Analytic Hypersurfaces and Local Parametrisation: Examples and Counterexamples

The germ $z_n$ is irreducible: if $z_n=ab$ with $a,b$ nonunits, then $a,b\in {\mathfrak{m}}_0$ and hence $z_n=ab\in {\mathfrak{m}}_0^2$, contradicting that $z_n\mathrm{\notin }{\mathfrak{m}}_0^2$ because its linear part is nonzero. By [F2] $z_n$ is therefore reduced, and $X=Z(z_n)$ is a hypersurface germ whose reduced defining germ is $z_n$ by [F1], with $Z(z_n)$ exactly the hyperplane $\{z_n=0\}$.

Every point $q\in X$ is regular. Indeed $X=\{z_n=0\}$ is the graph of the zero function over the $z^{\mathrm{\prime }}$-coordinates near $q$, so by the graph criterion of [F4] $q$ is regular; equivalently, $d{z}_n\mathrm{\ne }0$ everywhere and the reduced local equation $z_n$ has nonvanishing differential at $q$.

For the prepared equation $W=z_n$ of degree $1$ in the last variable, [F3] and [F5] give $D_W={\mathrm{Disc}}_{z_n}(z_n)=1\mathrm{\ne }0$. On the product representative $V\times D$, [F6] gives the local branch set $B_{{\pi }_W}=\{D_W=0\}=\mathrm{\varnothing }$ and the one-sheeted covering ${\pi }_W:X_W\to V$, where $X_W=Z(W)\cap (V\times D)=\{(z^{\mathrm{\prime }},0):z^{\mathrm{\prime }}\in V\}$. Separately, the global coordinate projection $\pi :X\to {\mathbb{C}}^{n-1}$ is the identity under the identification $X=\{z_n=0\}\cong {\mathbb{C}}^{n-1}$, so it is a one-sheeted covering over its whole base; the local map above is its restriction to $X_W$. When $n=1$ both bases are the single point $z^{\mathrm{\prime }}=0$.

For the same prepared equation $W=z_n$ as in step 2.2, the expansion [F9] in the last variable shows that the substitution $z_n=0$ gives a ring isomorphism ${\mathcal{O}}_{{\mathbb{C}}^n,0}\mathrm{/}(z_n)\cong {\mathcal{O}}_{{\mathbb{C}}^{n-1},0}$; combined with the definition of local dimension in [F8] this gives ${\dim }_{0}X=\dim {\mathcal{O}}_{{\mathbb{C}}^{n-1},0}=n-1$, where the numerical value is the dimension result [F7], the only use of the Axiom of Choice.

Assembling steps 2.1, 2.2 and 3.1: the hyperplane germ $X=\{z_n=0\}$ has the reduced equation $z_n$ with nonzero differential everywhere, its projection to $z^{\mathrm{\prime }}$ is one-sheeted with discriminant $1$ and branch set $\mathrm{\varnothing }$, and ${\dim }_{0}X=n-1$. At $n=1$ the curve is the point germ $\{0\}\subseteq \mathbb{C}$, the base is a point, and the dimension is $0=1-1$, so the degenerate case is covered.

Apply [F1] to $F(x,w)=w^2-(1+x)$ at the point $(x_0,w_0)=(0,1)$: $F(0,1)=0$ and ${\mathrm{\partial }}_{w}F(0,1)=2\mathrm{\ne }0$, so there is a holomorphic function $s$ on a neighbourhood of $0$ with $s(0)=1$ and $s(x{)}^2=1+x$ identically. In particular $s$ is a unit of ${\mathcal{O}}_{\mathbb{C},0}$ and $s(x)\mathrm{\ne }0$ for $\mathrm{\mid }x\mathrm{\mid }$ small.

Put $g_{\pm }:=y\mp xs(x)$, the two germs determined by the unit $s$ of step 1.1; each is a monic polynomial of degree $1$ in $y$ with coefficient $\mp xs(x)$ in ${\mathcal{O}}_{\mathbb{C},0}$ vanishing at $x=0$. Each $g_{\pm }$ is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$. Suppose $g_{+}=gh$ with $g,h$ nonunits. Since $g_{+}$ is regular in $y$ of order $1$, [F3] makes $g$ and $h$ regular with Weierstrass polynomials $G,H$ of positive degrees satisfying $W_{g_{+}}=GH$; but $W_{g_{+}}=g_{+}$ has degree $1$, while $\mathrm{deg}G+\mathrm{deg}H\ge 2$, a contradiction. The same argument applies to $g_{-}$.

