Let $V$ be an affine variety. The ideal of $V$ is $I(V) = \{f\in \bar K[X] \mid f(P)=0\;\forall P\in V \}$. If $I(V)$ is generated by elements in $K[X]$, the $V$ is said to be defined over $K$ and define $I(V/K) = I(V)\cap K[X]$. Define the coordinate rings
$$
\bar K[V] = \bar K[X] / I(V) \quad \text{ and } \quad K[V] = K[X]/I(V/K).
$$
The function field $K(V)$ is the field of fractions of $K[V]$, and the function field $\bar K(V)$ is the field of fractions of $\bar K[V]$.
Let $P$ be a point on the affine variety $V$. The ideal $M_P$ in the coordinate ring $\bar K[V]$ of $V$ is defined by $M_P = \{ f\in \bar K[V]\mid f(P) = 0\}$. So $M_P$ is a maximal ideal in $\bar K[V]$ since $\bar K[V]/M_P$ is isomorphic to the field $\bar K$ via $f \mapsto f(P)$.
The local ring of $V$ at $P$, denoted by $\bar K[V]_P$, is
$\bar K[V]_P = \{ F \in \bar K[V] \mid F =f/g$ for some $f,g \in \bar K[V]$ with $g(P) \neq 0\}$.
My question:
Why do not define $\ K[V]_P$?
Do we define local ring of variety over not necessarily closed field?
I think $\ K[V]_P = \{ F \in \ K[V] \mid F =f/g$ for some $f,g \in \ K[V]$ with $g(P) \neq 0\}$ is just natural definition, what is wrong with this definition?
Best Answer
Question: "My question: Why do not define K[V]P? Do we define local ring of variety over not necessarily closed field? I think K[V]P={F∈ K[V]∣F=f/g for some f,g∈ K[V] with g(P)≠0} is just natural definition, what is wrong with this definition?"
Answer: If $k$ is any field and $f:A\rightarrow B$ is a map of finitely generated $k$-algebras, it follows for any maximal ideal $\mathfrak{m} \subseteq B$ the inverse image $f^{-1}(\mathfrak{m}):=\mathfrak{n} \subseteq A$ is a maximal ideal. This is because the are inclusions
$$k \subseteq \kappa(\mathfrak{n}) \subseteq \kappa(\mathfrak{m}) $$
and since $k \subseteq \kappa(\mathfrak{m})$ is finite it follows $\kappa(\mathfrak{n})$ is a field. Hence if $Specm(A) \subseteq Spec(A)$ is the topological subspace consisting of maximal ideal in $A$, you get an induced continuous map (in the Zariski topology)
$$ f^*: Specm(B) \rightarrow Specm(A).$$
If $X$ is a scheme of finite type over a field (or a Dedekind domain, or a Hilbert-Jacobson ring) you may construct a locally ringed space $(X^*, \mathcal{O}_{X^*})$ using the closed points in $X$, and this construction captures "nilpotent elements".
Example: Let $A$ be a finitely generated $k$-algebra with $k$ any field. Let $X^*:=Specm(A) \subseteq Spec(A)$ be the set of maximal ideals in $A$, and let let $i: X^* \rightarrow X$ be the inclusion map. You define the structure sheaf $\mathcal{O}_{X^*}:=i^{-1}(\mathcal{O}_X)$ using the topological inverse $i^{-1}$.
By definition it follows $(X^*, \mathcal{O}_{X^*})$ is a ringed topological space with the following properties:
$$H^0(X^*, \mathcal{O}_{X^*})=A$$
and for any maximal ideal $\mathfrak{m} \subseteq A$ you get
$$ \mathcal{O}_{X^*,\mathfrak{m}} \cong A_{\mathfrak{m}}.$$
Hence $(X^*, \mathcal{O}_{X^*})$ is a locally ringed space. You may also prove that if $A,B$ are finitely generated $k$-algebras and if $f:A\rightarrow B$ is a map of unital $k$-algebras, there is a map of locally ringed spaces
$$(f,f^{\#}): (Y^*, \mathcal{O}_{Y^*}) \rightarrow (X^*, \mathcal{O}_{X^*}),$$
where $Y^*:=Specm(B)$. In some sense this definition is "more intuitive" and "more pedagogical" than the definition using prime ideals.
Question: "what is wrong with this definition?"
Answer: There is nothing wrong with this definition - you may construct a locally ringed space from the set of maximal ideals in $K[V]$ for any finitely generated $k$-algebra $K[V]$ and any field $k$. The construction using prime ideals is much more streamlined and this is why people prefer this approach. You should not let people call you a "fool" or a "retard" for asking such a question - it is a natural question to ask.
Here you find a discussion of this problem:
https://mathoverflow.net/questions/377922/building-algebraic-geometry-without-prime-ideals/378961#378961
Here is some history:
https://www.ams.org/notices/199908/fea-raynaud.pdf