When tensor reflects exact sequences?

Is there a characterization of modules $N$ for which the functor $N\otimes-$ reflects exact sequences?

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If a module $N$ both preserves and reflects exact sequences, it is called faithfully flat. – Alex Becker Jul 24 '12 at 22:23
MTS: This might or might not be "research level" but the answer is not immediately obvious to me. – Steven Landsburg Jul 25 '12 at 0:49
@Steven Landsburg: I agree with you. If you have an idea how to finish my proof below, please feel free to edit it and include that. I just can't do anything more without eating something. – David White Jul 25 '12 at 0:55
@Steven Landsburg: Ah, perhaps I was too hasty and misunderstood the question. I will vote to reopen if people think there is good reason to do so. (I still think the question would benefit from including some background, but that's a separate issue.) – MTS Jul 25 '12 at 3:36
So did I. I'm really sorry, +1 for this question. – Fernando Muro Jul 25 '12 at 9:39

As Alex Becker pointed out, $N$ is faithfully flat iff $N\otimes_R -$ preserves and reflects exact sequences. Since $N$ is flat iff $N\otimes_R -$ preserves exact sequences, you might think $N$ is faithful iff $N\otimes_R -$ reflects exact sequences. This is wrong, as we'll show below. These faithfully flat modules are extremely important, and probably the better place to do work if you can manage it. I think the concept of a module $N$ such that $N\otimes_R -$ reflects exact sequences (but $N$ is not faithfully flat) is an interesting one. However, this concept doesn't seem to appear in T.Y. Lam's Lectures on Modules and Rings, and that's my bible for these types of questions. So it seems the community hasn't found this concept interesting yet. Since no one seems to have coined a phrase for these yet, let's call them reflecting modules (not to be confused with reflexive modules).

When seeking to prove a module is flat you can use that $N\otimes_R -$ is always right exact, so all you need to show is that $N\otimes_R -$ preserves monomorphisms. To show $N$ is reflecting is harder. For example, you must show that if $N\otimes_R g$ is an epimorphism, then $g$ is an epimorphism. You must also show that if $N\otimes_R f$ is a monomorphism then $f$ is a monomorphism. These two follow if $N\otimes_R -$ is a faithful functor (this explains the terminology). According to Theorem 7.1 in Theory of Categories by Barry Mitchell, if $T:C\rightarrow D$ is faithful functor between exact categories which have zero objects, and if $T$ preserves the zero objects, then $T$ reflects exact sequences. Since $R$-mod is an exact category with a zero object, this tells us that $N$ is reflecting if $N\otimes_R -$ is faithful. We'll show the converse below. Consider the following two properties:

(1) For any left $R$-module $M$, $N\otimes_RM=0 \Rightarrow M=0$

(2) A homomorphism of left $R$-modules $\phi:M'\rightarrow M''$ is zero if the induced homomorphism $N\otimes_R M' \rightarrow N\otimes_R M''$ is zero. Note that this property is exactly saying $N\otimes_R -$ is faithful.

When $N$ is flat, these are equivalent, and also equivalent to the fact that $N$ is faithfully flat. Even without the flatness hypothesis, the theorem above shows (2) implies $N$ is reflecting. We can also see that (2) implies (1) because if $N\otimes_R M = 0$ then the map $N\otimes_R 0 \rightarrow N\otimes_R M$ is zero, so the map $0\rightarrow M$ is zero, proving $M=0$. Next, (1) implies (2) because if $N\otimes_R \phi = 0$ then $0=im(N\otimes_R \phi)=N\otimes_R im(\phi)$ so by (1), $im(\phi)=0$. Finally, $N$ reflecting implies (1) because knowing $N\otimes_R 0 \rightarrow N\otimes_R M\rightarrow N\otimes 0$ is exact implies $0\rightarrow M\rightarrow 0$ is exact, so we see $N\otimes_R M=0\Rightarrow M=0$. This completes the characterization:

$N$ is reflecting iff $N\otimes_R -$ is faithful iff $N\otimes_R M=0 \Rightarrow M=0$

