Normal Varieties, Normalization, and Zariski's Main Theorem — Examples

By [F1] the element $f$ is prime, so $R=k[x,y,z]\mathrm{/}(f)$ is a domain and $X=V(f)$ is an irreducible closed subset, hence a classical affine variety with coordinate ring $R$ [F2]; $R$ is Noetherian by [F6]. Since $(f)$ is a nonzero principal prime, $\mathrm{ht}(f)=1$ [F6]; the height formula [F6] then gives $\dim R=3-1=2$ and, for every point $x\in X$, ${\dim }_{x}X=\dim X=2$ [F6]. The Jacobian matrix of $f$ is the row $(y,x,-2z)$, so by [F3] a point $x\in X$ is regular exactly when that row has rank $3-2=1$, and singular exactly when $y=x=z=0$; as $\mathrm{char}k\mathrm{\ne }2$ this is the origin and nothing else. Hence every point $x\mathrm{\ne }(0,0,0)$ of $X$ is regular, and its local ring $R_{{\mathfrak{m}}_x}$ is a regular local ring.

The sequence $x,y$ is a regular sequence in $R$: $x$ is a nonzerodivisor because $R$ is a domain, and $R\mathrm{/}xR\cong k[y,z]\mathrm{/}(z^2)$, in which $y$ is again a nonzerodivisor. Hence $\mathrm{depth}{R}_{{\mathfrak{m}}_0}\ge 2=\dim R_{{\mathfrak{m}}_0}$, where ${\mathfrak{m}}_0=(x,y,z)$ is the maximal ideal of the origin [F5, F6].

$R$ satisfies condition $(R_1)$. Let $\mathfrak{p}$ be a prime with $\mathrm{ht}\mathfrak{p}\le 1$. If $\mathfrak{p}=0$ then $R_{\mathfrak{p}}$ is a field, hence regular. If $\mathrm{ht}\mathfrak{p}=1$, the quotient $R\mathrm{/}\mathfrak{p}$ has dimension $2-1=1$ by [F6], so the closed subvariety $V(\mathfrak{p})\subseteq X$, whose coordinate ring is the domain $R\mathrm{/}\mathfrak{p}$ [F2], has dimension $1$; a one-dimensional variety is not a single point, so $V(\mathfrak{p})$ contains a point $x\mathrm{\ne }(0,0,0)$. The corresponding maximal ideal ${\mathfrak{m}}_x$ contains $\mathfrak{p}$, and $R_{{\mathfrak{m}}_x}$ is regular by step 1.1, so $R_{\mathfrak{p}}=(R_{{\mathfrak{m}}_x}{)}_{\mathfrak{p}{R}_{{\mathfrak{m}}_x}}$ is regular by [F5]. Thus $(R_1)$ holds.

The origin is singular. At the origin the Jacobian row $(y,x,-2z)$ is the zero row, of rank $0$, while $3-{\dim }_{{\mathfrak{m}}_0}{R}_{{\mathfrak{m}}_0}=3-2=1$; by the criterion [F3] the local ring $R_{{\mathfrak{m}}_0}$ is not regular, so the origin is a singular point of $X$.

$R$ satisfies condition $(S_2)$. Let $\mathfrak{p}$ be a prime. If $\mathfrak{p}={\mathfrak{m}}_0$, step 1.2 gives $\mathrm{depth}{R}_{\mathfrak{p}}\ge 2=\dim R_{\mathfrak{p}}$. If $\mathfrak{p}\mathrm{\ne }{\mathfrak{m}}_0$ and $\mathrm{ht}\mathfrak{p}\le 1$, then $R_{\mathfrak{p}}$ is regular by step 2.1 and therefore Cohen--Macaulay, so $\mathrm{depth}{R}_{\mathfrak{p}}=\dim R_{\mathfrak{p}}=\min \{2,\dim R_{\mathfrak{p}}\}$. If $\mathrm{ht}\mathfrak{p}=2$ then $\mathfrak{p}$ is maximal, hence $\mathfrak{p}={\mathfrak{m}}_x$ for a point $x\mathrm{\ne }(0,0,0)$, and $R_{\mathfrak{p}}$ is regular by step 1.1, so again $\mathrm{depth}{R}_{\mathfrak{p}}=2=\min \{2,\dim R_{\mathfrak{p}}\}$. Thus every prime satisfies $\mathrm{depth}{R}_{\mathfrak{p}}\ge \min \{2,\dim R_{\mathfrak{p}}\}$, that is, $(S_2)$ holds.