With $s$ as in step 1.1, $f=y^2-x^{2}s(x{)}^2=(y-xs(x))(y+xs(x))$. The two factors $g_{\pm }:=y\mp xs(x)$ are Weierstrass polynomials of degree $1$ in $y$: each is monic of degree $1$ with its coefficient $\mp xs(x)$ lying in ${\mathcal{O}}_{\mathbb{C},0}$ and vanishing at $x=0$. The germ $f$ is itself a Weierstrass polynomial of degree $2$: it is monic of degree $2$ in $y$, its coefficients $-x^2(1+x)$ vanish at the origin, and $f(0,y)=y^2$, so it is regular in $y$ of order $2$ and is its own Weierstrass preparation by [F2].

The zero germs of $g_{+}$ and $g_{-}$ are distinct: $Z(g_{+})$ is the graph of ${\phi }_{+}(x)=xs(x)$ and $Z(g_{-})$ is the graph of ${\phi }_{-}(x)=-xs(x)$ over the $x$-coordinate. Since $s(x)\mathrm{\ne }0$ for $\mathrm{\mid }x\mathrm{\mid }$ small by step 1.1, the two graphs intersect exactly where $xs(x)=0$, that is, only at $x=0$, and for every small $x\mathrm{\ne }0$ the values $xs(x)$ and $-xs(x)$ differ; hence the two set germs at the origin are distinct, and neither is contained in the other.

The germ $f$ is reduced, and $X=Z(f)$ has exactly the two irreducible components $Z(g_{+})$ and $Z(g_{-})$. Indeed $f=g_{+}{g}_{-}$ exhibits $f$ as a product of two nonassociate irreducibles, each occurring once; by the uniqueness of factorisation in the UFD [F6], no irreducible germ divides $f$ twice, so $f$ is reduced, and the factorisation of a reduced germ into pairwise nonassociate irreducibles has all exponents one and is unique up to order and associates. Applying the decomposition statement [F4] to $f=g_{+}{g}_{-}$ gives $X=Z(g_{+})\cup Z(g_{-})$ with $Z(g_{+})$ and $Z(g_{-})$ the two irreducible components of $X$.

Each branch is a holomorphic hypersurface graph and carries an injective convergent parametrisation. For $\mathrm{\mid }t\mathrm{\mid }$ small, ${\gamma }_{+}(t)=(t,ts(t))$ is holomorphic with ${\gamma }_{+}(\mathrm{\Delta })=Z(g_{+})\cap \{\mathrm{\mid }x\mathrm{\mid }<\rho \}$ for a suitable radius $\rho >0$, since $y-xs(x)=0$ is exactly the graph $y=xs(x)$; similarly ${\gamma }_{-}(t)=(t,-ts(t))$ has image $Z(g_{-})\cap \{\mathrm{\mid }x\mathrm{\mid }<\rho \}$. Both maps are injective because their first coordinate is $t$, and both are restrictions of the holomorphic function $s$ of step 1.1, hence convergent. Every point of each branch is therefore regular as a point of that branch by [F5]. At a point other than the origin, step 2.3 separates the two graphs, so $X$ locally equals the branch through that point and is regular there. This does not assert regularity of their union at the origin.

The origin is the only singular point of $X$ near $0$. By step 3.1 the reduced defining germ of $X$ is $f$, and

Steps 2.1 to 4.1 establish the assertions: $y^2-x^2(1+x)$ is a reduced equation of $X$ whose irreducible components are the two smooth branches $y=\pm x\sqrt{1+x}$ described by the unit square root $s$ of $1+x$; the origin is their only intersection and the only singular point of the germ, with distinct tangent directions, so $X$ is an ordinary node; and each branch carries the convergent parametrisation $t\mapsto (t,\pm ts(t))$ with first coordinate $t$.

$f$ is irreducible in ${\mathcal{O}}_{{\mathbb{C}}^2,0}$. By [F1] and [F2] it suffices to show that the monic quadratic $f=y^2-x^3\in {\mathcal{O}}_{\mathbb{C},0}[y]$ is irreducible. Suppose $f=GH$ with $G,H$ nonunits. Their leading coefficients multiply to the leading coefficient $1$, so both are units by [F3]. Neither factor can have degree $0$, since a degree-zero factor is its leading coefficient and would be a unit. Since their degrees sum to $2$, both have degree $1$; rescaling by their unit leading coefficients, we may write $G=y-a$, $H=y-b$ with $a,b\in {\mathcal{O}}_{\mathbb{C},0}$. Comparing coefficients gives $a+b=0$ and $ab=-x^3$, hence $a^2=x^3$, contradicting [F4]. Thus $f$ is irreducible in the polynomial ring and, by [F2], in the germ ring; by [F5] $f$ is reduced and $I_0(X)=(f)$.

