Closed points in projective space correspond to which homogenous prime ideals in $k[x_0,...,x_n]$ I'm trying to think about exercise 4.5.O in Vakil's notes on Algebraic Geometry.
Before we defined the scheme $\mathbb{P}^n_k := \operatorname{Proj}(k[x_0,...,x_n])$ and showed that that for $k$ algebraically closed the closed points of $\mathbb{P}^n_k$ correspond to the points of "classical" projective space.
Now the question is: To which homogeneous prime ideal in $k[x_0,...,x_n]$ does the point $[a_0:\dots :a_n]$ correspond?
My thinking so far: Since there is an $a_i \ne 0$ we can pick the representative with $a_i = 1$. Then we get the homogeneous ideal
\begin{equation}
 I = (\sum_{j \ne i} a_j x_j + x_i)
\end{equation}
Is this the one we are looking for? If so, how do you prove all the details, e.g. why is $I$ prime, doesn't contain the inessential ideal and really corresponds to this point?
 A: To provide an answer along the lines of what you were thinking:
Take the classical point $[a_0: a_1 : \dots :a_n]$. For simplicity, let $a_0 \ne 0$.
Then $[a_0: a_1 : \dots :a_n]=[1: \frac{a_1}{a_0}: \dots: \frac{a_n}{a_0}]$.
This point corresponds to the maximal ideal $(\frac{x_1}{x_0}-\frac{a_1}{a_0}, \dots, \frac{x_n}{x_0} -\frac{a_n}{a_0}) \in k[\frac{x_1}{x_0}, \dots, \frac{x_n}{x_0}]=(k[x_0, \dots, x_n]_{x_0})_0$, which then corresponds to the homogeneous prime ideal $(a_0 x_1-a_1x_0, \dots, a_0 x_n-a_n x_0) \in k[x_0, \dots, x_n]$ (see exercise 4.5.E.).
The case when some $a_i \ne 0$ for $1\le i\le n$ is similar.
A: The classical point $[a_0:a_1:\dots:a_n]$ corresponds to the homogeneous ideal $$\langle a_ix_j-a_jx_i\vert i,j=0\dots n\rangle\subset k[x_0,\dots,x_n]$$ 1) Do not break the beautiful symmetry of this ideal by ugly choices (like $a_i=1$) for the homogeneous coordinates of your point.   
2) The formula is still valid for an arbitrary field $k$.
However if $k$ is not algebraically closed $\mathbb P^n_k$ will contain closed points not of the form above.
For example the ideal $\langle x_0^2+x_1^2\rangle\subset \mathbb R[x_0,x_1]$ also corresponds to a closed point of $\mathbb P^1_\mathbb R$.
