8
$\begingroup$

Let $R$ be a commutative unital ring and $M$ an $R$-module. Then $M$ is projective iff $\operatorname{Hom}(M,-)$ is exact, injective iff $\operatorname{Hom}(-,M)$ is exact, and flat iff $M\otimes-$ is exact. Furthermore, $M$ is faithfully flat when every chain complex is exact iff the induced $M\otimes-$ chain complex is exact.

Question 1: What are the modules with the property:

  • every chain complex is exact iff the induced $\operatorname{Hom}(-,M)$ chain complex is exact;
  • every chain complex is exact iff the induced $\operatorname{Hom}(M,-)$ chain complex is exact.

Is there a notion faithfully projective/injective, and does it coincide with projective/injective?

Question 2: Why is $M$ faithfully flat precisely when $(\ast)$ every map $A\to B$ is injective iff $A\!\otimes\!M\to B\!\otimes\!M$ is injective? I know that $-\!\otimes\!M$ is right exact, so it preserves epimorphisms, but if we assume $(\ast)$, how does $A\!\otimes\!M\to B\!\otimes\!M$ surjective imply $A\to B$ surjective?

$\endgroup$
1

2 Answers 2

Reset to default
3
$\begingroup$

Regarding question 2, consider the cokernel $C$ of $A \to B$. If $A\otimes M \to B\otimes M$ is surjective, you will easily check that $C \otimes M = 0$, and then deduce that $C = 0$.

$\endgroup$
2
  • $\begingroup$ If $A\overset{\alpha}{\to}B$ is a morphism and $A\!\otimes\!M\overset{\alpha\otimes1_M}{\to}B\!\otimes\!M$ is surjective, then exact $A\overset{\alpha}{\to}B\overset{\pi}{\to} Coker\alpha\to0$ and right exactness of $-\!\otimes\!M$ give us exact $A\!\otimes\!M\overset{\alpha\otimes1_M}{\to} B\!\otimes\!M\overset{\pi\otimes1_M}{\to} (Coker\alpha)\!\otimes\!M\to0$. Thus $\pi\otimes1_M$ is a surjective zero-map, so by exactness, $(Coker\alpha)\!\otimes\!M=0$. Hence $(Coker\alpha)\!\otimes\!M\overset{0\otimes1_M}{\to}0\!\otimes\!M$ is injective, and by assumption, $Coker\alpha\overset{0}{\to}0$ is $\endgroup$
    – Leo
    Sep 26, 2013 at 0:19
  • $\begingroup$ injective, i.e. $Coker\alpha=0$, meaning that $\alpha$ is surjective. However, your answer doesn't adress how we obtain $Ker \beta= Im \alpha$ from $Ker(\beta\!\otimes\!1_M)= Im(\alpha\!\otimes\!1_M)$. $\endgroup$
    – Leo
    Sep 26, 2013 at 0:20
3
$\begingroup$

  1. Observe that $0$ is a module with the property that $\mathrm{Hom}(0, -)$ is exact but not faithfully exact. I claim that a module $M$ has the property that $\mathrm{Hom}(M, -)$ is exact if and only if $M$ is a strong generator that is projective. Indeed, $M$ is a strong generator precisely if $\mathrm{Hom}(M, -)$ is (faithful and) conservative, and $M$ is projective precisely if $\mathrm{Hom}(M, -)$ is exact; but by abstract nonsense, a functor is faithfully exact if and only if it is conservative and exact.

    Dually, a module $M$ has the property that $\mathrm{Hom}(-, M)$ is faithfully exact if and only if $M$ is a strong cogenerator that is injective.

  2. Suppose $M$ has the property that ${-} \otimes M$ preserves and reflects monomorphisms. Then ${-} \otimes M$ preserves kernels and reflects the property of being $0$. Taking cokernels, we deduce that ${-} \otimes M$ reflects epimorphisms. Thus, ${-} \otimes M$ reflects isomorphisms, i.e. is conservative, so by abstract nonsense, ${-} \otimes M$ is faithfully exact.

$\endgroup$
5
  • $\begingroup$ Could you be a bit more explicit in (2)? Suppose that $A\!\otimes\!M \overset{\alpha\otimes1_M}{\to} B\!\otimes\!M\overset{\beta\otimes1_M}{\to} C\!\otimes\!M$ is exact, i.e. $Ker \beta\otimes1_M=Im \alpha\otimes1_M$. Then $0\to Ker \beta\otimes1_M\overset{\subseteq}{\to}B\!\otimes\!M$ and $C\overset{\pi}{\to}C/Im \alpha\otimes1_M\to0$ are exact. Now what? We have $Ker(\alpha\otimes1_M)\supseteq(Ker\alpha)\otimes M$ and $Im(\beta\otimes1_M)=(Im\beta)\otimes M$. What does "$-\otimes M$ preserves kernels" mean? $\endgroup$
    – Leo
    Sep 20, 2013 at 17:17
  • $\begingroup$ Preserves kernels means that $\ker(\alpha \otimes \mathrm{id}) = (\ker \alpha) \otimes \mathrm{id}$. $\endgroup$
    – Zhen Lin
    Sep 20, 2013 at 18:06
  • $\begingroup$ Since ${-} \otimes M$ is right exact, we may assume without loss of generality that $\beta$ is surjective with kernel $\alpha$. Right-exactness implies the image of $\alpha \otimes \mathrm{id}$ is the kernel of $\beta \otimes \mathrm{id}$, but if $\alpha \otimes \mathrm{id}$ is injective, then it must be the kernel. (Note, I am using "kernel" to refer to objects and morphisms interchangeably.) $\endgroup$
    – Zhen Lin
    Sep 20, 2013 at 21:13
  • $\begingroup$ I'm sorry, but your arguments are too quick for me to understand. Yes, $-\otimes M$ is right exact, i.e. preserves epimorphisms. Now suppose that $(\alpha\otimes1_M)(a\otimes m)=0$. Assuming (faithful?) flatness, how do I show $\alpha(a)=0$? $\endgroup$
    – Leo
    Sep 23, 2013 at 10:38
  • 1
    $\begingroup$ No, you misunderstand. ${-} \otimes M$ being right exact means that it sends a short exact sequence $0 \to A \to B \to C \to 0$ to a right exact sequence $A \otimes M \to B \otimes M \to C \otimes M \to 0$, which is much stronger than merely preserving epimorphisms. $\endgroup$
    – Zhen Lin
    Sep 23, 2013 at 11:20

You must log in to answer this question.

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