cup product in cohomology ring of a suspension Let $X$ be a CW-complex. Let $\Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of 
$$
H^*(\Sigma X;R)$$
trivial? How to prove? Where can I find the result?
 A: The answer depends on which homology theory you are using.
The statement fails for singular cohomology.
However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $\tilde{H}{\,}^\bullet(\Sigma X)=H^\bullet(\Sigma X,\text{pt})$.
Write $\Sigma X=\text{Cone}_+(X)\cup\text{Cone}_-(X)$ and let $\iota:\text{pt}\longrightarrow\text{Cone}(X)$ be the inclusion map. We get the following commutative diagram
$\require{AMScd}$
\begin{CD}
H^p(\Sigma X,\text{pt})\otimes H^q(\Sigma X,\text{pt}) @>{\smile}>> H^{p+q}(\Sigma X,\text{pt})\\
@A{H^p(\iota)\otimes H^q(\iota)}AA @AA{H^{p+q}(\iota)}A\\
H^p(\Sigma X,\text{Cone}_+(X))\otimes H^q(\Sigma X,\text{Cone}_-(X)) @>>{\smile}> H^{p+q}(\Sigma X,\Sigma X)
\end{CD}
Now note that $H^\bullet(\iota):H^\bullet(\Sigma X,\text{Cone}(X))\longrightarrow H^\bullet(\Sigma X,\text{pt})$ is an isomorphism because $\text{Cone}(X)$ is contractible.
Therefore we have found a factorisation of $a\smile b=H^{p+q}(\iota)\left(H^p(\iota)^{-1}(a)\smile H^q(\iota)^{-1}(b)\right)$ over $H^{p+q}(\Sigma X,\Sigma X)\cong0$, hence $a\smile b=0$ for any $a\otimes b\in H^p(\Sigma X,\text{pt})\otimes H^q(\Sigma X,\text{pt})$.
A: Yes. Actually there is a more general result. If $X=\bigcup_{i=1}^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $\Sigma X$ is a union of two cones, each of which is contractible, the result follows. 
To establish the general result, simply note that $H^*(X)\cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map \begin{eqnarray}H^*(X, A_1)\times \cdots \times H^*(X, A_n)\to H^*(X, \bigcup_{i=1}^n A_i)=H^*(X, X)=0\end{eqnarray}
