, in the $$\Bbb{Z}$$ module category, it has more morphsims and hence fails to be surjective.

I hope this can illustrate if I have understand the examples correctly or not. I am still verifying the examples of the cases for injective and flat module.

Thanks.