Regular functions on a principal open are the principal localization of the coordinate ring

A fraction $a\mathrm{/}{\bar{f}}^n\in k[X{]}_{\bar{f}}$ defines a function on $U$ by $x⟼\frac{a(x)}{\bar{f}(x{)}^n}\mathrm{.}$ If $a\mathrm{/}{\bar{f}}^n=b\mathrm{/}{\bar{f}}^m$, then after passing to the common denominator ${\bar{f}}^{n+m}$ their difference is $c\mathrm{/}{\bar{f}}^{n+m}$ with $c=a{\bar{f}}^m-b{\bar{f}}^n$. By [L2], some power of $\bar{f}$ annihilates $c$. Evaluating at a point of $U$, where $\bar{f}$ is nonzero, forces $c(x)=0$, so the two fractions give the same function. Thus there is a well-defined ring homomorphism $\mathrm{\Phi }:k[X{]}_{\bar{f}}\to {\mathcal{O}}_X(U)$.

If $U=\mathrm{\varnothing }$, then $\bar{f}(x)=0$ for every $x\in X$, so $\bar{f}=0$ in $k[X]$. By [L1], the localization $k[X{]}_{\bar{f}}$ is the zero ring. The only function $\mathrm{\varnothing }\to k$ is both additive and multiplicative identity, so ${\mathcal{O}}_X(U)$ is also the zero ring. Hence the statement is proved in this boundary case.

Let $\phi \in {\mathcal{O}}_X(U)$, given by a finite principal-open cover $U={\bigcup }_{i=1}^r{D}_X({\bar{h}}_i)$ and local formulas $\phi =\frac{{\bar{g}}_i}{{\bar{h}}_i}\text{on }{D}_X({\bar{h}}_i)\mathrm{.}$ Discard the indices with $D_X({\bar{h}}_i)=\mathrm{\varnothing }$; the remaining family still covers the nonempty open set $U$, and relabel it as $D_X({\bar{h}}_1),\dots ,D_X({\bar{h}}_r)$. For any $i,j$, the overlap $D_X({\bar{h}}_i)\cap D_X({\bar{h}}_j)=D_X({\bar{h}}_i{\bar{h}}_j)$ is open by [L4]. It is also nonempty because $D_X({\bar{h}}_j)$ is a nonempty open subset of $X$, while [L5] makes $D_X({\bar{h}}_i)$ dense in $X$. On that overlap the two local expressions for $\phi$ agree, so the class ${\bar{g}}_i{\bar{h}}_j-{\bar{g}}_j{\bar{h}}_i\in k[X]$ vanishes on a dense subset of $X$ and therefore on all of $X$. By [L3], ${\bar{g}}_i{\bar{h}}_j={\bar{g}}_j{\bar{h}}_i\text{in }k[X]\mathrm{.}$

Choose representatives $f,h_1,\dots ,h_r\in k[x_1,\dots ,x_n]$ of $\bar{f},{\bar{h}}_1,\dots ,{\bar{h}}_r$. Because $D_X(\bar{f})={\bigcup }_{i=1}^r{D}_X({\bar{h}}_i)$, a point of $X$ lies outside $U$ exactly when all $h_i$ vanish there. Thus $V(I(X)+(f))=V(I(X)+(h_1,\dots ,h_r))$ inside ${\mathbf{A}}_k^n$. Applying [L6] gives $f\in I(V(I(X)+(f)))=I(V(I(X)+(h_1,\dots ,h_r)))=\sqrt{I(X)+(h_1,\dots ,h_r)},$ so for some $M\ge 1$ there exist $b_1,\dots ,b_r\in k[x_1,\dots ,x_n]$ and $u\in I(X)$ with $f^M=u+b_1{h}_1+\cdots +b_r{h}_r\mathrm{.}$ Passing to $k[X]$ yields ${\bar{f}}^M={\bar{b}}_1{\bar{h}}_1+\cdots +{\bar{b}}_r{\bar{h}}_r\mathrm{.}$

Assume now that $U\mathrm{\ne }\mathrm{\varnothing }$. To prove injectivity, let $a\mathrm{/}{\bar{f}}^n\in k[X{]}_{\bar{f}}$ map to the zero function on $U$. Then $a(x)=0$ for every $x\in U$. On the complement $X\setminus U$ one has $\bar{f}(x)=0$. Therefore $(a\bar{f})(x)=0$ for every $x\in X$, so $a\bar{f}=0$ in $k[X]$. By [L2], this implies $a\mathrm{/}{\bar{f}}^n=0$ in $k[X{]}_{\bar{f}}$. Thus $\mathrm{\Phi }$ is injective.

Define $a:={\bar{b}}_1{\bar{g}}_1+\cdots +{\bar{b}}_r{\bar{g}}_r\in k[X]\mathrm{.}$ Fix $x\in D_X({\bar{h}}_j)$. Using the relations from step 1.3 and then step 1.4, we compute $a(x){\bar{h}}_j(x)={\sum }_{i=1}^r{\bar{b}}_i(x){\bar{g}}_i(x){\bar{h}}_j(x)={\bar{g}}_j(x){\sum }_{i=1}^r{\bar{b}}_i(x){\bar{h}}_i(x)={\bar{g}}_j(x)\bar{f}(x{)}^M\mathrm{.}$ Since ${\bar{h}}_j(x)\mathrm{\ne }0$, division in the field $k$ gives $\frac{a(x)}{\bar{f}(x{)}^M}=\frac{{\bar{g}}_j(x)}{{\bar{h}}_j(x)}=\phi (x)\mathrm{.}$ Because the $D_X({\bar{h}}_j)$ cover $U$, this holds for every $x\in U$. Hence $\phi =\mathrm{\Phi }(a\mathrm{/}{\bar{f}}^M)$, so $\mathrm{\Phi }$ is surjective.

Step 2.1 proves injectivity, step 2.2 proves surjectivity, and step 1.2 handles the empty-open boundary. Therefore $\mathrm{\Phi }:k[X{]}_{\bar{f}}\stackrel{\sim }{\to }{\mathcal{O}}_X(U)$ is a ring isomorphism.