By steps 2.1 and 3.1 the ring $R$ satisfies $(R_1)$ and $(S_2)$, so $R$ is normal by Serre's criterion [F4]; consequently each prime localisation $R_{\mathfrak{p}}$, in particular each local ring $R_{{\mathfrak{m}}_x}$ at a point of $X$, is an integrally closed domain, and $X$ is a normal variety [F7]. By step 2.2 the origin is a singular point. Therefore $X=V(xy-z^2)$ is a normal classical variety that is singular at the origin: normal does not imply nonsingular in dimension two, and the characteristic hypothesis $\mathrm{char}k\mathrm{\ne }2$ was used only to identify the singular locus with the origin.

By [F1] the ring $A$ is an integrally closed domain, so by the localisation criterion [F3] every maximal localisation $A_{\mathfrak{m}}$ is integrally closed; in particular, for each point $x\in X$ the local ring ${\mathcal{O}}_{X,x}\cong A_{{\mathfrak{m}}_x}$ of [F2] is an integrally closed domain.

Every point $x\in X$ has an integrally closed local ring by step 1.1, so by [F4] affine space is normal. This uses no characteristic or perfectness hypothesis: the input [F1] holds over every field.

The map $j$ is an open immersion and is quasi-finite: it is the inclusion of the principal open $U$ [F1], and its fibres are singletons over the points of $k^{\times }$ and empty over $0$, so every closed-point fibre is finite [F2].

The map $j$ is not finite. If it were finite, then by [F3] its image would be closed in ${\mathbf{A}}^1$; but its image is ${\mathbf{A}}^1\setminus \{0\}$, which by [F4] is infinite and hence not closed. Equivalently, the coordinate-ring inclusion $k[t]\hookrightarrow k[t,t^{-1}]$ would make $k[t,t^{-1}]$ a finite $k[t]$-module, which it is not: a finite generating set of Laurent polynomials has a bounded negative exponent, and no finite $k[t]$-span contains all powers $t^{-n}$.

The inclusion $j\text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\setminus \{0\}\hookrightarrow {\mathbf{A}}^1$ is therefore an open immersion with finite fibres that is not finite, refuting the claim; the missing hypothesis is properness (equivalently, closedness of the map), which is exactly what [F3] supplies for finite morphisms and what fails for this open immersion.

The substitution kills $P$: $\sigma (y{)}^2-\sigma (x{)}^2(\sigma (x)+1)=t^2(t^2-1{)}^2-(t^2-1{)}^2{t}^2=0$, because $\sigma (x)+1=t^2$. By [F1] there is a unique $k$-algebra homomorphism $\phi \text{\hspace{0.17em}⁣}:A\to k[t]$ with $\phi (\bar{x})=t^2-1$ and $\phi (\bar{y})=t(t^2-1)$.

The map $\phi$ is injective. By [F2] every element of $A$ is uniquely $a(x)+b(x)y$ with $a,b\in k[x]$, and $\phi (a+by)=a(t^2-1)+t(t^2-1)b(t^2-1)$ separates into an even part $a(t^2-1)$ and an odd part $t(t^2-1)b(t^2-1)$, so $\phi (a+by)=0$ forces both parts to vanish. If $c\in k[x]$ is nonzero then $c(t^2-1)\mathrm{\ne }0$: the term of highest degree $2\mathrm{deg}c$ has coefficient equal to the leading coefficient of $c$ by [F2]. Hence $a=0$, and since $k[t]$ is a domain and $t^2-1\mathrm{\ne }0$ we get $b(t^2-1)=0$, hence $b=0$. So $\phi$ is injective and $A$ is a domain, isomorphic to $k[t^2-1,t(t^2-1)]$. If a polynomial $Q(x,y)$ vanishes on $X$, its substitution vanishes at every $t\in k$ under the displayed parametrization. The field $k$ is infinite, so this substituted polynomial is zero. The kernel just computed is $(P)$, hence $I(X)=(P)$, justifying the coordinate-ring identification in the Given.

In $\mathrm{Frac}(A)$ one has $\bar{y}\mathrm{/}\bar{x}=(t^3-t)\mathrm{/}(t^2-1)=t$, so $k(t)\subseteq \mathrm{Frac}(A)$, and $A\subseteq k[t]$ gives the reverse inclusion; hence $\mathrm{Frac}(A)=k(t)$. Moreover $t^2=\bar{x}+1\in A$, so $t$ is integral over $A$, and since $t^n=t^{n-2}(\bar{x}+1)$ for $n\ge 2$ we get $k[t]=A+At$, a finite $A$-module.

