Extending a regular function across a normal point in dimension at least two

Local rings of varieties are Noetherian domains. The hypothesis that \(P\) is normal says exactly that \(R\) is integrally closed in its fraction field, so \(R\) is normal. By P-AGH312DIMLOCAL, \[ \dim R=\dim Y\ge2. \] Its fraction field is \(K(Y)=K\).

Choose an affine open neighborhood \(U\subseteq Y\) of \(P\), write \[ A=A(U), \qquad \mathfrak m=I_U(P), \] so that \(R=A_{\mathfrak m}\). The prime \(\mathfrak q\) has the form \[ \mathfrak q=\mathfrak p A_{\mathfrak m} \] for a prime \(\mathfrak p\subsetneq\mathfrak m\). The strict containment follows because \(\operatorname{ht}\mathfrak q=1\) whereas \[ \operatorname{ht}\mathfrak m=\dim R\ge2. \]

The irreducible closed set \(V(\mathfrak p)\subseteq U\) therefore contains a closed point \(Q\ne P\). The function \(f\) is regular at \(Q\). Hence, after shrinking inside the affine variety \(U\), it has a representation \[ f=\frac{a}{b} \] with \(a,b\in A\) and \(b(Q)\ne0\). Because \(Q\in V(\mathfrak p)\), the inequality \(b(Q)\ne0\) implies \[ b\notin\mathfrak p. \] Thus \(f\in A_{\mathfrak p}\). Since \(\mathfrak p\subseteq\mathfrak m\), \[ A_{\mathfrak p}\cong (A_{\mathfrak m})_{\mathfrak pA_{\mathfrak m}} =R_{\mathfrak q}, \] which proves the claim.

For a Noetherian normal domain, the height-one intersection theorem states that, inside its fraction field, \[ R=\bigcap_{\operatorname{ht}\mathfrak q=1}R_{\mathfrak q}. \] This is Matsumura, Commutative Ring Theory, Theorem 11.5. Step 2 places \(f\) in every ring in this intersection, so \(f\in R\).

By step 3, the germ \(f\in\mco_{P,Y}\) is represented by a regular function \(g\) on some open neighborhood \(V\) of \(P\). Both \(g\) and the original function on \(Y\setminus\{P\}\) represent the same element of \(K(Y)\). Hence they agree on the overlap \[ V\setminus\{P\}. \] They therefore glue to a regular function on \[ V\cup(Y\setminus\{P\})=Y \] extending the original \(f\).

Take \[ Y=\AA^1, \qquad P=0, \] and on \(Y\setminus\{0\}=D(x)\) take \[ f=\frac1x. \] This is regular on \(D(x)\). If it extended regularly across the origin, its germ there would lie in \[ \mco_{0,\AA^1}=k[x]_{(x)}. \] But \(1/x\) does not belong to this localization: every denominator of an element of \(k[x]_{(x)}\) is nonzero at \(x=0\), whereas a representation of \(1/x\) in reduced form has denominator divisible by \(x\). Indeed, if \[ \frac1x=\frac{a}{s} \] with \(a,s\in k[x]\) and \(s\notin(x)\), then \(s=ax\), so \(s(0)=0\), contradicting \(s\notin(x)\). Thus no extension exists, proving (b).

Steps <1>1–<1>4 prove (a), and step 5 proves (b).