Spectrum $E$ with $H^\bullet(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$ Let $\mathcal{A}$ be the Steenrod Algebra and $\mathcal{A}(n)$ be the subalgebra generated by $Sq^1, Sq^{2}, Sq^{2^2},\ldots, Sq^{2^n}$.
It is known that

*

*$H^*(H\mathbb{Z},\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(0)$,

*$H^*(ko,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(1)$,

*$H^*(tmf,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(2)$.

Now, I have read somewhere that it is known that there is no spectrum $E$ such that $H^*(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$ with $n\ge 3$. Where can I find the proof?
 A: The first two nonzero elements of the $A$-module $A /\!/ A(n)$ are the generator $e$ (in degree 0) and $Sq^{2^{n+1}} e$ (in degree $2^{n+1}$): all other elements in this degree and lower are a sum of products of lower squares, as a consequence of the Adem relations. When $n=3$ this bottom class is $Sq^{16}$.
However, Adams showed that these higher squaring operations can be decomposed in terms of secondary cohomology operations: there are "operations" $\Phi_{i,j}$ that are only defined on elements where certain Steenrod operations vanish, and relations that say that for $k > 3$ the element $Sq^{2^k}$ is a linear combination of Steenrod operations applied to $\Phi_{i,j}$. The exact relations aren't important here: the important part is that we get some relation
$$
Sq^{2^{n+1}}e = \sum Sq^{I_l} (\Phi_{i_l,j_l} e).
$$
The elements $\Phi_{i_l,j_l}$ are all in degrees strictly between 0 and $2^{n+1}$ and hence are forced to be zero, so $Sq^{2^{n+1}} e = 0$ for a contradiction.
The machinery behind this is in Adams' paper "On the nonexistence of elements of Hopf invariant one" (Ann. Math, 1960), doi:10.2307/1970147.
A: This is the same answer as Tyler's but formulated in a different way. In the Adams spectral sequence for the sphere, there is a differential $d_2(h_{n + 1}) = h_0 h_n^2$ for $n \geq 3$.
If such a spectrum $E$ existed, then the inclusion of the bottom cell $S \hookrightarrow E$ induces the obvious map $A/\!/A(n) \twoheadrightarrow k$. By inspecting the cobar complexes for $A$ and $A(n)$, we see that the image of $h_{n + 1}$ is trivial in $\operatorname{Ext}_A(A/\!/A(n), k) = \operatorname{Ext}_{A(n)}(k, k)$ but that of $h_0 h_n^2$ is not. This is a contradiction.
