If \(P \in Y\) and \(Q \in Z\) are analytically isomorphic plane curve singularities, show that the multiplicities \(\mu_P(Y)\) and \(\mu_Q(Z)\) are the same.
Let \(f = f_r + f_{r+1} + \cdots \in k[[x,y]]\), and suppose the leading form \(f_r\) of \(f\) factors as \(f_r = g_s h_t\), where \(g_s\) and \(h_t\) are homogeneous of degrees \(s\) and \(t\) respectively and have no common linear factor. Show that there are formal power series \[ g = g_s + g_{s+1} + \cdots, \qquad h = h_t + h_{t+1} + \cdots \] in \(k[[x,y]]\) such that \(f = gh\).
Let \(Y\) be defined by \(f(x,y) = 0\) in \(\AA^2\), and let \(P = (0,0)\) be a point of multiplicity \(r\) on \(Y\), so that \(f = f_r + (\text{higher terms})\). Call \(P\) an ordinary \(r\)-fold point if \(f_r\) is a product of \(r\) distinct linear factors. Show that any two ordinary double points are analytically isomorphic, and likewise for ordinary triple points. Show however that there is a one-parameter family of mutually nonisomorphic ordinary \(4\)-fold points.
Assume \(\operatorname{char} k \neq 2\). Show that any double point of a plane curve is analytically isomorphic to the singularity at \((0,0)\) of the curve \(y^2 = x^r\), for a uniquely determined \(r \geq 2\). If \(r = 2\) it is a node; if \(r = 3\) it is a cusp; if \(r = 4\) it is a tacnode.
Solution
Translate the chosen point to the origin. For a nonzero series \(F\in k[[x,y]]\), write \[ \operatorname{ord}(F)=\min\{i+j:[x^iy^j]F\ne0\}, \] and call its homogeneous term of that degree its initial form.
1. Analytic isomorphism preserves the multiplicity of a plane curve singularity.
Let \[ A=k[[x,y]]/(f),\qquad \mathfrak m=(x,y)A, \] where \(r=\operatorname{ord}(f)\) and \(f_r\) is the initial form. If \(r=1\), then \(A\) is a regular local ring of dimension one: after a linear change, the initial form is \(y\), and the formal inverse-function argument solves for \(y\) as a series in \(x\). Regularity is intrinsic to the complete local ring, so an analytically isomorphic plane curve point also has multiplicity one.
Assume now that \(r\ge2\). Then \(\dim_k\mathfrak m/\mathfrak m^2=2\), and the images of \(x,y\) give an identification \(\operatorname{Sym}_k(\mathfrak m/\mathfrak m^2)\cong k[x,y]\). For the \(\mathfrak m\)-adic filtration, \[ \operatorname{gr}_{\mathfrak m}A \cong k[x,y]/(f_r). \] Indeed, the initial ideal of the principal ideal \((f)\) is generated by \(f_r\): if \(u\) has initial form \(u_d\), then the initial form of \(uf\) is \(u_df_r\).
The smallest degree in which the natural map \[ \operatorname{Sym}_k(\mathfrak m/\mathfrak m^2) \longrightarrow \operatorname{gr}_{\mathfrak m}A \] has a nonzero kernel is therefore exactly \(r\). An isomorphism of complete local \(k\)-algebras carries the maximal ideal to the maximal ideal and induces an isomorphism of associated graded rings, preserving that smallest degree. For a plane curve, this order \(r\) is precisely the multiplicity at the point. Hence analytically isomorphic plane curve singularities have the same multiplicity, proving part (a).
2. If homogeneous forms \(g_s,h_t\in k[x,y]\) have no common linear factor, then for every \(d\ge s+t-1\),\[ k[x,y]_d=g_s k[x,y]_{d-s}+h_t k[x,y]_{d-t}. \]
Because every nonconstant homogeneous polynomial in two variables factors into linear forms over the algebraically closed field \(k\), the hypothesis says \(\gcd(g_s,h_t)=1\). Consequently the syzygies of \((g_s,h_t)\) are generated by \((h_t,-g_s)\), and there is an exact sequence \[ 0\longrightarrow k[x,y](-s-t) \xrightarrow{q\mapsto(qh_t,-qg_s)} k[x,y](-s)\oplus k[x,y](-t) \xrightarrow{(a,b)\mapsto ag_s+bh_t} (g_s,h_t)\longrightarrow0. \] Taking degree \(d\) and using \(\dim_k k[x,y]_m=m+1\) for \(m\ge0\) shows that the degree-\(d\) quotient by \((g_s,h_t)\) has dimension zero once \(d\ge s+t-1\). For \(d=s+t-1\) this is \[ (s+t)-(t)-(s)=0, \] and for \(d\ge s+t\) it is \[ (d+1)-(d-s+1)-(d-t+1)+(d-s-t+1)=0. \] Thus every form of degree at least \(s+t-1\) has the asserted expression.