If $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$, then a monic equation for $z$ over $A$ has coefficients in $k[t]$, so $z$ is integral over $k[t]$ and hence lies in $k[t]$ because $k[t]$ is integrally closed [F3]. Conversely every element of $k[t]=A+At$ is integral over $A$ by step 3.1. Therefore the integral closure of $A$ in $\mathrm{Frac}(A)$ is exactly $k[t]$.

By [F4] the normalization of $X$ is the affine variety with coordinate ring $k[t]$, namely ${\mathbf{A}}^1$, with the structure morphism $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$ induced by $A\hookrightarrow k[t]$; by [F4] applied to the parametrization this is $\nu (t)=(t^2-1,t(t^2-1))$, a finite birational map, and it is surjective.

The fibre of $\nu$ over the node $(0,0)$ consists of the parameters with $t^2-1=0$, namely $t=1$ and $t=-1$; these are distinct because $\mathrm{char}k\mathrm{\ne }2$, and both map to the origin because $t(t^2-1)=0$. Hence the node has a two-point normalization fibre, so it is not unibranch [F5]; since normal points are unibranch [F5], the node is not normal.

Summing up, the normalization of the nodal curve is the finite birational surjection $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$, $t\mapsto (t^2-1,t(t^2-1))$, whose fibre over the node is the two-point set $\{1,-1\}$; the node is therefore not unibranch and not normal.

By [F1] the fibre of the normalization over the node has the two distinct elements $1$ and $-1$, so by [F2] the set-theoretic preimage ${\nu }^{-1}((0,0))=\{1,-1\}$ has two elements.

The map $\nu$ therefore sends two distinct points of ${\mathbf{A}}^1$ to the same point of $X$, so it is not injective; the claim is refuted. In dimension one a normalization need not be injective; in contrast, when the target is normal the normalization is an isomorphism and hence injective.

The substitution kills $P$: $\sigma (y{)}^2-\sigma (x{)}^3=t^6-t^6=0$, so by [F1] there is a unique $k$-algebra homomorphism $\phi \text{\hspace{0.17em}⁣}:A\to k[t]$ with $\phi (\bar{x})=t^2$ and $\phi (\bar{y})=t^3$.

The map $\phi$ is injective. By [F2] every element of $A$ is uniquely $a(x)+b(x)y$ with $a,b\in k[x]$, and $\phi (a+by)=a(t^2)+t^{3}b(t^2)$ separates into an even part $a(t^2)$ and an odd part $t^{3}b(t^2)$, so $\phi (a+by)=0$ forces both to vanish. If $c\in k[x]$ is nonzero then $c(t^2)\mathrm{\ne }0$, because the term of highest degree $2\mathrm{deg}c$ has coefficient equal to the leading coefficient of $c$; hence $a=0$, and then $t^{3}b(t^2)=0$ with $k[t]$ a domain gives $b(t^2)=0$, hence $b=0$. So $A$ is a domain isomorphic to $k[t^2,t^3]$. If a polynomial $Q(x,y)$ vanishes on $X$, its substitution vanishes at every $t\in k$ under the displayed parametrization. The field $k$ is infinite, so this substituted polynomial is zero. The kernel just computed is $(P)$, hence $I(X)=(P)$, justifying the coordinate-ring identification in the Given.

In $\mathrm{Frac}(A)$ one has $\bar{y}\mathrm{/}\bar{x}=t^3\mathrm{/}{t}^2=t$, so $k(t)\subseteq \mathrm{Frac}(A)$ and, since $A\subseteq k[t]$, equality $\mathrm{Frac}(A)=k(t)$ holds. Moreover $t^2=\bar{x}\in A$, so $t$ is integral over $A$, and every even power of $t$ belongs to $A$ as a power of $t^2$, and every odd power belongs to $At$; hence $k[t]=A+At$ is a finite $A$-module.

If $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$, then a monic equation for $z$ over $A$ has coefficients in $k[t]$, so $z$ is integral over $k[t]$ and therefore lies in $k[t]$ because $k[t]$ is integrally closed [F3]; conversely every element of $k[t]=A+At$ is integral over $A$. Hence the integral closure of $A$ in its fraction field is exactly $k[t]$. By [F4] the normalization of the cusp is $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$ induced by $A\hookrightarrow k[t]$, namely $\nu (t)=(t^2,t^3)$, a finite birational map.

The map $\nu$ is bijective. If $u=0$ for a point $(u,v)$ of the cusp, then $v^2=u^3=0$, so $v=0$ and the unique parameter is $t=0$; if $u\mathrm{\ne }0$, then $t=v\mathrm{/}u$ satisfies $t^2=v^2\mathrm{/}{u}^2=u^3\mathrm{/}{u}^2=u$ and $t^3=v\cdot v^2\mathrm{/}{u}^3=v$, so $(u,v)$ is the image of the unique parameter $v\mathrm{/}u$. Hence every point of $X$ has exactly one preimage.

