Let \(k\) be an algebraically closed field, and let \(X\) be the surface in \(\AA_k^3\) defined by the equation \(x^2+y^3+z^5=0\). It has an isolated singularity at the origin \(P=(0,0,0)\).
Show that the affine ring \(A=k[x, y, z] /\left(x^2+y^3+z^5\right)\) of \(X\) is a unique factorization domain, as follows. Let \(t=z^{-1}\); \(u=t^3 x\), and \(v=t^2 y\). Show that \(z\) is irreducible in \(A\); \(t \in k[u, v]\), and \(A\left[z^{-1}\right]=k\left[u, v, t^{-1}\right]\). Conclude that \(A\) is a UFD.
Show that the singularity at \(P\) can be resolved by eight successive blowings-up. If \(\tilde{X}\) is the resulting nonsingular surface, then the inverse image of \(P\) is a union of eight projective lines, which intersect each other according to the Dynkin diagram \(\mathbf{E}_8\):
Solution
1. Part (a): \(A\) is a noetherian domain, and \(z\) is a prime element of\(A\).
Put \[ F=x^2+y^3+z^5. \] Regard \(F\) as a polynomial in \(x\) over \[ K=k(y,z). \] If \(F\) were reducible over \(K\), then, since it has degree \(2\) in \(x\), it would have a root in \(K\). Thus \[ -(y^3+z^5) \] would be a square in \(K\).
The polynomial \[ y^3+z^5\in k[y,z] \] is squarefree in every characteristic. Indeed, a repeated irreducible factor would divide both partial derivatives whenever both are nonzero. In characteristic \(3\) the \(z\)-derivative is \(5z^4\neq0\), and in characteristic \(5\) the \(y\)-derivative is \(3y^2\neq0\); in all other characteristics the two derivatives have no common nonconstant divisor with \(y^3+z^5\). Hence some irreducible factor of \(y^3+z^5\) occurs with valuation \(1\), whereas every square in \(K\) has even valuation at every irreducible polynomial. Therefore \(-(y^3+z^5)\) is not a square in \(K\).
So \(F\) is irreducible over \(K\). By Gauss’s lemma it is irreducible in the UFD \(k[x,y,z]\), hence prime there. Consequently \[ A=k[x,y,z]/(F) \] is a domain. It is noetherian because it is a finitely generated \(k\)-algebra.
Moreover, \[ A/(z)\cong k[x,y]/(x^2+y^3). \] The polynomial \(x^2+y^3\) is irreducible in \(k[x,y]\): over \(k(y)\) a factorization would give a root, hence would make \(-y^3\) a square, which is impossible because its \(y\)-adic valuation is \(3\). Thus \(A/(z)\) is a domain, so \((z)\) is prime in \(A\). In particular \(z\) is irreducible.
2. Part (a): with\[ t=z^{-1}, \qquad u=t^3x, \qquad v=t^2y, \]one has\[ \boxed{t=-(u^2+v^3)} \]and\[ \boxed{A[z^{-1}]=k[u,v,t^{-1}].} \]
In \(A[z^{-1}]\) the defining equation gives \[ u^2+v^3 = t^6(x^2+y^3) = -t^6z^5 = -t. \] Hence \[ t=-(u^2+v^3)\in k[u,v]. \]
Certainly \[ u,\ v,\ t^{-1}=z \] belong to \(A[z^{-1}]\), so \[ k[u,v,t^{-1}]\subseteq A[z^{-1}]. \] Conversely, \[ z=t^{-1}, \qquad x=t^{-3}u=(t^{-1})^3u, \qquad y=t^{-2}v=(t^{-1})^2v, \] and \(t\in k[u,v]\) by the preceding calculation. Thus every generator of \(A[z^{-1}]\) belongs to \(k[u,v,t^{-1}]\), proving equality.
