Finite spectrum annihilated by multiplication by two 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 "Spaces of finite characteristic.")
 A: 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.
