Consider $X\subseteq \mathbb A^n$ an affine variety, $p\in X$. Is it true that $\exists K_0\subseteq O_p(X)$ a subfield with $K_0\cong O_p(X)/m_p$ where $m_p$ is the maximal ideal of $O_p(X)$ ?
We have $O_p(X)\cong\Gamma(X)_{\overline{m_p}}$ where $\overline{m_p}$ is the image of $(x_1-p_1,\dots,x_n-p_n)$ by the quotient. I thought about taking the representatives of the equivalences classes of $\Gamma(X)_{\overline{m_p}}/\overline{m_p}.\Gamma(X)_{\overline{m_p}}$ together with $0$, but I can't manage to show that it's stable under addition.
If $X$ is an affine variety over an algebraically closed field, and $p$ is a closed point, then this is true. $O_{X, p} / m_p$ is a field extension of $k$. Since $p$ is closed, this is a finite degree extension of $k$, and since $k$ is algebraically closed, this is equal to $k$. $k$ can also be realized as a subset of $O_{X, p}$ since there is a morphism $k \rightarrow O_X(X) \rightarrow O_{X, p}$.
In general, this is false. Consider $X = Spec(\mathbb{Q}[x])$ and $p = (x^2 - 2)$. Then, $O_{X, p} / m_p = \mathbb{Q}[\sqrt{2}]$. But since $O_{X, p} \subseteq \mathbb{Q}(x)$, any such $K_0$ of $O_{X, p}$ is a subfield of $\mathbb{Q}(x)$, and since $\mathbb{Q}(x)$ doesn't contain $\sqrt{2}$, such a $K_0$ cannot be isomorphic to $\mathbb{Q}[\sqrt{2}]$.