Denominator ideal proof for global regular functions

A regular function $f$ on an entire affine variety is a polynomial function, even if the variety is reducible. Put $J=\{g\in A(X):gf\in A(X)\}$, an ideal of its coordinate ring. Near a point $p$, write $f=a/b$ with $b$ nonzero, and choose a principal open subset $D(h)$ containing $p$ inside that neighbourhood. Then $(hb)f=ha$ on all of $X$: on $D(h)$ this is the fraction identity, and off $D(h)$ both sides vanish. Hence $hb\in J$ and $(hb)(p)\ne 0$. No point is a common zero of $J$, so the weak Hilbert Nullstellensatz implies $J=A(X)$. Thus $1\in J$ and $f\in A(X)$. This proof does not require a fraction field for a reducible coordinate ring.