Understanding 'trivial' step in calculating the graded cohomology ring $H^*(\mathbb{R}P^n; \mathbb{Z})$ I have a text that says it is obvious that $H^*(\mathbb{R}P^1; \mathbb{Z}/2)$ is isomorphic to $\mathbb{Z}/2[x]/x^2$ where $x$ is of degree $1$. I do not understand why this is true. The cohomology modules are $H^0(\mathbb{R}P^1; \mathbb{Z}/2) \cong \mathbb{Z}/2$, $H^1(\mathbb{R}P^1; \mathbb{Z}/2) \cong \mathbb{Z}/2$ and all higher modules $0$ because $\mathbb{R}P^1$ is homeomorphic to the circle. So we have, we have that $H^*(\mathbb{R}P^1;\mathbb{Z}/2) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus 0 \oplus \dots$.
If I correctly understand what "$\mathbb{Z}/2[x]/x^2$ where $x$ is of degree $1$" means, this is the graded ring
$$(\mathbb{Z}/2 + (x^2))/(x^2) \oplus (\mathbb{Z}/2[x] + (x^2))/(x^2) \oplus (\mathbb{Z}/2[x^2] + (x^2))/(x^2) \oplus (\mathbb{Z}/2[x^3] + (x^2))/(x^2) \oplus \dots$$
where I denote $\mathbb{Z}/2[x^i]$ for the $\mathbb{Z}/2$ linearization of $x^i$.
But $ (\mathbb{Z}/2[x^3] + (x^2))/(x^2) \cong \mathbb{Z}/2[x]$, which is not trivial, while the fourth term of the graded cohomology ring of $H^*(\mathbb{R}P^2; \mathbb{Z}/2)$ is trivial, so then they cannot be isomorphic as graded rings can they?

I think your description of $\mathbb{Z}/2[x]/(x^2)$ is incorrect. It is a polynomial ring over $\mathbb{Z}/2$ with one variable, whose square vanishes. Explicitly, this ring consists of only the polynomials $p(x) = a + b x$ where $a, b\in \mathbb{Z}/2$, because $x^2 = 0$. As a graded ring this is simply $\mathbb{Z}/2 \oplus \mathbb{Z}/2\oplus 0 \oplus \dots$, which is isomorphic to $H^*(\mathbb{R}P^1;\mathbb{Z}/2)$.
A clean way to rigorously prove this is with the first isomorphism theorem for rings. Consider the ring homomorphism $\mathbb{Z}/2[x] \to H^*(\mathbb{R}P^1;\mathbb{Z}/2)$ which sends $x$ to the unique non-trivial element in the first cohomology. Then this is surjective, and its kernel is the two-sided ideal $(x^2)$.
