This post is part of a series (started here) of posts on the structure of the category ${\mathcal{U}}$ of unstable modules over the mod ${2}$ Steenrod algebra ${\mathcal{A}}$, 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 ${\mathcal{U}}$.

• First, using the (cocommutative) Hopf algebra structure on ${\mathcal{A}}$, we got a symmetric monoidal structure on ${\mathcal{U}}$, which was an algebraic version of the Künneth theorem.
• Second, we described a “Frobenius” functor

$\displaystyle \Phi : \mathcal{U} \rightarrow \mathcal{U},$

which was symmetric monoidal, and which came with a Frobenius map ${\Phi M \rightarrow M}$.

• We constructed an exact sequence natural in ${M}$,

$\displaystyle 0 \rightarrow \Sigma L^1 \Omega M \rightarrow \Phi M \rightarrow M \rightarrow \Sigma \Omega M \rightarrow 0, \ \ \ \ \ (4)$

where ${\Sigma}$ was the suspension and ${\Omega}$ the left adjoint. In particular, we showed that all the higher derived functors of ${\Omega}$ (after ${L^1}$) vanish.

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

Theorem 39 The category ${\mathcal{U}}$ is locally noetherian: the subobjects of the free unstable module ${F(n)}$ satisfy the ascending chain condition (equivalently, are finitely generated as ${\mathcal{A}}$-modules).

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

$\displaystyle 0 \rightarrow \Phi F(n) \rightarrow F(n) \rightarrow \Sigma F(n-1) \rightarrow 0,$

as we saw in the previous post. Observe that ${F(0) = \mathbb{F}_2}$ is clearly noetherian (it’s also not hard to check this for ${F(1)}$). Inductively, we may assume that ${F(n-1)}$ (and therefore ${\Sigma F(n-1)}$) is noetherian.

Fix a subobject ${M \subset F(n)}$; we’d like to show that ${M}$ is finitely generated.

1. Proof of the finite generation theorem

We now use the following Nakayama-type lemma:

Lemma 40 Let ${M \in \mathcal{U}}$ be connected (${M_0 =0}$). Then if ${\Omega M}$ is finitely generated, so is ${M}$.

In fact, we have a surjection ${M \rightarrow \Sigma \Omega M \rightarrow 0}$. Choose a finitely generated submodule ${F \subset M}$ such that ${F \rightarrow \Sigma \Omega M}$ is surjective. Then

$\displaystyle \Omega F \rightarrow \Omega M$

is surjective, so that (as ${\Omega}$ is right exact), ${\Omega (M/F) = 0}$. However, a connected object in ${\mathcal{U}}$ whose loop space vanishes is itself trivial, as one sees by the explicit description of ${\Omega}$. Therefore, ${F \rightarrow M}$ is a surjection and we are done.

So it suffices to show that ${\Omega M}$ is finitely generated. We’d like to say that ${\Omega M \subset F(n-1)}$ and use induction on ${n}$ to conclude that ${\Omega M}$ is finitely generated, but unfortunately ${\Omega }$ is not left exact. The strategy of the proof is to replace ${M}$ with something a little larger where ${L_1 \Omega }$ vanishes.

Namely, consider again the short exact sequence

$\displaystyle 0 \rightarrow \Phi F(n) \stackrel{V}{\rightarrow}F(n) \rightarrow \Sigma F(n-1) \rightarrow 0$

and let ${M_1}$ be the submodule of ${F(n)}$ corresponding to ${V^{-1}(M)}$. We thus get a short exact sequence

$\displaystyle 0 \rightarrow \Phi M_1 \stackrel{V}{\rightarrow} M \rightarrow \Sigma F(n-1).$

The fact that ${M_1}$ could be bigger than ${M}$ is a reflection of the failure of ${\Omega}$ to be left exact. Similarly, define ${M_2 \supset M_1}$ to be the submodule of ${F(n)}$ corresponding to ${V^{-1}(M_1) \subset \Phi F(n)}$. We get a diagram of exact sequences

Lemma 41 The inclusions ${M_1 \rightarrow M_2 \rightarrow \dots}$ stabilize eventually.

Proof: In fact, the images of ${M_i}$ in ${\Sigma F(n-1)}$ eventually stabilize, say starting at ${m-1}$. Then we have a diagram of exact sequence

We get a similar diagram with ${m}$ replaced by any ${k \geq m}$.

The claim is that the inclusion ${M_{k-1} \rightarrow M_k}$ is an isomorphism for all ${k \geq m}$, which will prove the lemma. In fact, suppose the contrary, and let ${t}$ be the smallest dimension in which ${(M_{k-1})_t \rightarrow (M_k)_t}$ fails to be surjective for some ${k \geq m}$. Then an application of the snake lemma shows that ${t}$ is even, and that ${(M_k)_{t/2} \rightarrow (M_{k+1})_{t/2}}$ fails to be a surjection. This is a contradiction, since ${t > 0}$. $\Box$

Now let ${M' }$ be the stable value of the ${M_k, k \gg 0}$. We have an exact sequence

$\displaystyle 0 \rightarrow \Phi M' \rightarrow M' \rightarrow \Sigma F(n-1),$

showing that ${L_1 \Omega M' = 0}$ and ${\Omega M' \subset F(n-1)}$. By induction on ${n}$, ${M'}$ is finitely generated. In other words, ${M_k}$ for ${k \gg 0}$ is finitely generated.

