Proof. This follows from the following local version applied repeatedly.

Proof. If A is smooth, then $\Omega ^2_{A/k}$ is free of rank $1$ and the assertion is clear. Suppose A is an RDP. Then it is a hypersurface isolated singularity, and it is well known that for such singularities, the canonical divisor is trivial, and then the former assertion follows. Since an RDP is a canonical singularity, the pullback of the canonical divisor to $\operatorname {\mathrm {Bl}}_{\mathfrak {m}} A$ is also trivial, and hence the latter assertion follows.

Proof. Let $B = \bigoplus _{i \in \mathbb Z/p^e\mathbb Z} B_i$ be the corresponding decomposition.

(1 $\iff $ 3) By the definition of $X^{\mu _p}$ , a closed point $w \in X$ is a (k-valued) point of $X^{\mu _p}$ if and only if $B \to B/\mathfrak {m}_w$ is compatible with the projections $\operatorname {\mathrm {pr}}_i$ to the ith summand $(-)_i$ for all i, where $B/\mathfrak {m}_w$ is equipped with the trivial decomposition.

(2 $\iff $ 3) If (3) holds, then we have $\operatorname {\mathrm {pr}}_i(\mathfrak {m}_w) \subset \mathfrak {m}_w$ for all i, and then each element x of $\mathfrak {m}_w$ is the sum of homogeneous elements $\operatorname {\mathrm {pr}}_i(x) \in \mathfrak {m}_w$ . Conversely, if $\mathfrak {m}_w$ is generated by homogeneous elements, then $\operatorname {\mathrm {pr}}_i(\mathfrak {m}_w) \subset \mathfrak {m}_w$ for all i, which implies (3).

(3 $\iff $ 4) Easy.

Assume $e = 1$ .

(2 $\iff $ 5) Assume $D(\mathfrak {m}_w) \subset \mathfrak {m}_w$ . Take a system of generators $(x_j)$ of $\mathfrak {m}_w$ . For each j, let $x_j = \sum _{i \in \mathbb F_p} x_{j,i}$ be the decomposition of $x_j$ in $B = \bigoplus _{i} B_i$ . Then $D^l(x_j) = \sum _i i^l x_{j,i}$ is also in $\mathfrak {m}_w$ . Since the matrix $(i^l)_{i,l=0}^{p-1}$ is invertible, this implies that $x_{j,i} \in \mathfrak {m}_w$ . Thus $\mathfrak {m}_w$ is generated by eigenvectors. The converse is clear.

(5 $\iff $ 6) This is clear since $\mathcal {O}_{X,w} = \mathfrak {m}_w + k$ and $D \rvert _{k} = 0$ .

(5 $\iff $ 7) Take coordinates $x_1, \ldots , x_n$ at a point w and write $D = \sum _j f_j \cdot (\partial / \partial x_j)$ . Then both of the conditions are equivalent to $(f_j) \subset \mathfrak {m}_w$ .

We show the final assertion assuming (2). If the maximal ideal $\mathfrak {m}$ is generated by homogeneous elements $x_j \in B_{i_j}$ , then, for each j, we can extend the action on the affine piece $\operatorname {\mathrm {Spec}} B[x_h/x_j]_h$ of $\operatorname {\mathrm {Bl}}_w X$ by declaring $x_h/x_j$ to be homogeneous of weight $i_h - i_j$ .

Proof. Recall that a subset of $\mathfrak {m}$ generates $\mathfrak {m}$ if and only if it generates $\mathfrak {m}/\mathfrak {m}^2$ .

Take elements $x^{\prime }_1, \dots , x^{\prime }_m$ generating $\mathfrak {m}$ , and let $x^{\prime }_j = \sum _{i \in \mathbb F_p} x^{\prime }_{j,i}$ be the decompositions to eigenvectors. By assumption, there exists a pair $(j,i)$ with $x^{\prime }_{j,i} \not \in \mathfrak {m}$ . We take such $j_0,i_0$ , and we may assume $i_0 \neq 0$ . We may assume $x^{\prime }_{j_0,i_0} - 1 \in \mathfrak {m}$ . Then $y = x^{\prime }_{j_0,i_0} - 1 $ satisfies $y \in \mathfrak {m}$ and $D(y) = i_0 (y + 1)$ . We have $y \not \in \mathfrak {m}^2$ , since $D(\mathfrak {m}^2) \subset \mathfrak {m}$ . By replacing y with $(y+1)^q - 1$ for an integer q with $q i_0 \equiv 1 \pmod p$ , we may assume $i_0 = 1$ . For each j, let $x_j = \sum _i (y+1)^{-i} x^{\prime }_{j,i}$ . Then we have $D(x_j) = 0$ and, since $x_j \equiv x^{\prime }_j \pmod {(y)}$ , the elements $x_j,y$ generate $\mathfrak {m}/\mathfrak {m}^2$ and hence generate $\mathfrak {m}$ . We can omit one of the $x_j$ ’s and then the remaining elements satisfies the required conditions (after renumbering).

