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!

  • 4
    $\begingroup$ 1) yes, 2) yes, 3) yes, consider $M=R=\mathbb Z/2$. $\endgroup$
    – Kenta S
    Commented Aug 10, 2022 at 6:00
  • $\begingroup$ @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. $\endgroup$
    – onRiv
    Commented Aug 10, 2022 at 14:33
  • 1
    $\begingroup$ I do not believe so, $M=R=\mathbb Z\times\mathbb Z/2$ works equally as well. $\endgroup$
    – Kenta S
    Commented Aug 10, 2022 at 15:04
  • $\begingroup$ @KentaS $\Bbb{Z} \times \Bbb{Z}/2$ is an excellent example. I think I maybe get it now. $\endgroup$
    – onRiv
    Commented Aug 10, 2022 at 16:25

1 Answer 1


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.



You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .