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