Let be a ring with 1, and let be a right unital -module.

- Prove that if is flat then for every ideal the mapping induced by the inclusion is injective.
- Suppose that for every finitely generated ideal , the mapping induced by the inclusion map is injective. Prove that in fact the induced map is injective for every ideal . Show further that if is any submodule of a finitely generated free module then is injective. Show that the same is true for any free module (not necessarily finitely generated).
- Under the hypothesis in part (2), suppose is an injective left -module homomorphism. Prove that the induced map is injective. Conclude that is flat.
- Let be a flat right unital -module and an -submodule of . Prove that is flat if and only if for every finitely generated ideal .

- This is true by our definition of flat module.
- Suppose the induced map is injective for all finitely generated ideals . Let be an arbitrary ideal and consider the induced map . Suppose . Now there is a finitely generated ideal which contains all of the . Moreover, restricted to is injective by our hypothesis. Thus , and so is injective.
Now let be a finitely generated free module, and let be a submodule. Now , where is a submodule (i.e. a left ideal). (Specifically, is the set of all th coordinates of elements of . This set is an -submodule of (an ideal) since is an -module.) Now . Thus the induced map is injective.

Finally, let be an arbitrary free module and let be an arbitrary submodule. Consider the induced map . If , then there is a finitely generated submodule of containing all the . Moreover, restricted to this submodule is injective, so that . Thus is injective.

- Let be an injective left module homomorphism. Now every module (for instance ) is isomorphic to a quotient of a free module. Suppose is a free module and a submodule such that . In particular, the sequence is short exact. Let be the subset consisting of precisely those elements such that , where is the natural projection. That is, . Define by , where . Now is total by definition and is well-defined because is injective. Suppose now that and . There exist such that and . So and . Now , so that . Thus we have , and so is a left -module homomorphism. We claim also that . It is certainly the case that , since . This gives the following diagram of module homomorphisms.
We claim that this diagram commutes and that the rows are exact. First we show exactness; the bottom row is exact by definition. For the top row, certainly , since is injective. Now suppose . Then , so that . Thus the rows are exact. Now we show that the diagram commutes. The left square certainly commutes, since the maps are all simply inclusions. Now let . Note that , where . Thus , and so the diagram commutes. In fact, the triple is a homomorphism of short exact sequences.

Recall that the functor is right exact. Thus the induced diagram of modules

commutes and has exact rows. By part (2) above, is injective, and certainly and are surjective. By part 4 of this previous exercise, is injective.

Thus is flat as a right -module.

- Suppose is flat, a submodule, and that for all finitely generated ideals we have . We wish to show that is flat. Let be an arbitrary finitely generated ideal. By the flatness criterion proved above, it suffices to show that is injective. To that end, consider the exact sequences shown in the rows of the following diagram.
With given by (and likewise ) and by (and likewise ), we claim that there exist suitable injective maps and so that this diagram commutes.

Define by . Certainly if , then , so that is well defined. Moreover, this map is clearly bilinear. Thus we have an -module homomorphism given by . Clearly the upper right hand square commutes.

Note that is surjective (obviously). Moreover, since is flat, the map is injective (by part (1) above). Now is an isomorphism and we have , so that in our diagram is injective. By part 1 of this previous exercise, is injective.

Now define by . First, if , then , so that is well defined. It is clear that is a homomorphism of modules, and that the lower right hand square commutes. We claim also that is injective. To see this, suppose . Now . Recall that via the “multiplication” homomorphism ; so in fact we have , so that . Now and , so that (by our hypothesis) . So , and in fact is injective.

Finally, we claim that . To see this, note that .

Thus is injective for all finitely generated ideals . By the flatness criterion, is flat as a right -module.

Conversely, suppose and are flat and let be a finitely generated ideal. Note that and , so that the inclusion always holds. Now consider the following commutative diagram of modules.

Again, (and and denotes the isomorphism given by . Let . (Letting and .) We will chase around the diagram to see that . Note that , so that in . Note that , and because is injective (by our hypothesis that is flat), in . In particular, . Since the top row is exact, we have for some and . Following these elements down to , we see that as desired.

Thus if and are flat, then for all finitely generated ideals .

## Comments

In the second paragraph of 2, why K = direct sum of the I_t ?

Good question. I’m about to go out, but I’ll get to this later today.

On further reflection, that line is clearly not true. (I confused this with the analogous situation for ideals in a product of rings.)

It will take more time than I thought to fix this, so I’ll have to mark this ‘incomplete’ and come back later.