We now use descending induction on ${k}$ to conclude that ${M = M_0}$ is finitely generated. The exact sequences

$\displaystyle 0 \rightarrow \Phi M_{k+1} \rightarrow M_k \rightarrow \Sigma F(n-1),$

together with the ascending chain condition for ${F(n-1)}$ inductively assumed, allow us to conclude by descending induction that ${M_0}$ is finitely generated.

2. An alternative description of $F(n)$

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

Proposition 42 ${F(n) \rightarrow ( F(1)^{\otimes n})^{\Sigma_n}}$ is an isomorphism.

This will be another example of the use of the functors ${\Phi}$ and a clever induction. Namely, suppose

$\displaystyle F(n-1) \rightarrow (F(1)^{\otimes (n-1)})^{\Sigma_{n-1}}$

is an isomorphism. We’d like to conclude the same, replacing ${n-1}$ by ${n}$. For this, we need a way of relating the problems at ${n-1}$ and ${n}$. We can do this using the morphisms

$\displaystyle F(n) \rightarrow \Sigma F(n-1) , \quad ( F(1)^{\otimes n})^{\Sigma_n} \rightarrow \Sigma (F(1)^{\otimes (n-1)})^{\Sigma_{n-1}}$

where the first map classifies ${\Sigma \iota_{n-1}}$ and the second map comes from the projection ${F(1) \simeq \mathbb{F}_2\left\{t, t^2, t^4, \dots\right\} \rightarrow \Sigma \mathbb{F}_2}$ that sends ${t \mapsto 1}$ and ${t^{2^i} \mapsto 0}$ for ${i \geq 1}$.

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

To clarify, we observe that ${\Phi F(1) \rightarrow F(1)}$ is injective (as we’ve seen, since ${F(1)}$ is projective), and since ${\Phi}$ is symmetric monoidal, the same holds for ${F(1)}$ replaced by ${F(n)}$. Since ${\Phi}$ is also exact and commutes with taking fixed points, it follows that the same is true for ${(F(1)^{\otimes n})^{\Sigma_n}}$. That verifies exactness in the first square. The exactness of the bottom row can now be checked by “hand” using the explicit description of ${F(1) = \mathbb{F}_2\left\{t, t^2, t^4, \dots\right\}}$.

By induction, the rightmost vertical arrow is an isomorphism. It thus follows by the snake lemma that if ${K, C}$ are the kernel and cokernel of ${F(n) \rightarrow (F(1)^{\otimes n})^{\Sigma_n}}$, then

$\displaystyle \Phi K \rightarrow K, \quad \Phi C \rightarrow C$

are isomorphisms. Using a straightforward argument with the grading (or Lemma… above), this implies that ${C, K = 0}$ as they are connected: ${C_0 = K_0 = 0}$.

3. Finite generation of tensor products

We now want to prove that the tensor product preserves the noetherian (i.e., finitely generated) objects in $\mathcal{U}$.

Theorem 43 Given finitely generated objects ${M, N \in \mathcal{U}}$, the (${\mathbb{F}_2}$-linear!) tensor product ${M \otimes N \in \mathcal{U}}$ is finitely generated.

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

Proof: It suffices to show that for each ${m, n}$, ${F(m) \otimes F(n)}$ is finitely generated. Using the noetherian property of ${\mathcal{U}}$, we can reduce to proving that

$\displaystyle F(1)^{\otimes (m+n)} \supset F(m) \otimes F(n)$

is finitely generated. So let’s prove that ${F(1)^{\otimes p}}$ is finitely generated by induction on ${p}$.

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

$\displaystyle \Phi F(1)^{\otimes p} \rightarrow F(1)^{\otimes p}$

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

$\displaystyle F(1)^{\otimes p} = \mathbb{F}_2\left\{ t_1^{2^{a_1}} \dots t_p^{2^{a_p}}, \quad a_1, \dots, a_p \geq 0\right\},$

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

$\displaystyle \mathbb{F}_2\left\{ t_1^{2^{a_1}} \dots t_p^{2^{a_p}}, \quad a_1, \dots, a_p > 0\right\} \hookrightarrow \mathbb{F}_2\left\{ t_1^{2^{a_1}} \dots t_p^{2^{a_p}}, \quad a_1, \dots, a_p \geq 0\right\},$

whose cokernel has a basis of monomials ${t_1^{2^{a_1}} \dots t_p^{2^{a_p}}}$ where at least one of ${a_1, \dots, a_p = 0}$.

As a result, we produce a map

$\displaystyle F(1)^{\otimes p} \rightarrow \bigoplus_{1 \leq i \leq p} F(1)^{\otimes (i-1)} \otimes \Sigma \mathbb{F}_2 \otimes F(1)^{\otimes p-i},$

such that the ${i}$th component is such that the ${i}$th tensor factor ${F(1)}$ maps to ${\Sigma \mathbb{F}_2}$, ${t_i \mapsto \Sigma 1}$. This map annihilates the image of the Frobenius, and in fact we have an exact sequence

$\displaystyle 0 \rightarrow \Phi F(1)^{\otimes p} \rightarrow F(1)^{\otimes p} \rightarrow \bigoplus_{1 \leq i \leq p} F(1)^{\otimes (i-1)} \otimes \Sigma \mathbb{F}_2 \otimes F(1)^{\otimes p-i}.$

By induction on ${n}$, the right-hand-side is finitely generated. We can now conclude that ${F(1)^{\otimes p}}$ is finitely generated. $\Box$