Cohomology ring $H^*(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)$. I am interested in computing the cohomology ring $H^*(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)$. Here # is the connected sum. Using a suggestion here on my earlier post, I computed the additive structure as
$$H^i(\mathbb{R}P^3 \# \mathbb{R}P^3; \mathbb{Z}_2)=\begin{cases}
\mathbb{Z}_2 & \mbox{if } i=0 \\
\mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=1 \\
\mathbb{Z}_2 \oplus \mathbb{Z}_2 & \mbox{if } i=2 \\
\mathbb{Z}_2 & \mbox{if } i=3 \\
\end{cases}$$
But I am having hard time in computing the ring structure. I actually want to compute $H^*(\mathbb{R}P^n \# \mathbb{R}P^n; \mathbb{Z}_2)$ for odd $n$ and think that the case $n=3$ should help me get the general case. Ths case $n=1$ is trivial as $\mathbb{R}P^1 \# \mathbb{R}P^1= \mathbb{S}^1$.
 A: This is a ellaboration of the above answer by Juan S. Further ellaborations are welcome.
Let $M,N$ be compact $d$-manifolds, let $B$ be a small ball along which we connect sum, and let $u\in H^p(M\# N), v\in H^q(M\# N)$ with $p,q\neq 0,d$, and -as you already have calculated the module structure of $H^*(M\# N)$- such that $u,v$ come from certain elements $u^*,v^*$ in $H^*(M)$ or $H^*(N)$. Moreover, as none of them has dimension $d$, we can assume $u$ (resp. $v$) has a representative $\bar{u}$ (resp. $\bar{v}$) supported in $M\backslash B$ or $N\backslash B$. Then you have the following cases:
1) $u^*,v^*\in H^*(M)$. Then the product of $\bar{u}$ and $\bar{v}$ has support contained in $M$, and it defines a representative $\overline{uv}$ which is in fact a extension by zero of a representative of $u^*\cup v^*$, so $u\cup v= w$, where $w$ comes from $u^*\cap v^*$.
2) $u^*,v^*\in H^*(N)$. The same as above.
3) $u^*\in H^*(M), v^*\in H^*(N)$. Then, the supports of $\bar{u}$ and $\bar{v}$ are dsjoint, hence $u\cup v=0$.
A: Do you know what the ring structure of $H^*(\mathbb{R}P^3)$ is?
Then the cohomology ring of a connected sum is the direct sum of the cohomology rings modulo identification of the 0 and $n$-th cohomology groups (Here is the precise statement). I guess you can kind of see this since you have worked out the additive structure anyhow. 
Note that in your case, since you are using $\mathbb{Z}/2\mathbb{Z}$ coefficients, everything is orientable. 
A reference for the result in the orientable case is Bredon - 'Geometric Topology'. This is Problem 1 on page 358.
