MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## What conditions are needed for $-\otimes_A B$ to be faithful?

For $A$ a (commutative) ring, $f:A\to B$ an $A$-algebra, what conditions do we need on $A$ and $B$ (and $f$) for the functor $-\otimes_A B:A-mod\to B-mod\quad$ to be faithful (i.e. injective on $Hom$-sets)?

I can't seem to come up with anything other than the rather obvious condition that tensoring with $B$ shouldn't kill anything (or at least not too much), but this is hardly satisfying. It seems like it would be faithful in general, but I fail to come up with an argument as to why this should be true, and precisely when it is (if at all). Dare I beg the aid of the MO?

Note: this is basically the same as $f^*$ being faithful for a morphism $f:X\to Y$ of schemes, which reduces to the above. Hence the algebraic geometry tag.

-
See also: mathoverflow.net/questions/47716/… . – Graham Leuschke Dec 2 2010 at 14:16
Certainly not faithful in general. Lot of examples when $A$ is local, but $f$ is not a local homomorphism. For example, it could be the inclusion of a domain into its quotient field. Then all the torsion stuff will die, definitely non-faithful. – Graham Leuschke Dec 2 2010 at 14:22

The functor you mention is faithful if and only if the functor $-\bigotimes_A B :A-mod\to A-mod$ is faithful, ie iff $B$ is a faithful $A$-module.

For a concrete counterexample take $f:\mathbb{Z}\to \mathbb{Q}$ and like Graham says this kills the torsion stuff.

-
Why must $B$ be flat? Seems like this is the content of the earlier question -- is it possible for $-\otimes_AB$ to be faithful without $B$ being flat. – Graham Leuschke Dec 2 2010 at 15:55
Yes, it is possible to be faithful without being flat. Let $A=k[x,y]$ and let $B= xA+yA$. – Greg Muller Dec 2 2010 at 17:29
Yes. I have now replaced "faithfully flat" to "faithful" in my answer. – Achilleas K Dec 2 2010 at 17:41
@Greg, do you know of any examples of commutative rings $B$ with unity? – Karl Schwede Dec 2 2010 at 17:57
I had missed the requirement that $B$ was a ring, rather than an $A$-module. Of course, I can replace $B$ by its symmetric tensor algebra to get a ring which is faithful but not fat. – Greg Muller Dec 2 2010 at 18:44

A morphism $f:X\rightarrow Y$ of schemes gives a faithful pullback functor $f^*$ exactly when the morphism $f$ is surjective on underlying sets. This can be seen in several steps.

1. Note that the functor is faithful iff it preserves zero; that is, $f^*(M)=0$ implies $M=0$.
a.To see this, note that if a right exact functor preserves zero, then any morphism which becomes the zero map was already the zero map (consider its cokernel). b. Then, if two maps $g,g'$ go to the same map, their difference goes to the zero map, and thus the difference was already zero.

2. A module $M$ on $Y$ is zero iff $Hom_Y(\mathcal{O}_Z,M)=0$ for every irreducible closed subscheme $Z$ (that is, for the closure of every point).

3. If $Hom_Y(\mathcal{O}_Z,f^*(M))=0$, then $Hom_X(\mathcal{O}_{f(Z)},M)=0$.

Now, if every point in $X$ is the image of a point in $Y$, and $f^*(M)=0$, then by 2 and 3, $Hom_X(\mathcal{O}_{Z'},M)=0$ for every point $Z'\in X$. Then, by $2$, we know that $M=0$, and so $f^*$ preserves zero. Thus, by 1, $f^*$ is faithful. If $f$ is not surjective on underlying sets, then pulling back the skyscraper sheaf of a missed point will be zero, and so by 1, it cannot be faithful.