Equivalently, \[ A[z^{-1}] \cong k[u,v,(u^2+v^3)^{-1}], \] via \[ t=-(u^2+v^3). \] Indeed, starting with independent variables \(U,V\) and putting \[ T=-(U^2+V^3), \qquad z=T^{-1}, \qquad x=T^{-3}U, \qquad y=T^{-2}V, \] gives \[ x^2+y^3+z^5 = T^{-6}(U^2+V^3+T) =0. \] This defines an inverse to the map sending \(U,V\) to \(u,v\), so the displayed isomorphism is genuine and in particular \(u,v\) are algebraically independent. This is a localization of the polynomial UFD \(k[u,v]\), hence is a UFD.
3. Part (a): \(A\) is a unique factorization domain.
We use the prime element \(z\) from step 1 and the UFD \[ A[z^{-1}] \] from step 2.
Because \(A\) is noetherian, every nonzero nonunit of \(A\) factors into irreducibles. It remains to prove that every irreducible of \(A\) is prime. Let \(p\in A\) be irreducible.
If \(p\) is associate to \(z\), then \(p\) is prime because \(z\) is prime. Assume therefore that \(p\) is not associate to \(z\).
First, \(p\) remains irreducible in \(A[z^{-1}]\). Suppose \[ p=\frac{a}{z^m}\frac{b}{z^n} \] in \(A[z^{-1}]\). Since \(A\) is a domain, \[ z^{m+n}p=ab. \] As long as a factor \(z\) remains on the left, primality of \(z\) implies that \(z\) divides \(a\) or \(b\); cancel such a factor. After \(m+n\) cancellations we obtain \[ p=a'b' \] with \(a',b'\in A\). Irreducibility of \(p\) makes one of \(a',b'\) a unit in \(A\). The corresponding factor \(a/z^m\) or \(b/z^n\) is therefore a unit in \(A[z^{-1}]\). Hence \(p\) is irreducible in \(A[z^{-1}]\).
Since \(A[z^{-1}]\) is a UFD, \(p\) is prime there. Now suppose \[ p\mid ab \] in \(A\). In \(A[z^{-1}]\), primality gives \[ p\mid a \qquad\text{or}\qquad p\mid b. \] Assume the first. Then for some \(c\in A\) and \(n\geq0\), \[ z^na=pc. \] If \(n>0\), the left side is divisible by \(z\). Since \(z\) is prime and \(z\nmid p\) (otherwise irreducibility of \(p\) would make \(p\) associate to \(z\)), one has \(z\mid c\). Cancel one factor of \(z\) and repeat. After \(n\) steps, \[ a=pc' \] for some \(c'\in A\). Thus \(p\mid a\) already in \(A\). The same argument applies if \(p\mid b\) in the localization.
Therefore every irreducible of \(A\) is prime. Since \(A\) is atomic, \(A\) is a UFD.
4. Part (b): after the first blowup, the only singular point of thestrict transform lies on one exceptional line \(E_1\).
Blow up the origin of \[ X:\quad x^2+y^3+z^5=0. \] The origin is the only singular point of \(X\). In characteristics different from \(2,3,5\) this follows immediately from the three partial derivatives. In characteristic \(2\), the \(y\)- and \(z\)-derivatives force \(y=z=0\), and then the equation forces \(x=0\); in characteristic \(3\), the \(x\)- and \(z\)-derivatives force \(x=z=0\) and then \(y=0\); in characteristic \(5\), the \(x\)- and \(y\)-derivatives force \(x=y=0\) and then \(z=0\).
In the \(x\)-chart, with \[ y=xy_1, \qquad z=xz_1, \] the strict transform is \[ 1+xy_1^3+x^3z_1^5=0, \] so it does not meet the exceptional divisor.
In the \(y\)-chart, with \[ x=yx_1, \qquad z=yz_1, \] the strict transform is \[ x_1^2+y+y^3z_1^5=0. \] Along the exceptional divisor \(y=0\) its derivative with respect to \(y\) is \(1\), so this chart is nonsingular along the exceptional locus.
