 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 idealIt 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 $h(p)\ne 0$ and $D(h)\subseteq V$. Thenas 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 $(hb)(p)\ne 0$.

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 provesThe 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.