To show the latter assertion, it suffices to show that D does not extend to $B' := B[x_j/y]_j$ . If it extends, then we have $D(x_j/y) = -x_j(y+1)/y^2 \in B'$ , hence $x_j/y^2 \in B'$ , and then on $\operatorname {\mathrm {Spec}} B'$ we have that $y = 0$ implies $x_j/y = 0$ , which is impossible since $\dim B' \geq 2$ .

Proof. The isomorphism for $n = 0$ is clear. It suffices to construct the isomorphism for $n = 1$ that is compatible with multiplication with $n = 0$ forms and with restriction to open subschemes.

Take a closed point $w \in X$ . Let $\varepsilon = 1$ (resp. $\varepsilon = 0$ ) if $w \not \in \Delta $ (resp. $w \in \Delta $ ). By Lemma 2.9 (resp. by [Reference OgusRS, Th. 1]), there are coordinates $x_1, \dots , x_m$ on a neighborhood of w with $D(x_j) = 0$ for $j < m$ and $D(x_m) = a (\varepsilon + x_m)$ for some $a \in \mathbb F_p^*$ . We define

$$ \begin{align*} \phi \colon (\pi_* (\Omega^m_{X/k}(\Delta)))_0 &\to \Omega^m_{X^D/k}(\pi_*(\Delta)) \\ f \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m) &\mapsto f \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m^p) \end{align*} $$

for f of weight $0$ (note that $dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log (\varepsilon + x_m)$ is a local generator of the left-hand side). We show that then $\phi $ sends

$$\begin{align*}f_0 \cdot df_1 \wedge \dots \wedge df_{m-1} \wedge d\log(f_m) \mapsto f_0 \cdot df_1 \wedge \dots \wedge df_{m-1} \wedge d\log(f_m^p) \end{align*}$$

for any $f_0, \dots , f_{m-1}$ and $f_m$ as in the statement. This implies that $\phi $ does not depend on the choice of the coordinates and hence that $\phi $ induces a well-defined morphism of sheaves. Then since $dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log (\varepsilon + x_m)$ (resp. $dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log (\varepsilon + x_m^p)$ ) is a local generator of $(\Omega ^m_{X/k}(\Delta ))_0$ (resp. $\Omega ^m_{X^D/k}(\pi _*(\Delta ))$ ), it follows that $\phi $ is an isomorphism and $\phi ^{\otimes n}$ are well-defined isomorphisms.

We may pass to the completion, so consider $f_h \in k[[x_1, \dots , x_m]]$ . By the assumption on the weight, we have $f_h \in k[[x_1, \dots , x_{m-1}, x_m^p]]$ for $h < m$ and $f_m \in (\varepsilon + x_m)^b k[[x_1, \dots , x_{m-1}, x_m^p]]$ for some $0 \leq b < p$ . Then we have $\partial f_h / \partial x_m = 0$ for $h < m$ and $\partial f_m / \partial x_m = b f_m/(\varepsilon + x_m)$ . Hence, we have

$$ \begin{align*} & f_0 \cdot df_1 \wedge \dots \wedge df_{m-1} \wedge d\log(f_m) \\ &= f_0 ((\varepsilon + x_m)/f_m) \det (\partial f_h / \partial x_j)_{1 \leq h,j \leq m} \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m) \\ &= b f_0 \det (\partial f_h / \partial x_j)_{1 \leq h,j \leq m-1} \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m) \\ &\stackrel{\phi}{\mapsto} b f_0 \det (\partial f_h / \partial x_j)_{1 \leq h,j \leq m-1} \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m^p). \end{align*} $$

On the other hand, in the invariant subalgebra $k[[x_1, \dots , x_{m-1}, x_m^p]]$ , we have $\partial f_m^p / \partial x_j = 0$ for $j < m$ and $\partial f_m^p / \partial x_m^p = b f_m^p/(\varepsilon + x_m^p)$ . Hence, we have

$$ \begin{align*} & f_0 \cdot df_1 \wedge \dots \wedge df_{m-1} \wedge d\log(f_m^p) \\ &= \dots = b f_0 \det (\partial f_h / \partial x_j)_{1 \leq h,j \leq m-1} \cdot dx_1 \wedge \dots \wedge dx_{m-1} \wedge d\log(\varepsilon + x_m^p). \end{align*} $$

The assertion follows.