$\gamma (t)=(t^2,t^3)$ is injective. Indeed $\gamma (t_1)=\gamma (t_2)$ means $t_1^2=t_2^2$ and $t_1^3=t_2^3$. The first equation gives $t_2=\pm t_1$; if $t_2=-t_1$, then $t_1^3=t_2^3=-t_1^3$, so $2t_1^3=0$ and $t_1=0=t_2$; otherwise $t_2=t_1$.

The image of $\gamma$ on ${\mathrm{\Delta }}_{\delta }=\{\mathrm{\mid }t\mathrm{\mid }<\delta \}$ is exactly the full representative $Z(f)\cap \{\mathrm{\mid }x\mathrm{\mid }<{\delta }^2\}$ of the germ $X$. First, $f(\gamma (t))=t^6-t^6=0$, so the image lies in $Z(f)$, and $\mathrm{\mid }{t}^2\mathrm{\mid }=\mathrm{\mid }t{\mathrm{\mid }}^2<{\delta }^2$. Conversely, let $(x,y)\in Z(f)$ with $\mathrm{\mid }x\mathrm{\mid }<{\delta }^2$; choose $t$ with $t^2=x$, so $\mathrm{\mid }t\mathrm{\mid }<\delta$. If $x=0$, then $y^2=0$ and $y=0=\gamma (0)$. If $x\mathrm{\ne }0$, then $y^2=x^3=x\cdot x^2$ gives $(y\mathrm{/}x{)}^2=x=t^2$, so $y\mathrm{/}x=\pm t$ and $y=\pm t^3$; replacing $t$ by $-t$ if necessary, we get $(x,y)=(t^2,t^3)=\gamma (t)$ with $t^2=x$. Hence every point of the representative is attained, and $h(t)=t^3={\sum }_{k>2}{a}_k{t}^k$ with $a_3=1$ has order $3>2$.

Every holomorphic parametrisation $s\mapsto (s^k,j(s))$ of the germ $X$ in these coordinates has $k\ge 2$. Such a parametrisation has image containing a full representative $X\cap U$ of the germ for some neighbourhood $U$ of $0$. Choose $\epsilon >0$ with $({\epsilon }^2,\pm {\epsilon }^3)\in U$; both points lie in $X$, because $(\pm {\epsilon }^3{)}^2={\epsilon }^6=({\epsilon }^2{)}^3$. So there are $s_1\mathrm{\ne }{s}_2$ with $s_i^k={\epsilon }^2$ and $j(s_1)={\epsilon }^3$, $j(s_2)=-{\epsilon }^3$; the two parameters are distinct since their images are. Thus the polynomial $T^k-{\epsilon }^2$ of degree $k$ has at least two distinct roots, so $k\ge 2$ by [F8].

$X$ is an irreducible hypersurface germ and its only singular point near $0$ is the origin. By step 1.1 the reduced defining germ $f$ is irreducible, so by [F5] the germ $X=Z(f)$ is irreducible: its decomposition has the single component $Z(f)=X$. The differential $df=(-3x^2,2y)$ vanishes at the origin and at no other point of $X$, because $x=0$ forces $y^2=x^3=0$ and then $y=0$. At a point $q\in X$ with $q\mathrm{\ne }0$ the translate of $f$ is a germ with nonzero differential, hence is not a product of two nonunits, that is, it is an irreducible and therefore reduced germ vanishing on $X$ near $q$; so it is a local reduced equation of $X$ at $q$ and [F6] makes $q$ a regular point.

Consequently $\gamma (t)=(t^2,t^3)$ is an injective convergent Puiseux parametrisation of $X$ in the standard coordinates: it is holomorphic on ${\mathrm{\Delta }}_{\delta }$, it is injective by step 1.2, the holomorphic function $h(t)=t^3$ satisfies $h(t)={\sum }_{k>2}{a}_k{t}^k$, and by step 1.3 its image germ is exactly $X=Z(f)$, matching the conclusion of [F7]; over a base value $x_0\mathrm{\ne }0$ the two branch values are $\pm x_0^{3\mathrm{/}2}$, the cusp's Puiseux exponent $3\mathrm{/}2$.

Steps 1.4, 2.1 and 3.1 prove all the assertions: $X$ is irreducible and singular only at the origin, $\gamma (t)=(t^2,t^3)$ is an injective convergent parametrisation of its germ, and since every parametrisation in the same coordinates has exponent $k\ge 2$ by step 1.4 while $\gamma$ has exponent $2$, the exponent is primitive (minimal), so this is the parametrisation the Puiseux theorem produces for the cusp in these coordinates.

