Show that analytically isomorphic curve singularities (I, 5.6.1) are equivalent in the sense of (3.9.4), but not conversely.
Solution
We first prove the forward implication for plane curve germs. For the counterexample to the converse we work over a field of characteristic zero, for example \(k=\CC\).
1. Let\[ A_f=k[[x,y]]/(f), \qquad A_g=k[[x,y]]/(g) \]be two singular plane-curve complete local rings. Any analytic isomorphism\[ A_f\xrightarrow{\sim}A_g \]lifts to an automorphism\[ \Phi:k[[x,y]]\xrightarrow{\sim}k[[x,y]] \]such that\[ \Phi(f)=u g \]for a unit \(u\in k[[x,y]]^\times\).
An analytic isomorphism is, by (I, 5.6.1), an isomorphism of completed local rings. Choose lifts \(X,Y\in k[[x,y]]\) of the images of the residue classes of \(x,y\) under the isomorphism. Since the induced map on cotangent spaces \[ \mathfrak m_f/\mathfrak m_f^2 \longrightarrow \mathfrak m_g/\mathfrak m_g^2 \] is an isomorphism, the linear parts of \(X,Y\) are linearly independent. The formal inverse-function theorem therefore makes \[ x\longmapsto X, \qquad y\longmapsto Y \] an automorphism \(\Phi\) of \(k[[x,y]]\) lifting the given quotient-ring isomorphism.
The kernel of the quotient map to \(A_f\) is the principal ideal \((f)\), and the kernel on the other side is \((g)\). Hence \[ \Phi((f))=(g). \] Two generators of the same principal ideal in the domain \(k[[x,y]]\) differ by a unit, so \[ \Phi(f)=ug. \]
2. A formal ambient automorphism as in step 1 identifies the completeembedded resolution data obtained by successive point blowups.
The automorphism \(\Phi\) carries the maximal ideal \((x,y)\) to itself and identifies the two curve ideals. The blowup of the closed point is the Proj of the Rees algebra of this maximal ideal. Therefore \(\Phi\) induces an isomorphism of the two formal blowups carrying exceptional divisor to exceptional divisor and strict transform to strict transform.
At every point where a further blowup is required, the induced isomorphism of completed local rings again identifies the maximal ideals and the strict transform ideals. Applying the same Rees-algebra construction recursively identifies the next blowups. Thus the entire sequence of infinitely near points, exceptional components, branch incidences, proximity relations, and multiplicities is preserved.
This is precisely the embedded-resolution data entering equivalence in (3.9.4). Hence analytically isomorphic curve singularities are equivalent.
3. Equivalently, in characteristic zero analytic isomorphism preservesthe Puiseux characteristic, and equal Puiseux characteristic gives the sameequivalence class in (3.9.4).
This is the standard Puiseux-resolution theorem cited by the retained source for V.3.6 (Wall, Singular Points of Plane Curves). The characteristic exponents of each branch, together with pairwise contact data, determine the successive point-blowup resolution data, and conversely that resolution data recovers the Puiseux characteristics. Step 2 gives the forward statement without needing this classification theorem; we record it here because it will identify the counterexample below as equivalent.
4. Consider over \(k=\CC\) the two irreducible plane-curve germs\[ C_1:\quad f_1=y^3+x^7=0, \]and\[ C_2:\quad f_2=y^3+x^5y+x^7=0. \]Both have the single Puiseux pair\[ \boxed{(3,7)}. \]
For \(C_1\) there is the parametrization \[ x=t^3, \qquad y=-t^7, \] so its unique branch has characteristic pair \((3,7)\).
For \(C_2\), substitute \[ x=t^3, \qquad y=t^7z(t). \] The equation becomes \[ t^{21}\bigl(z^3+tz+1\bigr)=0. \] At \(t=0\), the equation \(z^3+1=0\) has the root \(z=-1\), and \[ \frac{\partial}{\partial z}(z^3+tz+1)\bigg|_{(0,-1)}=3\ne0. \] The formal implicit-function theorem therefore gives a unique \(z(t)\in\CC[[t]]\) with \[ z(0)=-1, \qquad z(t)^3+t z(t)+1=0. \] Hence \[ x=t^3, \qquad y=-t^7+\text{higher-order terms}. \] Since \(\gcd(3,7)=1\), the gcd of the characteristic exponents has already dropped to \(1\) at exponent \(7\), so there are no further Puiseux pairs. The other two choices of the cube root of \(-1\) give the same branch after the reparametrizations \(t\mapsto\zeta t\) with \(\zeta^3=1\). Thus \(C_2\) is irreducible and also has Puiseux pair \((3,7)\).
