A question about an example of integral notherian scheme that is not a finite type over K While reading up on 'finite type over $k$' on Hartshorne, algebraic geometry, I have tried to understand the following example :

Example ) If $P$ is a point of a variety of $V$, with local ring $\mathcal{O}_P$, then $\operatorname{Spec} \mathcal{O}_P$ is an integral noetherian scheme, which is not in general of finite type over $k$.

However, I have not yet digested this example. The following texts are the way where I saw this example:
When considering the definition of point of variety $V$, I think that the local ring $\mathcal{O}_P $ is indeed a field of fractions of $V$. Since $\mathcal{O}_P $ is anyway a field (or, equivalently, a unique maximal ideal), I think that $X:=\operatorname{Spec} \mathcal{O}_P$ is merely singleton set (actually, $X=\{(0)\}$?) Then, my rough sketch is here,

i) $\mathcal{O}_X(U)$ is an integral domain.
ii) $X$ is covered by a finite collection of $\operatorname{Spec} A_i $ (where $A_i$'s
  are noetherian) 
iii) let $f:X \to Y$ be a morphism of schemes (  where $X=\operatorname{Spec} \mathcal{O}_P, Y=\operatorname{Spec} \mathcal{O}_Q$ ). 
  Then, for each i,
  $f^{-1}(\operatorname{Spec} {B}_i$) is cannot be covered finite number of
  $U_{ij}=\operatorname{Spec} A_{ij} $

However, everything is not easy now ... Above all, I wonder whether or not iii) is the right idea. When considering the definition 'a finite type over $k$', essentially there is a morphism of schemes. Thus, by setting another point $Q$ of variety of $V$, and then gave a morphism $f: \operatorname{Spec} \mathcal{O}_P \to \operatorname{Spec} \mathcal{O}_Q$  ... but I am not sure whether this is the right way...
 A: You have a lot of misconceptions! Let us try to sort some of them out.
First, $\mathcal{O}_{X,P}$ is NOT a field in general. Even the simplest examples of a closed point in positive-dimensional affine space are local rings which aren't fields! $\mathcal{O}_{\Bbb A^1,p} \cong k[x]_{(x)}$, and this is not a field because $x$ is not invertible. Actually, $\mathcal{O}_{X,x}$ is a field iff $x$ is a generic point of $X$, and as Hartshorne's varieties are irreducible, they have a unique generic point. So $\mathcal{O}_{X,x}$ is a field at exactly one point of a variety! 
Your parenthetical that a ring with a unique maximal ideal is a field is just flat wrong and you should discard this belief as soon as possible. A field is a ring with a unique proper ideal, but there are lots of rings with a unique maximal ideal which have more than one proper ideal. Again, $k[x]_{(x)}$ is good to keep in mind because it's a local ring which isn't a field: there is a unique maximal ideal $(x)$ but there is also the ideal $(0)$.
Next, the condition that $\mathcal{O}_{X,x}$ is not of finite type over $k$ is a purely ring-theoretic condition, and it exactly means that there is no $n\in\Bbb Z_{\geq 0}$ so we can find a surjective morphism of rings $k[x_1,\cdots,x_n]\to \mathcal{O}_{X,x}$. (Your sketch does not deal with this and probably cannot be salvaged - there are other issues with it.) Let us compute that a specific example of such a local ring cannot be finitely generated as a $k$-algebra.
Suppose we have a surjective morphism of $k$-algebras $k[t_1,\cdots,t_n]\to k[x]_{(x)}$. Then we can get a surjective morphism of $k$-algebras $k[t_1,\cdots,t_n,u]\to (k[x]_{(x)})[u]/(xu-1)$ and as $(k[x]_{(x)})[u]/(xu-1) \cong k(x)$, we have shown that $k(x)$ is finitely generated as a $k$-algebra. But by Zariski's lemma, this means that $k(x)$ must be a finite extension of $k$, and in particular, algebraic. On the other hand, it has transcendence degree one over $k$ and is therefore not purely algebraic, contradiction. So $k[x]_{(x)}$ is not finitely generated as a $k$-algebra.
