= Solution
<Polynomial functions> are regular, giving an <injective> map $A(X)\hookrightarrow\mathcal O_X(X)$; <injectivity> follows from quotienting by the full <vanishing ideal>. To prove <surjectivity>, let $f$ be globally regular and consider the denominator <ideal>
$$
J=\{g\in A(X):gf\text{ is a polynomial function on }X\}.
$$
It is an <ideal> because sums and <polynomial> multiples preserve this property.
Fix $p\in X$. Locally $f=a/b$ on a <neighbourhood> $V$ of $p$, with $b$ nonzero there. Principal affine opens form a <basis>, so choose $h\in A(X)$ with $h(p)\neq0$ and $D(h)\subseteq V$. Then
$$
(hb)f=ha
$$
as functions on all of $X$: on $D(h)$ it follows from the local fraction, and off $D(h)$ both sides vanish because $h=0$. Hence $hb\in J$ and $(hb)(p)\neq0$.
Thus $J$ has no common zero on $X$. By the weak Nullstellensatz for $A(X)$, every proper <ideal> lies in a <maximal ideal> belonging to a <point> of $X$; therefore $J$ cannot be proper. We get $1\in J$, so $f$ itself is <polynomial>. This proves
$$
\boxed{\mathcal O_X(X)\cong A(X).}
$$
The argument works for reducible <affine varieties> as well: it uses functions and denominator <ideals> rather than assuming a single <fraction field>.
This is the <denominator ideal proof for global regular functions>.
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