(EDIT: Thanks to Martin for pointing out a flaw in the previous version. Here is an argument, now superfluous and a bit fishy, that $N$ reflecting implies (2): suppose $N\otimes_R \phi$ is zero. Then, because $N\otimes_R ker(\phi) = ker(N\otimes_R \phi)$, we see $0\rightarrow N\otimes_R M' \rightarrow N\otimes_R ker(\phi)\rightarrow 0$ is exact, so $0\rightarrow M'\rightarrow ker(\phi)\rightarrow 0$ is exact, i.e. $\phi=0$. This is fishy because it's not clear to me that $N\otimes ker(-) = ker(N\otimes -)$, though it does seem to work for $im(-)$.)

It may also be equivalent to say that for all maximal ideals $I\subset R$, if $N\otimes_R R/I=0$ then $R/I=0$, but I haven't checked this.

EDIT (in response to requests from Will Sawin and Zhen Lin for examples): I'll end by remarking on a confusing bit of terminology and giving some examples. There is a notion of what it means for a module $N$ to be faithful, but it is not equivalent to the above. One says $N$ is faithful if $Ann_R(N)=0$ (this appears in Lam). Note that this website erroneously calls a module faithful if it satisfies (1) above, but the link is still useful as it gives some properties of such modules and examples of modules which satisfy (1) but are not flat. An additional example is $N=\mathbb{Z}\oplus \mathbb{Z}/2$ over $R=\mathbb{Z}$. This seems to show that if $M$ is faithfully flat and $L$ is any module, then $N = M\oplus L$ will be reflecting but not necessarily flat. It's easy to see that $N$ is faithful iff for all nonzero $a\in R$, $aN \neq 0$. So (1) implies $N$ is faithful, because we can take $M=(a)$ for each $a$ and use the fact that $N\otimes_R (a) = aN$.

Note that $N$ faithful does not imply that $N\otimes -$ is a faithful functor. For example, $\mathbb{Q}$ is faithful as a $\mathbb{Z}$-module (since it's torsion free), but $\mathbb{Q}$ clearly fails (1) because $\mathbb{Q}\otimes \mathbb{Z}/2=0$. Faithfully flat implies faithful and flat, but not conversely (again because of $\mathbb{Q}$). Indeed, a module can even be projective and faithful, but fail to be faithfully flat. Consider $P=\mathbb{Z}\oplus \mathbb{Z}\oplus \dots$ as a module over $R = \mathbb{Z}\times \mathbb{Z}\times \dots$. It's faithful because it's a subring, it's projective (e.g. by the dual basis lemma), but it's not faithfully flat because for any maximal ideal $m$ of $R$ containing $P$, $Pm=P$.

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How does right exactness have anything to do with it? Shouldn't you need "right faithful", a condition that $N\otimes -$ does not in general satisfy? Don't you mean the property that OP seeks to characterize is stronger than the two conditions listed, not too strong? – Will Sawin Jul 25 '12 at 14:45
What is an example of a module which is faithful and flat but not faithfully flat? – Will Sawin Jul 25 '12 at 16:32
I'm confused. Why is $\mathbb{Q}$ faithful? There is an evident non-zero morphism $\mathbb{Z} / 2\mathbb{Z} \to \mathbb{Z} / 2\mathbb{Z}$ that gets killed by tensoring with $\mathbb{Q}$, ergo $\mathbb{Q}$ does not satisfy condition (2)... – Zhen Lin Jul 25 '12 at 17:32
The proof of (1) => (2) is not correct (why is $N \otimes M \to N \otimes M''$ injective?). – Martin Brandenburg Jul 25 '12 at 18:35
I just completed what I hope is my last edit. It completes the classification by fixing the proof that (1)=>(2), and proving (2)=>reflecting. Furthermore, it spells out in detail the connection between these properties and the property that $N$ is faithful. I hope this clears up what Zhen Lin was asking about. Just so I have it somewhere, I'm going to write a couple more comments showing that (2) implies $N\otimes -$ reflects monos and epis, but they are contained in the more general result of Theory of Categories quoted above. Also, I'll include some observations for the $R=\mathbb{Z}$ case – David White Jul 31 '12 at 17:08