5. The singularities \(C_1\) and \(C_2\) are equivalent in the sense of(3.9.4).
By step 4, they have identical Puiseux characteristic. The Puiseux-resolution theorem of step 3 therefore gives identical embedded resolution data, hence equivalence in the sense of (3.9.4).
6. For a plane hypersurface germ \(f=0\), put\[ \tau(f) = \dim_k \frac{k[[x,y]]}{(f,f_x,f_y)}. \]The Tjurina number \(\tau(f)\) is invariant under analytic isomorphism of thecurve germ.
By step 1, an analytic isomorphism is induced by a formal coordinate automorphism \(\Phi\) together with multiplication of the equation by a unit: \[ \Phi(f)=ug. \]
The chain rule says that the two partial derivatives of \(\Phi(f)\) are invertible \(k[[x,y]]\)-linear combinations of \(\Phi(f_x),\Phi(f_y)\), because the Jacobian matrix of a formal coordinate automorphism is invertible. Thus \(\Phi\) carries \[ (f,f_x,f_y) \] to the Tjurina ideal of \(\Phi(f)\).
On the other hand, \[ (ug,(ug)_x,(ug)_y) = (ug,u g_x+u_xg,u g_y+u_yg) = (g,g_x,g_y), \] because \(u\) is a unit. Hence the two Tjurina algebras are isomorphic and have the same dimension.
7. The first germ has\[ \boxed{\tau(f_1)=12}. \]
Since \[ (f_1)_x=7x^6, \qquad (f_1)_y=3y^2, \] and the coefficients \(3,7\) are units over \(\CC\), one has \[ (f_1,(f_1)_x,(f_1)_y)=(x^6,y^2). \] Therefore the residue classes \[ x^i y^j, \qquad 0\le i\le5, \quad 0\le j\le1, \] form a basis of the Tjurina algebra. There are \(6\cdot2=12\) such monomials.
8. The second germ has\[ \boxed{\tau(f_2)=11}. \]
The derivatives are \[ (f_2)_x=5x^4y+7x^6, \qquad (f_2)_y=3y^2+x^5. \] Modulo the Tjurina ideal, therefore, \[ y^2=-\frac13x^5, \qquad x^4y=-\frac75x^6. \] Using the first relation in \(f_2=0\) gives \[ 2x^5y+3x^7=0, \] while multiplying \((f_2)_x=0\) by \(x\) gives \[ 5x^5y+7x^7=0. \] The coefficient matrix \[ \begin{pmatrix}2&3\\5&7\end{pmatrix} \] has determinant \(-1\), so \[ x^5y=x^7=0. \] Consequently the Tjurina ideal is generated by \[ y^2+\frac13x^5, \qquad x^4y+\frac75x^6, \qquad x^7. \] For lexicographic order with \(y>x\), these three generators have leading terms \[ y^2, \qquad x^4y, \qquad x^7, \] and their three \(S\)-polynomials reduce to zero using these same relations. They are therefore a Gröbner basis. Hence the standard monomials are \[ 1,x,\ldots,x^6, \qquad y,xy,x^2y,x^3y. \] There are \(7+4=11\) of them.
9. The equivalent singularities \(C_1\) and \(C_2\) are not analyticallyisomorphic.
If they were analytically isomorphic, step 6 would give \[ \tau(f_1)=\tau(f_2). \] But steps <1>7–<1>8 give \[ 12\ne11. \] Thus the germs are not analytically isomorphic. Together with step 5, this proves that equivalence in the sense of (3.9.4) does not imply analytic isomorphism.
10. Q.E.D.
Steps <1>1–<1>2 prove that analytic isomorphism implies equivalence. Steps <1>4–<1>9 give two equivalent singularities which are not analytically isomorphic, proving that the converse fails.