= Solution
Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, and
$$
R/(u,v)\cong\mathbb Z,
$$
so its <kernel> is the <prime ideal> $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb Z$-<module>, generated by the classes of $u,v$.
For integers $a,b$, including negative integers, reduction modulo $P^2$ gives
$$
(1+u)^a(1+v)^b\equiv1+au+bv\pmod{P^2}.
$$
For negative powers, this follows from $(1+u)^{-1}\equiv1-u$ and $(1+v)^{-1}\equiv1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a <Laurent polynomial> $f$ satisfies
$$
f\equiv\theta(f)+D_1(f)u+D_2(f)v\pmod{P^2},\qquad D_i(f)=\left.\frac{\partial f}{\partial X_i}\right|_{(1,1)}\in\mathbb Z.
$$
The <formal derivatives> obey the <product rule>, so $D_i(ab)=\theta(a)D_i(b)+\theta(b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a map
$$
P/P^2\longrightarrow\mathbb Z^2,\qquad [f]\longmapsto(D_1(f),D_2(f)).
$$
It sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the <conormal module> has the explicit basis
$$
\boxed{P/P^2\cong\mathbb Z^2,\quad [u],[v]\text{ a basis}.}
$$
Let $A=R_P$ be the <localization at a prime ideal>, with <maximal ideal> $\mathfrak m=PR_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The <residue field> is
$$
\boxed{A/\mathfrak m=\operatorname{Frac}(R/P)=\mathbb Q.}
$$
Indeed there is an explicit identification
$$
A\cong\mathbb Q[u,v]_{(u,v)}:
$$
inverting $1+u$ and $1+v$ has no effect in this <local ring>, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.
For the <Krull dimension>, the <prime ideal correspondence for localization> preserves the strict chain
$$
(0)\subsetneq(u)\subsetneq(u,v)
$$
of <prime ideals> in $R$. They are prime because the successive quotients are the <integral domains> $R$, $\mathbb Z[X_2,X_2^{-1}]$, and $\mathbb Z$. They remain distinct after <localization at a prime ideal> because they are all contained in $P$. Hence $\dim A\ge2$.
For the upper bound, $R$ is a <Noetherian ring> by the <Hilbert basis theorem> and <Localization of a Noetherian ring>. Since $P$ is generated by two elements, the <Krull height theorem> gives $\operatorname{ht}P\le2$. The <prime ideal correspondence for localization> identifies $\dim R_P$ with $\operatorname{ht}P$, so
$$
\boxed{\dim R_P=2.}
$$
A <Noetherian local ring> $(B,\mathfrak n)$ with <residue field> $\kappa$ is a <regular local ring> if
$$
\boxed{\dim_\kappa(\mathfrak n/\mathfrak n^2)=\dim B.}
$$
By the <Nakayama lemma>, the dimension on the left is the minimal number of generators of $\mathfrak n$; this is also called the <embedding dimension>.
The <localization of a module> is exact and commutes with quotients and products of ideals. Applying it to the already computed <conormal module> gives the <cotangent space of a local ring>
$$
\mathfrak m/\mathfrak m^2\cong(P/P^2)\otimes_{\mathbb Z}\mathbb Q\cong\mathbb Q^2.
$$
Its <dimension> is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb Z$, whereas the localized <cotangent space of a local ring> has a basis over the <residue field> $\mathbb Q$.
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