# Projective Space over an algebraically closed field

I want to show the following:

If $k$ is an algebraically closed field, then the closed points of $\mathbb{P}^n_k$ can be thought of projective coordinates $[x_0:x_1: ... :x_n]$ where not all the $x_i$ are zero and $[x_0:x_1: ... :x_n] = [\lambda x_0:\lambda x_1: ... :\lambda x_n]$ for every $\lambda \neq 0$.

The construction I should use to show that is as follows:

Define $U_i =$ Spec $k[x_{0/i},...,x_{n/i}]/(x_{i/i}-1), i=0...n$. Now we glue all those schemes together along $\varphi_{ij} : D(x_{j/i}) \to D(x_{i/j})$ given by the isomorphisms $x_{k/i} \mapsto \frac{x_{k/j}}{x_{i/j}}$.

Now I understand, that the assumption of $k$ being algebraically closed implies, that the closed points of every affine space used to glue $\mathbb{P}^n_k$ are especially easy. However, I'm struggling with the following: I thought that given $[a_0:...:a_n]$ with $a_i \neq 0$ I should think of that as the point $(x_{0/i}-\frac{a_0}{a_i},...,x_{n/i}-\frac{a_n}{a_i})$ (excluding $x_{i/i}$) in $U_i$. And I thought that if for example both $a_i \neq 0 \neq a_j$, then those points would be glued together via the isomorphisms given above in $\mathbb{P}^n_k$. But I don't really see why this is true, as $\varphi_{ij}(x_{k/i}-\frac{a_k}{a_i}) = \frac{x_{k/j}}{x_{i/j}} - \frac{a_k}{a_i}$.

I'd be very grateful for some hints on what I'm missing..

• What is the problem? $\frac{x_{k/j}}{x_{i/j}} - \frac{a_k}{a_i}$ belongs to the maximal ideal $(x_{0/j}-\frac{a_0}{a_j},\ldots ,x_{n/j}-\frac{a_n}{a_j})$, as one can see by considerong the map of the ring to the quotient ring by this maximal ideal (i.e. evaluation of coordinates). – danneks Feb 26 '18 at 11:16
• Wow. I honestly didn't notice that... Thank you very much! – lush Feb 26 '18 at 11:42