For each of the following singularities at \((0,0)\) in the plane, give an embedded resolution, compute \(\delta_P\), and decide which ones are equivalent.
\(x^3+y^5=0\).
\(x^3+x^4+y^5=0\).
\(x^3+y^4+y^5=0\).
\(x^3+y^5+y^6=0\).
\(x^3+x y^3+y^5=0\).
Solution
All blowups below are point blowups of the ambient plane or of the smooth surface obtained at the preceding stage. We keep blowing up after the strict transform becomes smooth whenever this is needed to make the total transform a simple normal-crossings divisor. For the delta invariant we use D-CRVPLSING: \[ \delta_P = \sum_Q\binom{m_Q}{2}, \] where only singular points \(Q\) of the successive strict transforms contribute, since a smooth point has multiplicity one.
1. For (a),\[ C_a:\quad x^3+y^5=0, \]the first blowup has strict-transform equation\[ u^3+v^2=0 \]in the chart\[ x=uv, \qquad y=v. \]
Substitution gives \[ x^3+y^5 = v^3(u^3+v^2). \] Thus the exceptional curve is \[ E_1:(v=0) \] and the strict transform is \[ C_{a,1}:(u^3+v^2=0). \] The point \[ Q_1=(0,0) \] is its unique point over the origin. It has multiplicity \(2\), and its tangent is \(v=0\), the exceptional curve \(E_1\).
2. Blowing up \(Q_1\) in (a), with\[ v=uw, \]gives\[ C_{a,2}:\quad u+w^2=0. \]This strict transform is smooth but passes through the intersection of thetwo exceptional curves and is tangent to the new exceptional curve.
Substitution gives \[ u^3+u^2w^2 = u^2(u+w^2). \] Hence the new exceptional curve is \[ E_2:(u=0), \] while the strict transform of \(E_1\) is \[ E_1':(w=0). \] The curve \(u+w^2=0\) is smooth at their intersection \((0,0)\), but its tangent is \(u=0=E_2\). Thus the total transform is not yet normal crossings.
3. The third blowup in (a) produces a point at which the strict transformand two exceptional curves meet with three distinct tangent directions; afourth blowup separates these three branches and completes an embeddedresolution.
At the point of step 2 use \[ u=st, \qquad w=s. \] Then \[ u+w^2 = s(t+s), \] so the strict transform is \[ C_{a,3}:(t+s=0). \] In this chart the new exceptional curve is \(E_3:(s=0)\) and the strict transform of \(E_2\) is \(E_2':(t=0)\). At \((s,t)=(0,0)\) the three curves \[ s=0, \qquad t=0, \qquad t+s=0 \] have distinct tangent directions. A fourth blowup at this triple point therefore makes their strict transforms meet the new exceptional \(E_4\) at three distinct points. All remaining intersections are transverse, so the total transform is simple normal crossings.
4. For (a),\[ \boxed{\delta_P(C_a)=4}. \]
The strict transform is singular only at the original point, of multiplicity \(3\), and at \(Q_1\), of multiplicity \(2\). All later centres lie on a smooth strict transform. Hence \[ \delta_P(C_a) = \binom32+\binom22 = 3+1 =4. \]
5. For (b),\[ C_b:\quad x^3+x^4+y^5=0, \]the first two strict transforms are\[ u^3+u^4v+v^2=0 \]and\[ u+w^2+u^3w=0. \]
The same first chart \(x=uv\), \(y=v\) gives \[ x^3+x^4+y^5 = v^3(u^3+u^4v+v^2). \] The unique point over the origin again has multiplicity \(2\) and tangent \(v=0\). Blowing it up with \(v=uw\) gives \[ u^3+u^5w+u^2w^2 = u^2(u+w^2+u^3w). \] Thus the second strict transform is smooth, passes through \(E_1'\cap E_2\), and is tangent to \(E_2:(u=0)\) exactly as in (a).
6. The third and fourth blowups for (b) have the same incidence patternas for (a), and\[ \boxed{\delta_P(C_b)=4}. \]
At the third centre put \(u=st\), \(w=s\). The second strict-transform equation becomes \[ s\bigl(t+s+s^3t^3\bigr)=0, \] so the third strict transform is \[ t+s+s^3t^3=0. \] Its tangent at the origin is \(t+s=0\), distinct from the two exceptional directions \(s=0\) and \(t=0\). Thus, exactly as in step 3, a fourth blowup of the triple point completes the embedded resolution.
The only singular strict transforms have multiplicities \(3\) and \(2\). Therefore \[ \delta_P(C_b)=\binom32+\binom22=4. \]
7. For (d),\[ C_d:\quad x^3+y^5+y^6=0, \]the first two strict transforms are\[ u^3+v^2+v^3=0 \]and\[ u+w^2+uw^3=0. \]
The first blowup gives \[ x^3+y^5+y^6 = v^3(u^3+v^2+v^3), \] so the infinitely near point again has multiplicity \(2\) and tangent \(v=0\). Under the second blowup \(v=uw\) one obtains \[ u^3+u^2w^2+u^3w^3 = u^2(u+w^2+uw^3). \] The second strict transform is smooth and tangent to \(E_2:(u=0)\) at \(E_1'\cap E_2\), just as in (a) and (b).
