Puiseux discs normalise a reduced plane curve germ

Fix for each branch the coordinates and injective map ${\gamma }_i(t)=(t^{m_i},h_i(t))$ supplied by the Puiseux theorem. In these coordinates $q_i(0,y)$ is not identically zero: otherwise its zero set would contain a vertical disc, whereas the image representative of ${\gamma }_i$ has only $(0,0)$ over $x=0$. Preparation therefore gives $q_i=v_i{W}_i$ with $W_i$ of some degree $d_i\ge 1$ in $y$. It is reduced and irreducible because it is associate to $q_i$. Apply [F15] to $W_i$. For all sufficiently small $x\mathrm{\ne }0$, all its $d_i$ roots lie in the neighbourhood where the branch agrees with the image representative of ${\gamma }_i$. Each such root is attained at a parameter satisfying $t^{m_i}=x$, so there are at most $m_i$ roots. Conversely all $m_i$ solutions of $t^{m_i}=x$ lie in the parameter disc when $x$ is small and their images lie in that same neighbourhood; they give $m_i$ distinct roots by injectivity. Thus $d_i=m_i$. Substitution sends $x$ to $t^{m_i}$ and $y$ to $h_i(t)$, and is defined on convergent germs by composition.

Each parametrisation is injective by the Puiseux theorem, and ${\gamma }_i^{\mathrm{\prime }}(t)=(m_i{t}^{m_i-1},h_i^{\mathrm{\prime }}(t))$ is nonzero for $t\mathrm{\ne }0$ because its first component is. At such a point the first coordinate $t\mapsto t^{m_i}$ has nonzero derivative, so its local inverse is holomorphic and the inverse of ${\gamma }_i$ on its image is the composition of that local inverse with the first-coordinate projection; by [F14] the injective parametrisation is therefore a biholomorphism from ${\mathrm{\Delta }}_{{\delta }_i}\setminus \{0\}$ onto its image with $p$ removed. If $m_i=1$ the first component is the identity, so $h_i$ is defined on the whole disc and ${\gamma }_i$ is a biholomorphism of ${\mathrm{\Delta }}_{{\delta }_i}$ onto its image.

For each $i$ the substitution ${\sigma }_i$ annihilates $q_i$, so it factors through ${\bar{\sigma }}_i:A_i\to \mathbb{C}\{t\}$ by [F3]. This factor is injective: if ${\bar{\sigma }}_i(h)=0$ for a class $h$, then a representative of $h$ vanishes at every point of the full representative ${\gamma }_i({\mathrm{\Delta }}_{{\delta }_i})$ of $Z(q_i)$; hence $h\in I_p(Z(q_i))=(q_i)$ by [F4], so $h=0$ in $A_i$.

By [F5] applied to $W_i$, the branch ring $A_i$ is a free $\mathbb{C}\{x\}$-module with basis the classes of $1,y,\dots ,y^{m_i-1}$; write $\bar{y}$ for the class of $y$. By [F7] $W_i$ is irreducible in $\mathbb{C}\{x\}[y]$ and it is monic, hence primitive, so by [F6] it is irreducible in $L_i[y]$ for $L_i:=\mathrm{Frac}(\mathbb{C}\{x\})$. Since $W_i(\bar{y})=0$ and $W_i$ is monic of degree $m_i$ and irreducible over $L_i$, it is the minimal polynomial of $\bar{y}$ over $L_i$.

Under ${\bar{\sigma }}_i$ the class of $x$ goes to $t^{m_i}$, so the image of $L_i=\mathrm{Frac}(\mathbb{C}\{x\})$ is the subfield $\mathbb{C}((t_i^{m_i})):=\mathrm{Frac}(\mathbb{C}\{t_i^{m_i}\})$ of $\mathbb{C}((t_i))$ consisting of convergent Laurent germs in $t_i^{m_i}$. We claim $[\mathbb{C}((t_i)):\mathbb{C}((t_i^{m_i}))]=m_i$: every element of $\mathbb{C}((t_i))$ is $g\mathrm{/}{t}^N$ with $g\in \mathbb{C}\{t\}$, and splitting the exponents of the expansion of $g$ by their residue modulo $m_i$ writes it as ${\sum }_{j<m_i}{t}^j{f}_j(t^{m_i})$ with $f_j\in \mathrm{Frac}(\mathbb{C}\{t^{m_i}\})$; each grouped series converges for $\mathrm{\mid }{t}^{m_i}\mathrm{\mid }$ sufficiently small by absolute convergence of the original series. Thus the $m_i$ elements $1,t,\dots ,t^{m_i-1}$ span, and they are linearly independent because $t$-expansions are unique and terms of distinct residues modulo $m_i$ cannot cancel [F13].

