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?



Spectrum $E$ with $H^\bullet(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$
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](https://doi.org/10.2307/1970147).


