Let $X$ be a finite spectrum. Say that $X$ has **characteristic two** if multiplication by two on $X$ is nullhomotopic. 


Does there exist a noncontractible finite spectrum of characteristic two? 


(This is mentioned as an open problem in Barratt's 1959 [paper](http://www.maths.ed.ac.uk/~aar/papers/barratt1.pdf%E2%80%8E) "Spaces of finite characteristic.")



Finite spectrum annihilated by multiplication by two
This doesn't seem very hard. Am I missing something?


Let $X$ be a non-trivial finite spectrum of characteristic $2$. Then let $R=\mathrm{Hom}(X,X)$, the function spectrum of maps $X$ to $X$. This $R$ is an associative $S$-algebra, with $0=2$ in $\pi\_0R$; furthermore, $R$ is finite.


If $X\neq0$, then $H\_\*(X;F)\neq0$, where $F=\mathbb{F}\_2$, and hence $H\_\*(R;F)\neq0$. Note that in this case the unit map $S\to R$ has non-trivial image in homology (which is the unit of the ring $H\_\*(R;F)$), and thus $H^0(R;F)\to F$ is surjective.


Consider the forgetful functor $\mathrm{Ass}\_S\to \mathrm{Unital}\_S$ from associative $S$-algebras to unital spectra. This has a (homotopical) left adjoint $T\_\*$: given a unital spectrum $M=(M,f\colon S\to M)$, let $T\_\*(M)$ be the homotopy pushout in $\mathrm{Ass}\_S$ of 
$$
T(M) \xleftarrow{T(f)} T(S)\to S,
$$
where $T$ is the free associative algebra functor.


You can also build $T\_\*(M) = \mathrm{colim}\_n T\_\*^n(M)$ iteratively a la the James construction: to get $T\_\*^n(M)$, glue the $n$-fold smash product $M^{\wedge n}$ to $T\_\*^{n-1}(M)$ along a suitable map $F^n(M)\to M^{\wedge n}$. [I.e., consider the $n$-cubical diagram obtained from smashing $S\to M$ with itself $n$-times; $F^n(M)\to M^{\wedge n}$ is the map from the hocolim of the deleted $n$-cube to the object at the terminal position of the $n$-cube.]


Let $M=S^0\cup\_2 e^1\to R$ be the map of unital spectra which exists by our hypothesis on $R$. The map extends to an associative algebra map
$$ 
T\_\*(M) \to R.
$$
In homology, $H\_\*(T\_\*(M);F) \approx F[x]$ with $|x|=1$, since $M^{\wedge n}/F\_n(M) \approx S^n$.


In mod 2 cohomology, the generator $u\in H^0(T\_\*(M);F\_2)$ must satisfy $Sq^1(u)\neq0$, and an easy argument shows that therefore $Sq^n(u)\neq0$ for all $n$. But $H^0(R;F) \to H^0(T\_\*(M);F) \xrightarrow{\sim} H^0(S;F)=F$ is surjective, so there exists $v\in H^0(R;F)$ such that $Sq^n(v)\neq0$ for all $n$, contradicting finiteness.