-
I'd give this an extra +1 if I could, as it also answers the problem I had which motivated the thread. Unfortunately, the answer was negative. – Ketil Tveiten Dec 2 2010 at 18:19
I don't believe that 2. is true. Since $\mathcal{O}_Z$ is a quotient of $\mathcal{O}_Y$, already $\Gamma(X,M) = 0$ implies $\mathrm{Hom}_X(\mathcal{O}_Z,M)=0$. Or do you mean the Hom-sheaf? – Martin Brandenburg Nov 12 2011 at 12:13
Also the last step of the proof seems to be problematic if the point is not closed: How do you prove that the skyscraper sheaf pulls back to $0$? – Martin Brandenburg Nov 12 2011 at 12:40
Here is a counterexample to the statement "f surjective implies $f^*$ faithful": It is enough to find a nil ideal $I$ of some ring $A$ and an $A$-module $M$ such that $M = IM$, but $M \neq 0$; then $\mathrm{Spec}(A/I) \to \mathrm{Spec}(A)$ is a homeomorphism, but the corresponding pullback functor is not faithful. One specific example is $A = k[x_1,x_2,\dotsc]/(x_i^i = 0, x_i = x_{i+1} x_{i+2})_{i \geq 1}, M = I = (x_1,x_2,\dotsc)$. – Martin Brandenburg Nov 12 2011 at 15:44
A more easy counterexmaple: The zero section $\mathrm{Spec}(k) \to \mathrm{Spec}(k[e]/e^2)$. – Martin Brandenburg Nov 15 2011 at 8:28

A sufficient condition is that the embedding $A \to B$ splits in the category of $A$-modules, i.e. $B \cong A \oplus B'$ as $A$-module. In this case the functor $-\otimes_A B$ has $\id$ as a direct summand, hence faithful. An example $A = {\mathbb Z}$, $B = {\mathbb Z} \oplus {\mathbb Z}/2{\mathbb Z}$ shows that one does not need to require that $B$ is flat.

On the other hand, if you want $-\otimes_A B$ to be faithful also on $Ext$'s, then the above condition is also necessary. Indeed, consider exact sequence $$0 \to A \to B \to B/A \to 0.$$ It gives an extension of $B/A$ by $A$. If we tensor it with $B$ we obtain $$B \to B\otimes_A B \to (B/A)\otimes_A B \to 0.$$ Note that the first map is a split monomorphism (because of the multiplication $B\otimes_A B \to B$), hence the corresponding extension is trivial. So, if we want the functor to be injective on extensions the initial exact sequence should be split, hence $B = A \oplus (B/A)$.

-
 What do you mean exactly by "injective on Exts" or "injective on extensions"? – Martin Brandenburg Nov 12 2011 at 15:56

The answer is given in the following - very interesting - paper:

Bachuki Mesablishvili, Descent Theory for Schemes, Applied Categorical Structures 12: 485–512, 2004.

The author generalizes Grothendieck's descent theory from faithfully flat morphisms to pure morphisms. These are morphisms of schemes $f$ such that every base change $f'$ of it is schematically dense in the sense that $f'$ does not factor through a proper subscheme. In the affine case, $\mathrm{Spec}(A) \to \mathrm{Spec}(R)$ is pure iff $R \to A$ is "stable injective" in the sense that for every $R$-algebra $B$ the map $B \to A \otimes_R B$ is injective.

One of the main results (Theorem 5.15) states that for a quasi-compact morphism $f : X \to Y$ the following are equivalent:

1) $f$ is pure

2) $f^* : \mathrm{Qcoh}(Y) \to \mathrm{Qcoh}(X)$ faithful

3) $f$ is a stable effective descent morphism for quasi-coherent sheaves

In the case that $X,Y$ are affine there are even more characterizations (Theorem 4.18), for example we can also add 4) $f^*$ is conservative, and 5) $f^*$ is comonadic.

For flat morphisms, it is well-known that $f^*$ is faithful iff $f$ is surjective iff $f$ is faithfully flat. In general, every pure morphism is surjective (since otherwise some base change $\emptyset \to \mathrm{Spec}(\text{field})$ is not schematically dense). But there are lots of surjective morphisms which are not pure, for example the zero section $s: \mathrm{Spec}(k) \to \mathrm{Spec}(k[\epsilon]/\epsilon^2)$.

-