Definition of $\mathbb{P}_A^n$ I feel confused with these words: "$B=A[x_0,\dots,x_n]$. Then $B_+=(x_0,\dots,x_n)$ but this will certainly not give a point on $\textrm{Proj }B=\mathbb P^n_A$. (The point $(0:\cdots:0)$ is not defined)."
What does the point $(0:\cdots:0)$ mean in $\mathbb P^n_A$? By definition every point in $\textrm{Proj }B=\mathbb P^n_A$ is a prime ideal which doesn't contain $B_+$ right? How can we get them in the coordinate form?
 A: The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $\mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,\dots,a_n)$ and maximal ideals $(x_1 - a_1,\dots,x_n - a_n)$.
For $\mathbb{P}^n_k$, a point $a = (a_0: \cdots : a_n)$ corresponds to the line
$$ \ell_a = \{ (ta_0, \dots, ta_n) : t \in k \} \subseteq k^{n + 1}$$
Each line contains $(0,\dots,0)$ and $\ell_a = \ell_{a'}$ if and only if $(a_0,\dots,a_n) = t(a_0',\dots,a_n')$ for some $t \in k^*$.
So the closed points of $\mathbb{P}^n_k$ correspond to elements of $k^{n + 1} \setminus \{(0,\dots,0)\}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:\dots:a_n)$ is: it's the equivalence class of $(a_0,\dots,a_n)$ modulo non-zero scalar multiplication. This is how $\mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $\ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $\ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 \ne 0$. Then the ideal is generated by $x_j = \frac{a_j}{a_0}x_0$ for $j \in \{1,\dots,n\}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $\mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,\dots,x_n$, the variety it defines in $\mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,\dots,a_n) \in V(I)$ then $ta \in V(I)$ for any $t \in k$ and hence $\ell_a \subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,\dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,\dots,0)$ and $\ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is $\{0\}$. By the Nullstellensatz, $V(I) = \{0\}$ if and only if $\sqrt{I} = (x_0,\dots,x_n)$. Hence the condition of not containing $(x_0,\dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $\mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,\dots,x_n]$ which are maximal with respect to not containing $(x_0,\dots,x_n)$.