8. The third and fourth blowups for (d) again have the same incidencepattern as (a), and\[ \boxed{\delta_P(C_d)=4}. \]
With \(u=st\), \(w=s\), the second strict-transform equation becomes \[ s\bigl(t+s+s^3t\bigr)=0. \] Thus the third strict transform has tangent \(t+s=0\) at the triple point of the two relevant exceptional components, and one final blowup separates the three distinct directions. The singular multiplicity sequence is again \[ 3,2, \] so \[ \delta_P(C_d)=4. \]
9. The singularities (a), (b), and (d) are equivalent.
Steps <1>1–<1>3, <1>5–<1>6, and <1>7–<1>8 exhibit the same succession of centres, exceptional-component incidences, tangencies, and multiplicities: the curve has multiplicities \(3,2,1,1\) at the four centres; after the first blowup it is a double point tangent to \(E_1\); after the second it is smooth, tangent to \(E_2\), and passes through \(E_1'\cap E_2\); after the third it passes through a triple point with a third distinct tangent direction; the fourth blowup separates the three branches.
This is the same embedded-resolution data in the sense of (3.9.4), so the three singularities are equivalent. In characteristic zero this also follows from the formal changes \[ x\mapsto x(1+x)^{1/3} \] for (b) and \[ y\mapsto y(1+y)^{1/5} \] for (d), which analytically identify both germs with (a).
10. For (c),\[ C_c:\quad x^3+y^4+y^5=0, \]the first blowup gives the smooth strict transform\[ u^3+v+v^2=0, \]which is tangent to the exceptional curve.
With \(x=uv\), \(y=v\), \[ x^3+y^4+y^5 = v^3(u^3+v+v^2). \] The strict transform is smooth at the origin because the coefficient of \(v\) in its linear term is \(1\). Its tangent is \(v=0=E_1\).
11. Three further blowups give an embedded resolution of (c), and\[ \boxed{\delta_P(C_c)=3}. \]
Blow up the tangency in step 10 by setting \(v=uw\). Then \[ u^3+uw+u^2w^2 = u(u^2+w+uw^2), \] so the second strict transform is \[ u^2+w+uw^2=0. \] It is smooth, tangent to the strict transform \(E_1':(w=0)\), and passes through \(E_1'\cap E_2\).
Blow up this point with \(w=us\). The equation becomes \[ u\bigl(u+s+u^2s^2\bigr)=0, \] so the third strict transform has tangent \(u+s=0\), distinct from the two exceptional directions through the resulting triple point. A fourth blowup separates these directions and gives simple normal crossings.
The curve itself was singular only at the original point, where its multiplicity is \(3\). Hence \[ \delta_P(C_c)=\binom32=3. \]
12. The singularity (c) is not equivalent to any of (a), (b), or (d).
After the first blowup, (c) is already smooth, whereas the strict transforms of (a), (b), and (d) have multiplicity \(2\). Thus their infinitely near multiplicity data differ. Equivalently, the delta invariant of (c) is \(3\) while that of (a), (b), and (d) is \(4\).
13. For (e),\[ C_e:\quad x^3+xy^3+y^5=0, \]the first blowup gives\[ u^3+uv+v^2=0. \]At the infinitely near point this is an ordinary node with tangent cone\[ v(u+v)=0. \]
With \(x=uv\), \(y=v\), \[ x^3+xy^3+y^5 = v^3(u^3+uv+v^2). \] At \((u,v)=(0,0)\) the lowest-degree part is \[ uv+v^2=v(u+v), \] the product of two distinct linear forms. Hence the strict transform has an ordinary node of multiplicity \(2\) there. One branch is tangent to the old exceptional curve \(E_1:(v=0)\) and the other has tangent \(u+v=0\).
14. Two further blowups give an embedded resolution of (e), and\[ \boxed{\delta_P(C_e)=4}. \]
Blow up the node with \(v=uw\). Then \[ u^3+u^2w+u^2w^2 = u^2(u+w+w^2). \] Thus the two branches meet the new exceptional curve \(E_2:(u=0)\) at the two distinct points \[ w=0, \qquad w=-1. \] The point \(w=-1\) is already transverse and lies away from \(E_1'\). At \(w=0\), the strict transform passes through \(E_1'\cap E_2\) with tangent \(u+w=0\), distinct from both exceptional directions. One final blowup at this triple point separates the three components and completes the embedded resolution.
The singular strict-transform multiplicities are \(3\) at the origin and \(2\) at the node after the first blowup. Hence \[ \delta_P(C_e) = \binom32+\binom22 =4. \]
15. The singularity (e) is not equivalent to any of (a)–(d).
The first strict transform of (e) is a node and therefore has two analytic branches. In contrast, (a), (b), (c), and (d) have a single branch through every infinitely near point in their displayed resolutions. Correspondingly, (e) reaches simple normal crossings after three blowups, while each of (a)–(d) requires four in the embedded resolution described above.
Branch incidence and the exceptional-resolution graph are part of the equivalence data in (3.9.4), so (e) is inequivalent to all four others.
16. The complete answer is\[ \boxed{ \begin{array}{c|ccccc} & (a)&(b)&(c)&(d)&(e)\\ \hline \delta_P&4&4&3&4&4 \end{array}} \]and the equivalence classes are\[ \boxed{\{(a),(b),(d)\},\qquad\{(c)\},\qquad\{(e)\}.} \]
14.Step 9 proves the equivalence of (a), (b), and (d); steps 12 and15 separate (c) and (e) from all other classes.
17. Q.E.D.
Steps <1>1–<1>16 give an embedded resolution, delta invariant, and equivalence classification for every singularity in the exercise.