Proof. Suppose C is integral. Then $C \cong \mathbb P^1$ and the result is well known (indeed, $T_{\mathbb P^1} \cong \mathcal {O}_{\mathbb P^1}(2)$ ).

Suppose C is reducible. We may assume $C = (xy = 0)$ . Let $U = \operatorname {\mathrm {Spec}} k[x,y] / (xy)$ . Then $T_U = ( x \frac {d}{dx} - y \frac {d}{dy}) \cdot \mathcal {O}_U$ and hence the origin belongs to $\operatorname {\mathrm {Fix}}(D)$ .

Suppose C is non-reduced. We may assume $C = (X_3^2 = 0)$ . If $p \neq 2$ , then D induces a derivation $D_{\mathrm {red}}$ on $C_{\mathrm {red}} \cong \mathbb P^1$ , and we have $\operatorname {\mathrm {Fix}}(D) \approx \operatorname {\mathrm {Fix}}(D_{\mathrm {red}}) \neq \emptyset $ . Suppose $p = 2$ . It is easy to see that $H^0(C, T_C) \stackrel {\sim }{\to } H^0(C, T_{\mathbb P^2} \rvert _{C}) \stackrel {\sim }{\leftarrow } H^0(\mathbb P^2, T_{\mathbb P^2})$ . Hence, there exist $f_1, f_2, f_3 \in H^0(\mathbb P^2, \mathcal {O}(1)) = \bigoplus _{i=1}^3 k X_i$ such that $D\bigl (\frac {X_i}{X_j}\bigr ) = \frac {f_i}{X_j} - \frac {X_i f_j}{X_j^2}$ . If $f_3 \in k X_3$ , then D induces a derivation $D_{\mathrm {red}}$ on $C_{\mathrm {red}} \cong \mathbb P^1$ , and we conclude as above. Suppose $f_3 \not \in k X_3$ . By a coordinate change, we may assume $f_3 - X_2 \in k X_3$ . Letting $x_i = X_i/X_1$ ( $i = 2,3$ ) and restricting to $\operatorname {\mathrm {Spec}} k[x_2,x_3]/(x_3^2) = (X_1 \neq 0) \subset C$ , we have $D(x_3) - x_2 \in (x_3)$ , in particular $D(x_3) \in \mathfrak {m} := (x_2, x_3)$ . If $D(x_2) \in \mathfrak {m}$ , then the origin is a fixed point. Suppose $D(x_2) \not \in \mathfrak {m}$ and $D^2 = h D$ . Then $h = D^2(x_2) / D(x_2) \in \mathcal {O}_{\mathfrak {m}}$ , and hence $D(x_2) \equiv D^2(x_3) = h D(x_3) \equiv 0 \pmod {\mathfrak {m}}$ , contradiction.

Proof. The action extends to the blowup by Proposition 2.8. Let $D'$ be the induced derivation on $\operatorname {\mathrm {Bl}}_w X$ . Let $C \subset \operatorname {\mathrm {Bl}}_w X$ be the (possibly non-reduced) exceptional divisor, which is a quadratic curve in $\mathbb P^2$ since w is an RDP. Since $D'(\mathcal I_C) \subset \mathcal I_C$ , $D'$ induces a derivation $D^{\prime }_{C}$ (of multiplicative type) on C. By Lemma 2.11, $D^{\prime }_C$ has at least one fixed point, and that point is also a fixed point of $D'$ .

Proof. The first assertion follows from Proposition 2.8.

(1) $H^0(\operatorname {\mathrm {Spec}} B \setminus \{\mathfrak {m}\}, \Omega ^2_{B/k})$ is generated by $dx \wedge dy$ , which is of weight $a+b$ .

