specific modules over the Steenrod algebra with one generator I'd be happy to clarify the following. Consider the module which is a quotient of the Steenrod algebra mod $2$ by the left ideal generated by $\operatorname{Sq}^1, \operatorname{Sq}^2, \operatorname{Sq}^3, \dots, \operatorname{Sq}^k$. Is it true that there is a space (or spectrum) with such a cohomology? I found in the book by Margolis a statement that these modules (or may be at least one) are related to the connective K-theory.
 A: This realization problem is well-known in algebraic topology.  Let $A$ denote the mod $2$ Steenrod algebra.  For $k=0$ the module $A$ is the mod $2$ cohomology of $H\mathbb{Z}/2$, and for $k=1$ the module $A/ASq^1$ is the mod $2$ cohomology of $H\mathbb{Z}$.  For $k=2$ and $k=3$, the module $A/A(Sq^1,Sq^2) = A/A(Sq^1, Sq^2, Sq^3)$ is the mod $2$ cohomology of $ko$ (connective, real $K$-theory).  This follows from calculations of Stong.
For $4 \le k < 8$ the module $A/A(Sq^1, \dots, Sq^k) = A/A(Sq^1, Sq^2, Sq^4)$ is the mod $2$ cohomology of $tmf$, the connective topological modular forms spectrum.  This construction of $tmf$ is due to Hopkins, Mahowald and Miller, and explained by Behrens in the Talbot proceedings book on "Topological Modular Forms".  The claim that its cohomology is $A/A(Sq^1, Sq^2, Sq^4)$ is proved in a paper by Mathew.  For $8 \le k < \infty$ there is no spectrum with cohomology $A/A(Sq^1, \dots, Sq^k)$, due to Adams' factorization of $Sq^{2^i}$ for $i\ge4$ as a composition of secondary cohomology operations, as part of his proof that $Sq^{2^i}$ does not detect a class in $\pi_{2^i-1}(S)$.  All of the modules discussed so far are not unstable, hence cannot be the cohomology of a space. If you allow $k=\infty$, then $A/A(Sq^1, \dots) = \mathbb{Z}/2$ is the cohomology of a point, or the reduced cohomology of two points.
