TITLE: Spectrum $E$ with $H^\bullet(E,\mathbb{Z}/2)=\mathcal{A}//\mathcal{A}(n)$
QUESTION [10 upvotes]: 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?

REPLY [11 votes]: 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.