For $i\mathrm{\ne }j$ the function $q_j\circ {\gamma }_i$ is holomorphic on ${\mathrm{\Delta }}_{{\delta }_i}$ and vanishes at $0$ because ${\gamma }_i(0)=p\in Z(q_j)$. It is not identically zero: otherwise the representative $q_j$ would vanish on the full representative ${\gamma }_i({\mathrm{\Delta }}_{{\delta }_i})$ of $Z(q_i)$, so $q_j\in I_p(Z(q_i))=(q_i)$ by [F4], making the irreducible germs $q_j$ and $q_i$ associate and contradicting [F1]. By [F12] the zeros of $q_j\circ {\gamma }_i$ are isolated, so after shrinking ${\delta }_i$ we may assume $q_j\circ {\gamma }_i$ has no zero in the punctured disc; then ${\gamma }_i({\mathrm{\Delta }}_{{\delta }_i}\setminus \{0\})$ is disjoint from $Z(q_j)$, hence from ${\gamma }_j({\mathrm{\Delta }}_{{\delta }_j}\setminus \{0\})$. Doing this for the finitely many ordered pairs and shrinking once more so that every $Z(q_i)$ agrees near $p$ with ${\gamma }_i({\mathrm{\Delta }}_{{\delta }_i})$ and $X$ agrees with ${\bigcup }_{i}Z(q_i)$, the punctured images are pairwise disjoint and their union with $p$ is a full representative of $X$.

The quotient maps $A\to A_i$, $a\mapsto a+(q_i)$, are well defined because $f=u{q}_1\cdots q_r\in (q_i)$, and composing with the injections ${\bar{\sigma }}_i$ gives ${\mathrm{\Phi }}_i:A\to \mathbb{C}\{t_i\}$ and $\mathrm{\Phi }=({\mathrm{\Phi }}_1,\dots ,{\mathrm{\Phi }}_r):A\to {\prod }_i\mathbb{C}\{t_i\}$. If $\mathrm{\Phi }(a)=0$, then $a\in (q_i)$ for every $i$. Since the $q_i$ are pairwise nonassociate irreducibles in the unique factorisation domain ${\mathcal{O}}_{{\mathbb{C}}^2,p}$ [F1], each $q_i$ divides $a$ and the pairwise coprime factors $q_i$ have product dividing $a$; hence $q_1\cdots q_r$, which is associate to $f$, divides $a$ and $a=0$ in $A$. Thus $\mathrm{\Phi }$ is injective.

Every element of $K_i=\mathrm{Frac}(A_i)$ lies in $L_i\cdot A_i$, so $K_i$ is generated as an $L_i$-vector space by $1,\bar{y},\dots ,{\bar{y}}^{m_i-1}$ and $[K_i:L_i]\le m_i$; indeed for $0\mathrm{\ne }b\in A_i$ the $L_i$-linear map $x\mapsto bx$ on the finite-dimensional $L_i$-space $L_i\cdot A_i$ is injective (a domain), hence bijective, so $b$ is invertible in $L_i\cdot A_i$. On the other hand $\bar{y}$ has degree $m_i$ over $L_i$ by step 2.2, so $[K_i:L_i]\ge m_i$ and therefore $[K_i:L_i]=m_i$ and $K_i=L_i[\bar{y}]$.

