This question was asked in my assignment in algebraic geometry and I am struck on it.
Question: Let $K$ be an algebraically closed field, $K[X_1, \dotsc, X_n]$ the polynomial ring in $n$ indeterminates $X_1, \dotsc, X_n$ over $K$ and let $X = \operatorname{Spec} K[X_1, \dotsc, X_n]$. Show that:
(a) The set of closed points in $X$ can canonically be identified with $K^n$.
(b) If $n = 1$, then there is exactly one non-closed point in $X$, namely the generic point of $X$.
(c) If $n = 2$, then the non-closed points in $X$ that are different from the generic point are given by principal ideals $\langle f \rangle$ where $f \in K[X_1, X_2]$ is irreducible and the closure $\bar{y}$ of such points consists if $y$ as the generic point and of the curve $\{ x \in K^2 \mid f(x) = 0 \}$.
Attempt: In a topological space $(X, T)$, an element $x \in X$ is called a closed point if the singleton set $\{ x \} \subset X$ is a closed subset of $X$.
I am assuming that it is defined on Zariski topology.
(a) $x = V(S) = \{ x \in \mathbb{A}^n \mid \text{$f(x) = 0$ for all $f \in S$} \}$. But I am confused about what map should I define from the set of closed points in $X$ to $K^n$ to prove that isomorphism exists. Can you please help with this?
(b) Assume that there exists no non-closed points, i.e., all points are closed. What result should I use to get a contradiction?
Similarly, if I assume that let there exists more than one no-closed points. Again I am not able to move forward.
(c) I would like to do it by myself.
Can you please help me with this by giving a few hints?
I shall be really grateful.
a) Hint: Show that there is a correspondence between maximal ideals in $K[x_1,...,x_n]$ and closed points of $X$ and use the fact that common zeros of $(x-a_1,...,x-a_n)$ consists of a unique element $(a_1,...,a_n)\in K^n$.
For part b), consider that all prime ideals in $K[x]$ are maximal except $(0)$, hence in $X$ all elements are closed except the member related to prime ideal $(0)$. You can make sure it's a generic point ($\bar{(0)}=X$) by checking that it's contained in any open subset $ D(f)\subset X \quad \forall f\in K[x]$.
c) Hint: prime ideals of $K[x_1,x_2]$ are $(0)$, maximal ideals and others generated by irreducible elements. Then you can consider an irreducible element such as $x_1^2-x_2$ and check $(x_1-2,x_2-4)$ lies in the closure of $(x_1^2-x_2)$.
Hint 2: For this example, $p=(x_1^2-x_2)\subset (x_1-2,x_2-4)$ hence from $f\notin (x_1-2,x_2-4)$, where $f\in K[x_1,x_2]$ we have $f\notin p.$ Therefore any open subset $D(f)$ containing $(x_1-2,x_2-4)$ also contains $p$. Hence $(x_1-2,x_2-4)$ is in the closure of $p$.