Normal varieties and the normalization of an affine variety

Let \(\xi\in\operatorname{Frac}A\) be integral over \(S^{-1}A\). Choose a monic relation \[ \xi^n+\frac{a_{n-1}}{s_{n-1}}\xi^{n-1}+\cdots+\frac{a_0}{s_0}=0, \] with \(a_i\in A\) and \(s_i\in S\), and put \(s=s_0\cdots s_{n-1}\). Multiplying by \(s^n\) rewrites the equation as a monic equation for \(s\xi\) with coefficients in \(A\). Thus \(s\xi\) is integral over \(A\), hence lies in \(A\). Therefore \(\xi=(s\xi)/s\in S^{-1}A\).

If \(A\) is integrally closed, then for every point \(P\in Y\), \[ \mco_{P,Y}\cong A_{\mathfrak m_P} \] is integrally closed by step 1, so \(Y\) is normal.

Conversely, suppose every local ring \(A_{\mathfrak m}\) at a maximal ideal is integrally closed. Let \(\xi\in\operatorname{Frac}A\) be integral over \(A\). Then \(\xi\) is integral over every \(A_{\mathfrak m}\) and hence belongs to every such localization. We claim \[ A=\bigcap_{\mathfrak m\text{ maximal}}A_{\mathfrak m} \subseteq\operatorname{Frac}A. \] Indeed, if \(\xi\notin A\), the ideal \[ I=\{a\in A:a\xi\in A\} \] is proper and lies in some maximal ideal \(\mathfrak m\). But \(\xi\in A_{\mathfrak m}\) would give \(s\notin\mathfrak m\) with \(s\xi\in A\), so \(s\in I\subseteq\mathfrak m\), a contradiction. Hence \(\xi\in A\), proving that \(A\) is integrally closed.

By P-AGH31CONICS, every projective conic is isomorphic to \(\PP^1\). The standard affine charts of \(\PP^1\) are copies of \(\AA^1\), whose coordinate ring \(k[t]\) is a UFD and hence integrally closed. By step 2 each chart is normal, and the local ring of \(\PP^1\) at a point is the same as the local ring in any open chart containing that point. Thus \(\PP^1\) is normal. Normality is preserved by isomorphism, so every conic is normal.

Define \[ R\longrightarrow k[s,t], \qquad x\longmapsto s^2, \quad y\longmapsto t^2, \quad z\longmapsto st. \] Every class in \(R\) has a representative of the form \[ f(x,y)+z g(x,y), \] because powers \(z^r\) reduce using \(z^2=xy\). Its image is \[ f(s^2,t^2)+st\,g(s^2,t^2). \] The first summand contains only monomials whose two exponents are even, and the second only monomials whose two exponents are odd. Their supports are disjoint, and \(s^2,t^2\) are algebraically independent, so the displayed map is injective. This also proves that the displayed representatives are unique. Its image is \[ k[s^2,st,t^2], \] which is exactly the subring of \(k[s,t]\) spanned by monomials of even total degree.

Write \(K=\operatorname{Frac}R\subseteq k(s,t)\). We first show \[ k[s,t]\cap K=R. \] Let \(h\in k[s,t]\cap K\) and write \(h=a/b\) with nonzero \(a,b\in R\). Decompose \(h=h_{\mathrm{ev}}+h_{\mathrm{odd}}\) according to parity of total monomial degree. Since every monomial of \(b\) has even total degree, multiplication by \(b\) preserves this parity decomposition. The equation \(hb=a\), whose right side has only even total degree, forces \[ h_{\mathrm{odd}}b=0. \] Because \(k[s,t]\) is a domain, \(h_{\mathrm{odd}}=0\). Thus \(h\) has only even-total-degree monomials and hence belongs to \(R\).

Now let \(\alpha\in K\) be integral over \(R\). The same monic equation shows that \(\alpha\) is integral over \(k[s,t]\). Since the UFD \(k[s,t]\) is integrally closed, \[ \alpha\in k[s,t]. \] The intersection equality then gives \(\alpha\in R\). Therefore \(R\) is integrally closed.

The only point of \(Q_2\) at which \(x=y=z=0\) is the vertex \[ V=[0:0:0:1]. \] Every other point lies in \(D_+(x)\) or \(D_+(y)\): if \(z\ne0\), the equation \(xy=z^2\) forces both \(x\) and \(y\) to be nonzero. On \(D_+(x)\), setting \(x=1\) gives \(y=z^2\), so the chart is \(\AA^2\) with coordinates \(z,w\); similarly for \(D_+(y)\).

Its coordinate ring is \[ A(C)=k[x,y]/(y^2-x^3)\cong k[t^2,t^3], \] under \(x\mapsto t^2\) and \(y\mapsto t^3\). The element \(t\) belongs to the fraction field, since \[ t=\frac{t^3}{t^2}, \] and it is integral over \(A(C)\) because it satisfies \[ T^2-t^2=0. \] But \(t\notin k[t^2,t^3]\): every positive exponent occurring in that subring belongs to the semigroup generated by \(2\) and \(3\), which does not contain \(1\). Thus \(A(C)\) is not integrally closed, and step 2 implies that \(C\) is not normal.

Dominance makes the pullback \[ \phi^*:A\longrightarrow\mco(Z) \] injective. Indeed, if \(0\ne a\in A\) had \(a\circ\phi=0\), the dense set \(\phi(Z)\) would lie in the proper closed zero set \(Z_Y(a)\), impossible. Thus \(\phi^*\) extends to an embedding \[ K=\operatorname{Frac}A\hookrightarrow K(Z). \]

Take \(b\in B\) and regard its image as an element of \(K(Z)\). Because \(b\) is integral over \(A\), at every point \(z\in Z\) its image satisfies a monic equation with coefficients in the local ring \(\mco_{z,Z}\). Normality of \(Z\) makes that local ring integrally closed in \(K(Z)\), so the image of \(b\) belongs to \(\mco_{z,Z}\) for every \(z\). Hence it is a global regular function on \(Z\): the local representatives supplied by these germs agree on overlaps because they represent the same rational function. We obtain a \(k\)-algebra homomorphism \[ B\longrightarrow\mco(Z) \] extending \(\phi^*:A\to\mco(Z)\). Choosing affine coordinates for \(\widetilde Y\) from generators of \(B\), these global regular functions define a morphism \[ \theta:Z\to\widetilde Y \] [@Har10a, Lemma I.3.6]. The equality of pullbacks on \(A\) gives \[ \phi=\pi\circ\theta. \]

For uniqueness, let \(\theta'\) be another such factorization. Its pullback \(B\to\mco(Z)\) agrees with \(\phi^*\) on \(A\). For any \(b\in B\subseteq K=\operatorname{Frac}A\), write \(b=a/c\) with \(a,c\in A\) and \(c\ne0\). Then in the domain \(K(Z)\) every extension of \(\phi^*\) must satisfy \[ \theta'^*(b)=\frac{\phi^*(a)}{\phi^*(c)}=\theta^*(b), \] since injectivity gives \(\phi^*(c)\ne0\). Thus the two pullbacks on \(B\) coincide, so the two morphisms to the affine variety \(\widetilde Y\) coincide. This is the required universal property.

Steps 3, <1>4–<1>6, 7, 2, and <1>8–<1>9 prove parts (a)–(e), respectively.