The germs $x$ and $y$ are irreducible. Neither lies in ${\mathfrak{m}}_0^2$, because both have nonzero linear part; if $x=ab$ with $a,b$ nonunits, then $a,b\in {\mathfrak{m}}_0$ by [F3] and hence $x=ab\in {\mathfrak{m}}_0^2$, a contradiction. So $x$ is not a product of two nonunits, and the same argument applies to $y$. The two germs are not associates: $x=uy$ with $u$ a unit would give $x(t,0)=u(t,0)\cdot 0=0$ for every small $t$, contradicting $x(t,0)=t\mathrm{\ne }0$ for $t\mathrm{\ne }0$. Hence $f=xy$ is a product of two pairwise nonassociate irreducibles, each occurring once, so $f$ is reduced and its reduced defining germ is $f$ itself, with $I_0(X)=(f)$ by [F1].

Let $\mathrm{\Delta }\subseteq \mathbb{C}$ be a connected open set containing the origin and let $\gamma =(u,v):\mathrm{\Delta }\to {\mathbb{C}}^2$ be holomorphic with $\gamma (\mathrm{\Delta })\subseteq X$, that is, $u(t)v(t)=0$ for every $t\in \mathrm{\Delta }$. By [F6] applied to the connected domain $\mathrm{\Delta }$, one of the two coordinate functions vanishes identically, so the image of $\gamma$ is contained in a single axis: either $\gamma (\mathrm{\Delta })\subseteq Z(x)$ or $\gamma (\mathrm{\Delta })\subseteq Z(y)$.

By [F4] applied to the factorisation $f=x\cdot y$ of step 1.1, $X=Z(x)\cup Z(y)$, and the two branches $Z(x)=\{x=0\}$ and $Z(y)=\{y=0\}$ are exactly the irreducible components of $X$; they are distinct as set germs and neither contains the other.

Every point of $Z(x)$ is regular: $Z(x)$ is the graph $\{(x,y):x=0\}$ of the zero function over the $y$-coordinate near each of its points, hence a holomorphic hypersurface graph near every such point, so [F5] gives regularity. The same argument exhibits $Z(y)=\{(x,y):y=0\}$ as the graph of the zero function over the $x$-coordinate, so every point of $Z(y)$ is regular as well.

The origin is a singular point: by step 1.1 the reduced defining germ of $X$ is $xy$, and $d(xy)=ydx+xdy$ vanishes at $0$. So $0$ is not regular by [F5]. Since $Z(x)\cap Z(y)=\{0\}$, every point of $X$ other than the origin lies on exactly one of the two branches and is regular by step 3.1; hence the origin is the only singular point of $X$ in a neighbourhood of $0$, and it is exactly the crossing of the two branches.

Suppose first that $\gamma (\mathrm{\Delta })\subseteq Z(x)$. The set germ of the image of $\gamma$ at the origin is then contained in $Z(x)$, which is a proper subgerm of $X$: for every small $\epsilon \mathrm{\ne }0$ the point $(\epsilon ,0)$ belongs to $X$ but not to $Z(x)$. Hence the image of $\gamma$ cannot contain a full representative of $X$, so its image germ is not $X$; the case $\gamma (\mathrm{\Delta })\subseteq Z(y)$ is the same with the roles of $x$ and $y$ exchanged. Therefore no holomorphic map of a connected disc has image germ $X$, injective or not, and in particular no single injective branch parametrisation covers both components. The individual branches are parametrised by the injective holomorphic maps $t\mapsto (t,0)$ and $t\mapsto (0,t)$, whose images are full representatives of $Z(y)$ and $Z(x)$ respectively.

The germ $x$ is irreducible by [F1] and hence reduced by [F2]; its factorisation has the single irreducible factor $x$, so the square-free reduction of $x^2$ is $x$ and the square-free reduction of $x$ is itself. By [F2] we have $Z(x)=Z(x^2)$ near $0$, so the two equations define the same hypersurface germ $X$, with reduced defining germ $x$ by [F3]; the equation $x^2$ is not reduced, because the irreducible germ $x$ divides it twice.

The germ $X=\{x=0\}$ is the graph $X=\{(x,y):x=0\}$ of the zero function over the $y$-coordinate, hence is a holomorphic hypersurface graph near each of its points; by [F4] every point of $X$ is regular, and $X$ is smooth.

On the one hand $dx$ is the constant nonzero covector $dx$, so the differential of the reduced equation $x$ never vanishes and the criterion [F4] is satisfied at every point of $X$. On the other hand $d(x^2)=2xdx$ vanishes at every point of $X$, because $x=0$ there. Thus the raw differential of the nonreduced equation $x^2$ vanishes on the very hypersurface on which the reduced equation $x$ has nonzero differential, and the regularity criterion must specify a reduced equation.