The local ring at a point of an affine variety is the localization at its maximal ideal

Let $f_x\in {\mathcal{O}}_{X,x}$ be a germ, represented by a regular function $f$ on an open neighbourhood $U$ of $x$. By [L4], choose a principal open neighbourhood $D_X(s)$ with $x\in D_X(s)\subseteq U$. Then [L5] identifies $f{\mathrm{\mid }}_{D_X(s)}$ with some fraction $a\mathrm{/}{s}^n\in k[X{]}_s$. Because $x\in D_X(s)$, the denominator $s$ does not vanish at $x$, so [L2] and [L3] give $s\mathrm{\notin }{\mathfrak{m}}_x$. Define $\mathrm{\Phi }(f_x):=a\mathrm{/}{s}^n\in k[X{]}_{{\mathfrak{m}}_x}$. This gives a candidate image for the germ $f_x$.

Conversely, let $a\mathrm{/}s\in k[X{]}_{{\mathfrak{m}}_x}$. By [L2] and [L3], $s(x)\mathrm{\ne }0$, so $x\in D_X(s)$. On that principal open, the function $y\mapsto a(y)\mathrm{/}s(y)$ is regular by [L5], hence defines a germ at $x$. Put $\mathrm{\Psi }(a\mathrm{/}s):=(a\mathrm{/}s{)}_x$. This gives the inverse candidate on localisation fractions.

The value in step 1.1 is independent of the representative. Indeed, if the same germ is also represented by $g$ on a neighbourhood $V$, choose by [L4] a principal open $D_X(t)$ with $x\in D_X(t)\subseteq U\cap V$ on which $f=g$. By [L5], the restrictions of $f$ and $g$ give equal fractions in $k[X{]}_t$. Since $t\mathrm{\notin }{\mathfrak{m}}_x$ by [L2] and [L3], the equality criterion [L6] makes their images equal in the further localisation $k[X{]}_{{\mathfrak{m}}_x}$. Thus $\mathrm{\Phi }$ is well defined.

If $a\mathrm{/}s=b\mathrm{/}u$ in $k[X{]}_{{\mathfrak{m}}_x}$, then [L6] gives $t\mathrm{\notin }{\mathfrak{m}}_x$ with $t(au-bs)=0$ in $k[X]$. On the principal open $D_X(stut)$ every factor $s$, $u$, and $t$ is nonzero. Evaluating the relation there gives $t(y)(a(y)u(y)-b(y)s(y))=0,$ and division by $t(y)\mathrm{\ne }0$ in the field $k$ yields $a(y)u(y)=b(y)s(y)$. Therefore $a\mathrm{/}s=b\mathrm{/}u$ as ordinary functions on $D_X(stut)$. Since $x\in D_X(stut)$ by [L2] and [L3], the two regular functions have the same germ at $x$. Thus $\mathrm{\Psi }$ is well defined.

Start with a germ $f_x$. Choosing a principal-open representative as in step 1.1, the construction of $\mathrm{\Psi }(\mathrm{\Phi }(f_x))$ recovers exactly the germ of that same representative, so $\mathrm{\Psi }(\mathrm{\Phi }(f_x))=f_x$.

Start with a fraction $a\mathrm{/}s\in k[X{]}_{{\mathfrak{m}}_x}$. The germ constructed in step 1.2 is represented on $D_X(s)$ by the regular function corresponding under [L5] to the same fraction $a\mathrm{/}s$. Applying $\mathrm{\Phi }$ therefore returns $a\mathrm{/}s$. Hence $\mathrm{\Phi }(\mathrm{\Psi }(a\mathrm{/}s))=a\mathrm{/}s$.

Steps 2.3 and 2.4 show that $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse ring homomorphisms. By [L7], the target is local, so ${\mathcal{O}}_{X,x}$ is canonically identified with the localisation $k[X{]}_{{\mathfrak{m}}_x}$ as a local ring.