3. The factors \(g\) and \(h\) required in part (b) can be constructed degree by degree.
Since \(f_r=g_sh_t\), we have \(r=s+t\). Suppose for some \(n\ge1\) that homogeneous terms \[ g_s,\ldots,g_{s+n-1},\qquad h_t,\ldots,h_{t+n-1} \] have been chosen so that \[ f-\left(\sum_{i=0}^{n-1}g_{s+i}\right) \left(\sum_{j=0}^{n-1}h_{t+j}\right) \] has no homogeneous term of degree below \(r+n\). Let \(R_{r+n}\) be its degree-\((r+n)\) term. By step 2, because \(r+n=s+t+n\ge s+t-1\), there are forms \[ g_{s+n}\in k[x,y]_{s+n},\qquad h_{t+n}\in k[x,y]_{t+n} \] such that \[ R_{r+n}=g_{s+n}h_t+g_sh_{t+n}. \] Adding these two terms therefore kills the discrepancy in degree \(r+n\) without changing lower degrees.
Induction defines \[ g=\sum_{i\ge0}g_{s+i},\qquad h=\sum_{j\ge0}h_{t+j}. \] For each total degree only finitely many summands contribute, so these are well-defined formal power series. The construction makes every homogeneous component of \(f-gh\) zero; hence \(f=gh\). This proves part (b).
4. An ordinary double point is analytically isomorphic to \(xy=0\), and an ordinary triple point is analytically isomorphic to \(xy(x-y)=0\).
For an ordinary \(r\)-fold point, repeatedly apply step 3, separating one tangent factor from the product of the others. This gives \[ f=g_1\cdots g_r, \] where each \(g_i\) has a distinct nonzero linear initial form \(L_i\).
For \(r=2\), the two linear forms are independent. Thus \((X,Y)=(g_1,g_2)\) is a formal change of coordinates: its linear part is invertible, so the inverse is constructed recursively degree by degree. In these coordinates \((f)=(XY)\).
For \(r=3\), first use \((g_1,g_2)\) as coordinates \(x,y\). The third factor then has the form \[ g_3=ax+by+q, \] where \(a,b\ne0\) and \(q\in(x,y)^2\), because its tangent line is distinct from both coordinate axes. Write \(q=xq_1+yq_2\) and put \[ X=x(1+q_1/a),\qquad Y=y(1+q_2/b). \] This is another formal coordinate change, and \(g_3=aX+bY\). Up to multiplication by a unit, the equation is therefore \(XY(aX+bY)\). A linear change of \(X,Y\) carries its three distinct tangent lines to \(x=0\), \(y=0\), and \(x-y=0\). Thus all ordinary double points, respectively all ordinary triple points, have the same analytic type.
5. Ordinary quadruple points have a one-dimensional family of analytic types.
For \(\lambda\in k\setminus\{0,1\}\) consider \[ A_\lambda =k[[x,y]]/\bigl(xy(x-y)(x-\lambda y)\bigr). \] Each is an ordinary quadruple point. An isomorphism \(A_\lambda\cong A_\mu\) induces an isomorphism of associated graded rings by step 1, hence an invertible linear map of the tangent space carrying the unordered four tangent lines of \(A_\lambda\) to those of \(A_\mu\). After identifying lines through the origin with points of \(\PP^1\), this says that the unordered configurations \[ \{0,\infty,1,\lambda\},\qquad \{0,\infty,1,\mu\} \] are projectively equivalent.
The cross ratio of four ordered points is invariant under \(\operatorname{PGL}_2(k)\). Permuting the four points changes \(\lambda\) only among \[ \lambda,\quad 1-\lambda,\quad \lambda^{-1},\quad (1-\lambda)^{-1},\quad \frac{\lambda}{\lambda-1},\quad \frac{\lambda-1}{\lambda}. \] Therefore \(A_\lambda\cong A_\mu\) can hold only when \(\mu\) lies in this finite orbit of \(\lambda\). The parameter line \(k\setminus\{0,1\}\) modulo these finite orbits is one-dimensional; choosing one representative from each orbit gives a one-parameter collection of mutually nonisomorphic ordinary quadruple points. This proves part (c).