In the \(z\)-chart, with \[ x=zx_1, \qquad y=zy_1, \] the strict transform is \[ X_1:\quad x_1^2+zy_1^3+z^3=0. \] Its reduced exceptional curve is \[ E_1:\quad z=x_1=0, \] in this affine chart; together with its piece in the \(y\)-chart it is the projective line \(E_1\cong\PP^1\). Along \(E_1\), the derivative with respect to \(z\) is \(y_1^3\), so the only singular point on \(E_1\) is \[ x_1=y_1=z=0. \] Away from the exceptional divisor the blowup is an isomorphism, so this is the only singular point of the first strict transform.
5. After the second blowup, the reduced exceptional locus is\[ E_1\cup E_2, \]and its unique singular point is \(E_1\cap E_2\).
Rename the coordinates in the singular chart of step 4 as \[ x,y,z, \] so the equation is \[ x^2+zy^3+z^3=0. \] Blow up its origin.
The \(x\)-chart again misses the exceptional locus of the strict transform. In the \(z\)-chart the strict transform is \[ x_2^2+z+z^2y_2^3=0, \] which is nonsingular along \(z=0\) because the \(z\)-derivative is \(1\) there.
In the \(y\)-chart, with \[ x=yx_2, \qquad z=yz_2, \] the strict transform is \[ X_2:\quad x_2^2+y^2z_2+yz_2^3=0. \] The new reduced exceptional curve is \[ E_2:\quad y=x_2=0, \] in this affine chart; globally it is a projective line \(E_2\cong\PP^1\). and the strict transform of \(E_1\) is \[ E_1:\quad z_2=x_2=0. \] They meet at the origin. Along \(E_2\), the derivative of the equation with respect to \(y\) is \(z_2^3\) at \(y=0\), so the only singular point there is \(z_2=0\). Thus the unique remaining singular point is precisely \[ E_1\cap E_2. \]
6. The third blowup separates \(E_1\) and \(E_2\), introduces a projectiveline \(E_3\), and leaves exactly two singular points: an ordinary doublepoint at \(E_1\cap E_3\) and a point with local equation\(X^2+zY(Y+z)=0\) at \(E_2\cap E_3\).
Rename the coordinates in step 5 as \(x,y,z\), so \[ X_2:\quad x^2+y^2z+yz^3=0. \] Blow up the origin.
In the \(y\)-chart, \[ x=yX, \qquad z=yZ, \] and the strict transform is \[ X^2+yZ+y^2Z^3=0. \] The only singular point on the exceptional curve \(y=X=0\) is the origin. Its quadratic tangent cone is \[ X^2+yZ=0, \] which is a nonsingular conic in \(\PP^2\). Hence this is an ordinary double point. This point is the intersection of the strict transform of \(E_1\) with the new exceptional line \(E_3\).
In the \(z\)-chart, \[ x=zX, \qquad y=zY, \] and the strict transform is \[ X^2+zY^2+z^2Y = X^2+zY(Y+z) =0. \] Again the only singular point on the exceptional curve \(z=X=0\) is the origin, with exact local equation \[ X^2+zY(Y+z)=0. \] It is the intersection of the strict transform of \(E_2\) with \(E_3\).
Thus after the third blowup the incidence graph of the old exceptional curves is \[ E_1-E_3-E_2, \] with the two displayed intersection points still singular.
7. Blowing up the ordinary double point \(E_1\cap E_3\) once resolvesthat point and inserts a projective line \(E_4\) between \(E_1\) and \(E_3\).
At this point the local equation has quadratic tangent cone \[ X^2+yZ=0, \] a nonsingular conic in \(\PP^2\). The blowup of an ordinary double point with nonsingular projectivized tangent cone is nonsingular, and its exceptional curve is exactly that conic. Hence the fourth blowup resolves this singularity and introduces \[ E_4\cong\PP^1. \]
The strict transforms of \(E_1\) and \(E_3\) meet the conic in the two distinct tangent directions \[ [0:1:0] \qquad\text{and}\qquad [0:0:1]. \] Thus this blowup replaces the edge \(E_1-E_3\) by \[ E_1-E_4-E_3. \]
8. Blowing up the point \(E_2\cap E_3\) with local equation\(X^2+zY(Y+z)=0\) introduces a projective line \(E_5\) carrying exactlythree ordinary double points.
