I am trying to understand the Proj $R$ construction, in particular why Proj R only contains the graded prime ideals that do not contain the irrelevant ideal. I was reading this post: Why does $\operatorname{Proj}(B)$ not contain ideals containing the irrelevant ideal?.
In the answer we consider the example of $R = K[x_1, \dots, x_{n+1}]$ being the polynomial ring with the irrelevant ideal $R_+ = (x_1, \dots, x_{n+1})$. And then Proj R $= \mathbb{P}^n$ but I do not understand why. I did not find any proof for this or don't know if this is obvious.
Intuitively, it makes a bit sense to me, since the elements of Proj $R$ are the prime ideals generated by homogeneous polynomials, which satisfy $f(\lambda x_1, \dots, \lambda x_{n+1}) = 0 \Longleftrightarrow f(x_1, \dots, x_{n+1}) = 0$, which reminds me of the coordinates of the projective space.
Is there any proof for this equality ?
Edit: Here is my confusion Proj $R = \{ p \in Spec R | ~ p$ graded ideal, p $\not\supset R_+ \}$, and $\mathbb{P}^n = (K \setminus \{0\})^{n+1} / \sim $ with the equivalence relation $x \sim y \Leftrightarrow$ there exists a $\lambda \in K$, s.t. $x_i = \lambda y_i$ for all $i$.
How can they be equal if the elements from these two sets are of different type: elements of Proj $R$ are ideals of polynomials and elements of $\mathbb{P}^n$ are equivalence classes of points of $(K \setminus \{0 \})^{n+1}$.
Let me denote $\mathbf{P}(K^{n+1})$ what you defined as $\mathbf{P}^n$ in your comment.
The result you are looking for is the following: there is a bijection between $\mathbf{P}(K^{n+1})$ and the set of rational points of $\mathrm{Proj}(R)$. See for example lemma 2.3.43 of Qing Liu's Algebraic geometry and arithmetic curves.
Furthermore, whenever $X$ is a $K$-variety, you can endow $X(K)$ (which might be empty!) with a structure of $K$-variety (in the naive sense): here what you get is that the aforementioned bijection is an isomorphism of $K$-varieties.