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A propositional geometric theory has primitive propositions and sequents between formulas built with finite conjunction and arbitrary set-indexed disjunction. Start with the free frame on the primitives, identify conjunction with meet and disjunction with join, and impose each sequent as an inequality. The resulting quotient frame is $\mathcal{O}(X_T)$. In the sheaf topos $\mathrm{Sh}(X_T)$, interpret each primitive as the subterminal sheaf corresponding to its generating open. Operations on these subterminal sheaves are precisely the frame operations, so every imposed sequent is valid. This constructs the canonical model.

For a commutative ring $R$, denote the generator attached to $r$ by $U_r$. The presentation impliesThe product equality uses the multiplication implication in one direction and both factor implications in the other; the latter follow by commuting $r,s$. A ring homomorphism $\phi :R\to S$ sends these relations to the corresponding ones in $S$. It therefore gives a frame homomorphism $U_r\mapsto U_{\phi (r)}$ and hence a map $X_S\to X_R$ of locales.

If $ar+bs=1$, thenso the two opens cover. Conversely let $S=R/(r,s)$ and pull back a putative cover along $X_S\to X_R$. Both generators become $U_0$, so $1=0$ in $\mathcal{O}(X_S)$. The allowed degeneracy criterion gives $0=1$ in $S$, equivalently $1\in (r,s)$. ThusThis necessity argument never selects a prime ideal.

Finite intersections of the generating opens are again generators, so they form a basis. Assign $R[r^{-1}]$ to $U_r$. The localization of a ring carries its ordinary addition and multiplication of fractions; equality of $a/r^n$ and $b/r^m$ means that some power of $r$ annihilates the cross-multiplied difference. On $U_r\cap U_s=U_{rs}$, restriction is further localization.

We also justify restriction for any inclusion of basic opens. Adding the condition $U_t=1$ to the presentation of $X_R$ gives the open $U_t$ and the presentation of $X_{R[t^{-1}]}$: the generator for $a/t^n$ corresponds to $U_a\wedge U_t$. This is well-defined since equality of fractions, after multiplication by a power of the now-invertible $t$, makes the two membership propositions equivalent. More explicitly, $U_t=1$ implies $U_{t^{k}z}=U_z$. If $t^{k}z=0$, this gives $U_z=0$. If $x-y$ is killed by a power of $t$, then $U_{x-y}=U_{y-x}=0$ (since $U_{-1}=1$), and the additive inequalities for $x=y+(x-y)$ and $y=x+(y-x)$ give $U_x=U_y$. A zero element in the localized ring has false membership, multiplication gives intersections, and the additive relation follows by clearing denominators. The reverse assignment on $a\in R$ proves that these maps are inverse. Thus $X_{R[t^{-1}]}\cong U_t$. If $U_t\subseteq U_r$, then in this localized theory $U_r=1$. Apply the quotient argument above, with the second generator $0$, to conclude $1\in (r)$ in $R[t^{-1}]$. Therefore $r$ is invertible there, and the universal property of localization gives the restriction $R[r^{-1}]\to R[t^{-1}]$. These restrictions compose uniquely.

It remains to glue, and the allowed basis criterion reduces this to a whole-locale cover $U_r\vee U_s=1$. Let $\alpha \in R[r^{-1}]$ and $\beta \in R[s^{-1}]$ have equal restrictions to $R[(rs{)}^{-1}]$. Choose a common initial exponent $n$, writing $\alpha =a/r^n$, $\beta =b/s^n$. Equality on the overlap supplies $k\ge 0$ withSet $N=n+k$, $A=r^{k}a$ and $B=s^{k}b$. Thenwith the last equality in $R$ itself. This clearing step is necessary when the commutative ring has zero divisors.

Since $(r,s)=R$, also $(r^N,s^N)=R$. Choose $\lambda r+\mu s=1$. For $N\ge 1$, expand $(\lambda r+\mu s{)}^{2N-1}=1$; each term contains either $r^N$ or $s^N$, producing $C,D\in R$ with $C{r}^N+D{s}^N=1$. For $N=0$ choose $C=1,D=0$. DefineIn $R[r^{-1}]$,and similarly in $R[s^{-1}]$,Thus $t$ glues the matching pair. If $t^{\prime }$ is another gluing, some $r^p$ and $s^q$ annihilate $t-t^{\prime }$. The same binomial argument gives $(r^p,s^q)=R$, so $t-t^{\prime }=0$. This proves uniqueness without cancellation or a domain assumption.

The exactness of the unit-ideal localization Čech complex proves the sheaf axiom for the required basic covers, and hence the assignment with its restrictions defines a sheaf of rings $\tilde{R}$ with $\tilde{R}(U_r)=R[r^{-1}]$. On the empty basic open, localization is the zero ring, whose underlying set has one section, as required. The construction and proof remain valid for the zero ring.