The pullback $\phi \text{\hspace{0.17em}⁣}:k[t^2,t^3]\hookrightarrow k[t]$ is not surjective: its image consists of sums of monomials $t^n$ with $n\in ⟨2,3⟩=\{0,2,3,4,\dots \}$, and $t$ has exponent $1$, so $t$ is not in the image. By the anti-equivalence [F4] the map $\nu$ is not an isomorphism, although it is bijective. Its fibre over the cusp is the singleton $\{0\}$, because $t^2=0$ forces $t=0$ in the field $k$; hence the cusp is unibranch [F6].

The conductor is the maximal ideal $\mathfrak{m}=(t^2,t^3)$ of $A$: indeed $t^{2}k[t]\subseteq A$ and $t^{3}k[t]\subseteq A$ because every exponent $\ge 2$ lies in $⟨2,3⟩$, so $\mathfrak{m}\subseteq \mathfrak{c}$; conversely an $a\in A$ outside $\mathfrak{m}$ has nonzero constant coefficient $c$. All its other monomials have exponent at least two, so $at$ has a nonzero coefficient $c$ at exponent one and cannot belong to $A$, so $\mathfrak{c}\subseteq \mathfrak{m}$. By [F5] $V(\mathfrak{c})$ is the non-normal locus, and $V(t^2,t^3)$ is the single point $(0,0)$; hence the cusp is the unique non-normal point of $X$.

Summing up: the normalization of the cusp is $\nu \text{\hspace{0.17em}⁣}:{\mathbf{A}}^1\to X$, $t\mapsto (t^2,t^3)$, a finite birational bijection which is not an isomorphism; the cusp is its unique non-normal point, with singleton fibre $\{0\}$, so it is unibranch but not normal; and the conductor of the extension is the maximal ideal $(t^2,t^3)$ of $k[t^2,t^3]$.

By [F1] the normalization $\nu$ is finite and birational, and by [F2] it is bijective but not an isomorphism: its pullback $k[t^2,t^3]\hookrightarrow k[t]$ misses $t$, so this pullback is not an isomorphism.

The target of $\nu$ is the cusp, which is not normal at its singular point by [F1]; the isomorphism criterion [F3] therefore does not apply, exactly because its target-normality hypothesis fails.

Hence finite plus birational plus bijective does not imply isomorphism: the cusp normalization is a counterexample, and bijectivity cannot replace the normality hypothesis in the finite-birational-to-normal isomorphism lemma. The failure is isolated precisely at the non-normal point of the target.

$t^{2}k[t]\subseteq A$: for $n\ge 0$ the monomial $t^{n+2}$ has exponent at least $2$, and every exponent $\ge 2$ lies in the numerical semigroup generated by $2$ and $3$, so $t^{n+2}\in A$; the same argument gives $t^{3}k[t]\subseteq A$. Hence $t^2,t^3\in \mathfrak{c}$, and since $\mathfrak{c}$ is an ideal, $(t^2,t^3)\subseteq \mathfrak{c}$.

The semigroup $⟨2,3⟩=\{0,2,3,4,\dots \}$ contains every integer $n\ge 2$ and misses $1$, so the least exponent $m$ with $t^n\in A$ for all $n\ge m$ is $m=2$. The element $t^3$ is not a multiple of $t^2$ inside $A$: an equation $t^3=t^{2}a$ with $a\in A\subseteq k[t]$ would force $a=t\mathrm{\notin }A$ since $k[t]$ is a domain. Hence $t^2$ alone does not generate the conductor, and the two monomials $t^2,t^3$ are both needed.

Conversely let $a\in A$ satisfy $ak[t]\subseteq A$. If $a\mathrm{\notin }(t^2,t^3)$, it has a nonzero constant coefficient $c$ and all remaining terms have exponent at least two. Thus $at$ has nonzero coefficient $c$ at exponent one, so $at\mathrm{\notin }A$, a contradiction; $a=0$ lies in $(t^2,t^3)$. Hence $a\in (t^2,t^3)$ and $\mathfrak{c}\subseteq (t^2,t^3)$; with step 1.1, $\mathfrak{c}=(t^2,t^3)$.

Therefore the conductor of the cusp extension $k[t^2,t^3]\subseteq k[t]$ is the maximal ideal $(t^2,t^3)$ of the semigroup ring, whose generators are the first two positive elements of the semigroup: the least exponent from which all monomials lie in $k[t^2,t^3]$ is $2$, and neither generator alone suffices.