Let $\mathbb{R}P^{\infty}$ be the infinite real projective space. Its integral homology is given by $$ H_{i}(\mathbb{R}P^{\infty})=\left\{\begin{array}{ll} \mathbb{Z} & \text{if }i=0 \\ \mathbb{Z}/2 & \text{if }i\text{ is odd} \\ 0 & \text{if }i>0\text{ and is even}\\ \end{array} \right. $$
Then by the universal coefficient theorem, we can get that $H^{i}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)=\mathbb{Z}/2$ for all $i\geqslant0$. I want to know the ring structure of this $\mathbb{Z}/2$-coefficient cohomology.
Since $\mathbb{R}P^{\infty}$ is a model of $K(\mathbb{Z}/2,\,1)$, we can use the fibration $$K(\mathbb{Z},\,1)\rightarrow K(\mathbb{Z}/2,\,1)\rightarrow K(\mathbb{Z},\,2)$$ and cohomology Leray-Serre spectral sequence to study $H^{\bullet}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)$. We apply $S^{1}$ to be a model of $K(\mathbb{Z},\,1)$ and $\mathbb{C}P^{\infty}$ to be a model of $K(\mathbb{Z},\,2)$. Then after some argument on $E_{2}$ page, we can get that as a $\mathbb{Z}_{2}$-algebra, $H^{\bullet}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)$ is generated by two elements $a$ and $b$ where $\deg(a)=1$ and $\deg(b)=2$. In fact, one has $$H^{2i+1}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)=\mathbb{Z}/2\cdot ab^{i}$$ and $$H^{2i}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)=\mathbb{Z}/2\cdot b^{i}$$
However, I don't know that whether $a\cdot a=0$ or $b$.
If $a\cdot a=0$, then we have $$H^{\bullet}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)\cong \Lambda(a)\otimes \mathbb{Z}_{2}[b]$$
Otherwise, it is just generated by one element $a$ and $$H^{\bullet}(\mathbb{R}P^{\infty},\,\mathbb{Z}/2)\cong \mathbb{Z}_{2}[a]$$
Can anyone give me some help for this problem? Thank you very much.