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?

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 $\mathbb P^1_\mathbb R$.