${\prod }_i\mathbb{C}\{t_i\}$ is a finitely generated $A$-module. Indeed, splitting by residue modulo $m_i$ gives $\mathbb{C}\{t_i\}={\sum }_{j<m_i}\mathbb{C}\{t_i^{m_i}\}{t}_i^j$, and $\mathbb{C}\{t_i^{m_i}\}={\bar{\sigma }}_i(\mathbb{C}\{x\})$ is contained in ${\mathrm{\Phi }}_i(A)$ because ${\sigma }_i(x)=t_i^{m_i}$; hence the $m_i$ elements $1,t_i,\dots ,t_i^{m_i-1}$ generate $\mathbb{C}\{t_i\}$ over ${\mathrm{\Phi }}_i(A)$, and ${\mathrm{\Phi }}_i(A)$ is a quotient of $A$. Placing these finitely many generators in their respective coordinates and zero in the other coordinates generates the finite product over $A$.

The injection ${\bar{\sigma }}_i$ extends to an injective field homomorphism $K_i\to \mathbb{C}((t_i))$ whose image contains $\mathbb{C}((t_i^{m_i}))$ and has degree $[K_i:L_i]=m_i$ over it by steps 2.3 and 3.2. Since $[\mathbb{C}((t_i)):\mathbb{C}((t_i^{m_i}))]=m_i$, the image is all of $\mathbb{C}((t_i))$; thus ${\bar{\sigma }}_i$ induces an isomorphism $K_i\to \mathbb{C}((t_i))$.

By [F9] there is an isomorphism $Q(A)\to {\prod }_i{K}_i$ restricting to the componentwise quotient maps on $A$; composing with the componentwise isomorphisms $K_i\to \mathbb{C}((t_i))$ of step 4.1 gives an isomorphism $\mathrm{\Psi }:Q(A)⟶{\prod }_i\mathbb{C}((t_i))$ whose restriction to $A$ is exactly $\mathrm{\Phi }$.

${\prod }_i\mathbb{C}\{t_i\}$ is integral over $\mathrm{\Phi }(A)$: it is a finitely generated $\mathrm{\Phi }(A)$-module by step 3.3, so [F8] applies with $R^{\mathrm{\prime }}=\mathrm{\Phi }(A)\mathrm{\ne }0$. Consequently, if $b\in Q(A)$ has $\mathrm{\Psi }(b)\in {\prod }_i\mathbb{C}\{t_i\}$, then $\mathrm{\Psi }(b)$ satisfies a monic equation with coefficients in $\mathrm{\Phi }(A)$, and applying ${\mathrm{\Psi }}^{-1}$ exhibits $b$ as integral over $A$.

Conversely, let $b\in Q(A)$ be integral over $A$ and write $\mathrm{\Psi }(b)=(c_i)$. Applying $\mathrm{\Psi }$ to a monic equation for $b$ over $A$ gives a monic equation for $(c_i)$ with coefficients in $\mathrm{\Phi }(A)\subseteq {\prod }_i\mathbb{C}\{t_i\}$; by [F11] each component $c_i$ satisfies a monic equation over ${\mathrm{\Phi }}_i(A)\subseteq \mathbb{C}\{t_i\}$ and is therefore integral over $\mathbb{C}\{t_i\}$, hence $c_i\in \mathbb{C}\{t_i\}$ because $\mathbb{C}\{t_i\}$ is integrally closed in $\mathbb{C}((t_i))$ [F10]. Therefore $\mathrm{\Psi }(b)\in {\prod }_i\mathbb{C}\{t_i\}$.

By steps 6.1 and 6.2 the integral closure of $A$ in $Q(A)$ is ${\mathrm{\Psi }}^{-1}({\prod }_i\mathbb{C}\{t_i\})$; identifying $Q(A)$ with ${\prod }_i\mathbb{C}((t_i))$ along the isomorphism $\mathrm{\Psi }$, the normalisation of $A$ is exactly ${\prod }_i\mathbb{C}\{t_i\}$. This proves the finite, integral, birational and normalisation assertions.

Steps 3.1, 3.3, 5.1 and 7.1 give the finite birational integral embedding and the identification of the normalisation with ${\prod }_i\mathbb{C}\{t_i\}$; steps 1.2 and 2.4 give the separation of the branches and the local biholomorphism statement. ∎