6. Assume \(\operatorname{char}k\ne2\). Every double point is analytically isomorphic to \(y^2=x^r\) for some \(r\ge2\).
Let \(f=f_2+f_3+\cdots\) define the double point. If \(f_2\) is a product of two distinct linear forms, step 4 gives the node \(xy=0\). Since \(k\) is algebraically closed and \(2\) is invertible, the linear change \[ x=Y-X,\qquad y=Y+X \] identifies \(xy=0\) with \(Y^2=X^2\), so this is the case \(r=2\).
Otherwise \(f_2\) is the square of one linear form. After a linear change and multiplication of \(f\) by a nonzero scalar, write \[ f=y^2+\text{terms of total degree at least }3. \] We now eliminate every higher-degree term containing \(y\). Suppose inductively that through degree \(N-1\) the equation has the form \[ y^2-a(x), \] and let \(R_N(x,y)\) be the remaining homogeneous term of degree \(N\). Write uniquely \[ R_N(x,y)=yB_{N-1}(x,y)+c_Nx^N. \] Replacing \(y\) by \(y-\tfrac12B_{N-1}(x,y)\) changes \(y^2\) by \(-yB_{N-1}\) in degree \(N\); all other new terms have degree greater than \(N\). Thus it kills the \(y\)-divisible part and absorbs \(c_Nx^N\) into \(a(x)\). The successive coordinate changes have identity linear part and orders tending to infinity, so their infinite composition converges in the \((x,y)\)-adic topology to a formal automorphism. It transforms the equation into \[ y^2-a(x)=0 \] for a series \(a(x)\in x^3k[[x]]\).
The series \(a\) is nonzero. Indeed, the local ring of the plane curve variety is reduced and essentially of finite type over \(k\). Its completion map has geometrically regular formal fibres, as recorded in the regularity of completion for rings essentially of finite type over a field, so reducedness ascends to the completion. If \(a=0\), the transformed completed local ring would be \(k[[x,y]]/(y^2)\), which is not reduced. Write \[ a(x)=x^ru(x),\qquad r\ge3, \] with \(u(0)\ne0\). Because \(k\) is algebraically closed, choose a square root of \(u(0)\). Since \(2\) is invertible, recursively comparing coefficients gives a unit \(v(x)\in k[[x]]^\times\) with \(v(x)^2=u(x)\). The coordinate change \(Y=y/v(x)\) gives \[ Y^2=x^r. \] Together with the distinct-tangent case, this proves existence for every \(r\ge2\) that occurs.
7. The exponent \(r\) in step 6 is an invariant of the completed local ring.
Put \[ A_r=k[[x,y]]/(y^2-x^r). \] If \(r=2m\) is even, \(A_r\) has exactly two minimal primes \[ \mathfrak p_+=(y-x^m),\qquad \mathfrak p_-=(y+x^m). \] Their intersection multiplicity is the intrinsic length \[ \operatorname{length}\bigl(A_r/(\mathfrak p_++\mathfrak p_-)\bigr) =\operatorname{length}\bigl(k[[x]]/(2x^m)\bigr)=m. \] An isomorphism permutes the minimal primes and preserves this length, so it determines \(m\) and hence \(r\).
If \(r\) is odd, \(A_r\) is a domain and its normalization is \(k[[t]]\) under \[ x=t^2,\qquad y=t^r. \] Thus the set of orders of nonzero elements of \(A_r\) is the numerical semigroup \[ \{2a+rb:a,b\ge0\}. \] In particular, \(r\) is its least odd positive element. Any isomorphism of the domains \(A_r\) and \(A_{r'}\) extends to an isomorphism of their normalizations. An automorphism of the DVR \(k[[t]]\) preserves its \(t\)-adic valuation, so it preserves this value semigroup and hence its least odd element. Therefore \(r=r'\) in the odd case as well.
Finally, an even \(r\) cannot be isomorphic to an odd \(r'\) because the former ring has two minimal primes and the latter is a domain. Hence the exponent is uniquely determined.
8. Q.E.D.
Step 1 proves part (a), steps <1>2–<1>3 prove part (b), steps <1>4–<1>5 prove part (c), and steps <1>6–<1>7 prove existence and uniqueness in part (d), including the node, cusp, and tacnode cases \(r=2,3,4\).