# Mod 3 Moore spectrum

I only know through stories that mod 3 moore spectrum is not associative. I do not know of any proof. I have been informed that Toda had proved it in the paper "Extended $p^{th}$ power". I was not able to follow it. Can anybody give me a proof that mod 3 Moore spectrum is not associative?

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This is also Lemma 6.2 in Toda's On spectra realizing exterior parts of the Steenrod algebra (sciencedirect.com/science/article/pii/0040938371900176), which you might find helpful. (I don't know if this proof is any different from the one in the Toda reference you mention, which I don't have available.) –  Eric Peterson Jan 15 '13 at 20:47
That paper of Toda's is a little concentrated. The nature of the answer to this question will probably depends pretty heavily on whether you know what a Massey product is. You can show that there must be a map from $M$ to $H\mathbb{Z}/3$ which preserves the unit and multiplication. The resulting map $H_* M \to H_* H\mathbb{Z}/3$ has as image a square-zero class in the dual Steenrod algebra ($\tau_0$) whose triple Massey product $\langle \tau_0, \tau_0, \tau_0 \rangle$ is not in the image. –  Tyler Lawson Jan 16 '13 at 4:42
Hi @TylerLawson. Could you say a bit more about the shape of the argument? You are saying that if $M$ had an $A_{\infty}$-structure, and the above map is induced from an $A_{\infty}$-map then for all elements in the image, their triple Massey products (a representative of which is given by the $A_{\infty}$-structure) must always be in the image because - is this correct? –  Elden Elmanto Jul 11 at 20:39