Use the exact local equation from step 6: \[ X^2+zY(Y+z)=0. \] Blow up its origin.
In the \(Y\)-chart, with \[ X=YU, \qquad z=YZ, \] the strict transform is \[ U^2+YZ(1+Z)=0. \] Along the exceptional line \(Y=U=0\), singular points occur exactly when \[ Z(1+Z)=0, \] namely at \[ Z=0 \qquad\text{and}\qquad Z=-1. \]
In the \(z\)-chart, with \[ X=zU, \qquad Y=zV, \] the strict transform is \[ U^2+zV(1+V)=0. \] Along the exceptional line \(z=U=0\), singular points occur at \[ V=0 \qquad\text{and}\qquad V=-1. \] The points \(Z=-1\) and \(V=-1\) are the same point on the overlap. Hence there are exactly three singular points on the new exceptional line \[ E_5\cong\PP^1. \]
Near each of them the factor multiplying the transverse coordinate is a unit times a local parameter, so the quadratic tangent cone is \[ U^2+ab=0. \] Thus all three are ordinary double points.
The point \(Z=0\) is where the strict transform of \(E_3\) meets \(E_5\), the point \(V=0\) is where the strict transform of \(E_2\) meets \(E_5\), and the third point is a free point of \(E_5\).
9. Three further blowups resolve the three ordinary double points on\(E_5\) and produce projective lines \(E_6,E_7,E_8\).
Blow up the ordinary double point \[ E_2\cap E_5. \] As in step 7, this resolves that point and inserts a smooth conic \[ E_6\cong\PP^1 \] between \(E_2\) and \(E_5\).
Next blow up \[ E_3\cap E_5. \] This inserts \[ E_7\cong\PP^1 \] between \(E_3\) and \(E_5\).
Finally blow up the third ordinary double point, which lies only on \(E_5\). This introduces \[ E_8\cong\PP^1 \] meeting \(E_5\) once.
Each ordinary double point is resolved by its single blowup, and the three points are distinct. Therefore after these three blowups the strict transform is nonsingular.
10. The resolution uses exactly eight successive point blowups, andthe reduced inverse image of \(P\) has dual graph \(\mathbf E_8\).
The number of blowups is \[ 3+1+1+3=8: \] three blowups reach the two remaining singular points, one resolves the ordinary double point, one blows up the point with equation \(X^2+zY(Y+z)=0\), and three resolve the resulting ordinary double points.
Tracking incidences through the blowups gives precisely the edges \[ E_1-E_4-E_3-E_7-E_5-E_6-E_2 \] and the additional edge \[ E_5-E_8. \] Thus the seven curves \[ E_1,E_4,E_3,E_7,E_5,E_6,E_2 \] form a chain, and \(E_8\) meets the fifth curve \(E_5\) in that chain. There are no other intersections. Hence the reduced inverse image of \(P\) is a union of eight projective lines with incidence diagram \[ \begin{tikzcd} \circ & \circ & \circ & \circ & \circ & \circ & \circ \\ &&&& \circ \arrow[dash, from=1-1, to=1-2] \arrow[dash, from=1-2, to=1-3] \arrow[dash, from=1-3, to=1-4] \arrow[dash, from=1-4, to=1-5] \arrow[dash, from=1-5, to=1-6] \arrow[dash, from=1-6, to=1-7] \arrow[dash, from=1-5, to=2-5] \end{tikzcd} \] which is the Dynkin diagram \(\mathbf E_8\).
11. Q.E.D.
Steps <1>1–<1>3 prove part (a). Steps <1>4–<1>10 give the eight successive blowups, prove smoothness after the eighth, and identify the exceptional incidence graph required in part (b).