= 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 $(hb)(p)\neq0$. 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>.
Back to article page
