This post is part of a series (started here) of posts on the structure of the category of unstable modules over the mod Steenrod algebra , which plays an important role in the proof of the Sullivan conjecture (and its variants).

In the previous post, we introduced some additional structure on the category .

- First, using the (cocommutative) Hopf algebra structure on , we got a symmetric monoidal structure on , which was an algebraic version of the Künneth theorem.
- Second, we described a “Frobenius” functor
which was symmetric monoidal, and which came with a Frobenius map .

- We constructed an exact sequence natural in ,
where was the suspension and the left adjoint. In particular, we showed that all the higher derived functors of (after ) vanish.

The first goal of this post is to use this extra structure to prove the following:

Theorem 39The category is locally noetherian: the subobjects of the free unstable module satisfy the ascending chain condition (equivalently, are finitely generated as -modules).

In order to prove this theorem, we’ll use induction on and the technology developed in the previous post as a way to make Nakayama-type arguments. Namely, the exact sequence (4) becomes

as we saw in the previous post. Observe that is clearly noetherian (it’s also not hard to check this for ). Inductively, we may assume that (and therefore ) is noetherian.

Fix a subobject ; we’d like to show that is finitely generated.

**1. Proof of the finite generation theorem**

We now use the following Nakayama-type lemma:

Lemma 40Let be connected (). Then if is finitely generated, so is .

In fact, we have a surjection . Choose a finitely generated submodule such that is surjective. Then

is surjective, so that (as is right exact), . However, a connected object in whose loop space vanishes is itself trivial, as one sees by the explicit description of . Therefore, is a surjection and we are done.

So it suffices to show that is finitely generated. We’d like to say that and use induction on to conclude that is finitely generated, but unfortunately is not left exact. The strategy of the proof is to replace with something a little larger where vanishes.

Namely, consider again the short exact sequence

and let be the submodule of corresponding to . We thus get a short exact sequence

The fact that could be bigger than is a reflection of the failure of to be left exact. Similarly, define to be the submodule of corresponding to . We get a diagram of exact sequences

Lemma 41The inclusions stabilize eventually.

*Proof:* In fact, the images of in eventually stabilize, say starting at . Then we have a diagram of exact sequence

We get a similar diagram with replaced by any .

The claim is that the inclusion is an isomorphism for all , which will prove the lemma. In fact, suppose the contrary, and let be the smallest dimension in which fails to be surjective for some . Then an application of the snake lemma shows that is even, and that fails to be a surjection. This is a contradiction, since .

Now let be the stable value of the . We have an exact sequence

showing that and . By induction on , is finitely generated. In other words, for is finitely generated.

We now use descending induction on to conclude that is finitely generated. The exact sequences

together with the ascending chain condition for inductively assumed, allow us to conclude by descending induction that is finitely generated.

**2. An alternative description of **

Our next goal is to prove the following loose end, which was claimed earlier:

Proposition 42is an isomorphism.

This will be another example of the use of the functors and a clever induction. Namely, suppose

is an isomorphism. We’d like to conclude the same, replacing by . For this, we need a way of relating the problems at and . We can do this using the morphisms

where the first map classifies and the second map comes from the projection that sends and for .

We have, in fact, a commutative and exact diagram:

To clarify, we observe that is injective (as we’ve seen, since is projective), and since is symmetric monoidal, the same holds for replaced by . Since is also exact and commutes with taking fixed points, it follows that the same is true for . That verifies exactness in the first square. The exactness of the bottom row can now be checked by “hand” using the explicit description of .

By induction, the rightmost vertical arrow is an isomorphism. It thus follows by the snake lemma that if are the kernel and cokernel of , then

are isomorphisms. Using a straightforward argument with the grading (or Lemma… above), this implies that as they are connected: .

**3. Finite generation of tensor products**

We now want to prove that the tensor product preserves the noetherian (i.e., finitely generated) objects in .

Theorem 43Given finitely generated objects , the (-linear!) tensor product is finitely generated.

Note that this really uses the unstable condition. The tensor product of with itself in -modules is an infinite direct sum of copies of and is very far from being finitely generated.

*Proof:* It suffices to show that for each , is finitely generated. Using the noetherian property of , we can reduce to proving that

is finitely generated. So let’s prove that is finitely generated by induction on .

As above, we need to show that the cokernel of the Frobenius

is finitely generated, and this will be sufficient. Observe that

and that the Frobenius can be identified with the inclusion of vector spaces

whose cokernel has a basis of monomials where at least one of .

As a result, we produce a map

such that the th component is such that the th tensor factor maps to , . This map annihilates the image of the Frobenius, and in fact we have an exact sequence

By induction on , the right-hand-side is finitely generated. We can now conclude that is finitely generated.

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