k-valued point of $\mathbb{P}_k^n $ Let $ k $ be a field. A k-valued point of $\mathbb{P}_k^n $ is define to be a scheme morphism
$$ \operatorname{Spec}k \rightarrow \mathbb{P}_k^n$$
Is this the same as just specifying a maximal homogeneous ideal of $k[x_ 0,…,x_n] $? If so, is such an ideal always generated by n linear homogeneous polynomials?
 A: By Stacks Project Lemma 01NA, morphisms from $\operatorname{Spec}(k)$ to $\operatorname{Proj}k[x_0,\ldots,x_n]$ correspond to strict equivalences (see the discussions before the cited lemma) of pairs of an invertible $k$-module $\mathcal L$ and a graded ring morphism $\psi:k[x_0,\ldots,x_n]\rightarrow\Gamma_*(\operatorname{Spec}k,\mathcal L)$ such that $\mathcal L$ is generated by the global sections $\psi(f)$ for $f\in k[x_0,\ldots,x_n]$ of degree $1$.
Since an invertible $k$-module is isomorphic to $k$, a $k$-point is determined by a morphism $\psi:k[x_0,\ldots,x_n]\rightarrow\Gamma_*(\operatorname{Spec}k,k)$ such that $\psi(x_i)$ is not all $0$ for every $i=0,\ldots,n$. Since $\psi$ is a graded ring homomorphism, its kernel $K$ is a homogeneous ideal of $k[x_0,\ldots,x_n]$. Since $k[x_0,\ldots,x_n]/K\cong k$, $K$ is a maximal ideal.
But not every maximal ideal corresponds to a $k$-valued point: only those with residue field $k$. If $k$ is algebraically closed, then indeed $k$-valued points are the same as a maximal homogeneous ideals of $k[x_0,\ldots,x_n]$.
Remark:
There are definitely easier ways of showing this, but maybe this can be useful.
Edit:
As to the added question whether such a maximal homogeneous ideal is generated by $n$ linear homogeneous ideals, consider an index $i_0$ such that $\psi(x_{i_0})\ne0$. Let $a_i$ denote the image of $x_i$ in $k$ under $\psi$. Then we can show that the kernel $K$ is generated by $X_i-(a_i/a_{i_0})X_{i_0}$, where $i\ne i_0$, which is left as an exercise. ;P

A reference for the statement in the edit is this stack exchange question.


If there are any errors, please point it out.
Thanks in advance.