(2) Take an element $H \in k[[x,y,z]]$ such that $\hat {B} = k[[x,y,z]] / (H)$ , and let $H = \sum _{i \in \mathbb Z/n\mathbb Z} H_i$ be the decomposition with respect to the $\mu _n$ -action. Since $H = 0$ in $\hat {B}$ , we have $H_i = 0$ in $\hat {B}$ , and hence there are $f_i \in k[[x,y,z]]$ such that $H_i = f_i H$ . Since $\sum f_i = 1$ , there exists $d \in \mathbb Z/n\mathbb Z$ with $f_d \in k[[x,y,z]]^*$ . We can take $F = H_d$ , which is of weight d.

Then $H^0(\operatorname {\mathrm {Spec}} \hat {B} \setminus \{\mathfrak {m}\}, \Omega ^2_{\hat {B}/k})$ is generated by $\omega = F_x^{-1} dy \wedge dz = F_y^{-1} dz \wedge dx = F_z^{-1} dx \wedge dy$ (this means that the restriction of $\omega $ to the open subscheme $\operatorname {\mathrm {Spec}} \hat {B}[F_x^{-1}]$ is equal to $F_x^{-1} dy \wedge dz$ , and so on), and we have $\operatorname {\mathrm {wt}}(\omega ) = a + b + c - d$ since $\operatorname {\mathrm {wt}}(F_x^{-1}) = -(d - a)$ and $\operatorname {\mathrm {wt}}(dy \wedge dz) = b + c$ , and so on.

Proof of Theorem 4.6(3)

By Proposition 2.10, we have an isomorphism

$$\begin{align*}(H^0(\operatorname{\mathrm{Spec}} \mathcal{O}_{X,w} \setminus \{w\}, \Omega^2))_0 \cong H^0(\operatorname{\mathrm{Spec}} \mathcal{O}_{Y,\pi(w)} \setminus \{\pi(w)\}, \Omega^2) \end{align*}$$

preserving the zero loci of $2$ -forms. If $\pi (w)$ is either a smooth point or an RDP, then the right-hand side has a nonvanishing $2$ -form and hence there is a nonvanishing form $\omega $ on $\operatorname {\mathrm {Spec}} \mathcal {O}_{X,w} \setminus \{w\}$ of weight $0$ . Being nonvanishing, $\omega $ is a generator of $H^0(\operatorname {\mathrm {Spec}} \mathcal {O}_{X,w} \setminus \{w\}, \Omega ^2)$ . However, this contradicts the non-symplecticness assumption.

Proof of Theorem 4.6(1) and Proposition 4.7

If w is a smooth point, then taking coordinates $x,y$ as in Lemma 2.9 (i.e., $D(x) = 0$ and $D(y) = 1+y$ ), we have $\hat {\mathcal {O}}_{Y,\pi (w)} \cong k[[x,y^p]]$ , and hence $\mathcal {O}_{Y,\pi (w)}$ is smooth.

Assume w is an RDP. By Lemma 2.9, we have coordinates $x,y,z$ satisfying $D(x) = D(y) = 0$ and $D(z) \neq 0$ . We have $\hat {\mathcal {O}}_{X,w} \cong k[[x,y,z]] / (F)$ for some $F \in k[[x,y,z]]$ such that $D(F) \in (F)$ , and we may assume $F \in k[[x,y,z^p]]$ . We show that, after replacing F with a multiple by a unit, and after a coordinate change of $k[[x,y,z]]$ that preserves the subring $k[[x,y,z^p]]$ , F coincides with one in Table 3. (Such coordinate changes are given by $x',y',z' \in \mathfrak {m}$ that are linearly independent in $\mathfrak {m}/\mathfrak {m}^2$ and satisfy $x',y' \in \mathfrak {m} \cap k[[x,y,z^p]]$ .) A similar classification is given in [Reference Ekedahl, Hyland and Shepherd-BarronEH+, Prop. 3.8], but they missed the case of $E_7^0$ in characteristic $3$ .

Assume the classification for the moment. Then, in each case, we observe that $\pi (w)$ is either a smooth point or an RDP, and it is straightforward to check that $X \times _Y \tilde {Y}$ is an RDP surface crepant over X. (In Table 3, the entries of the singularities of $X \times _Y \tilde {Y}$ are omitted if Y is already smooth.) For example, consider $X = \operatorname {\mathrm {Spec}} k[x,y,z]/(F)$ , $F = xy + z^{mp}$ with $m \geq 2$ . Then $X' := X \times _Y \operatorname {\mathrm {Bl}}_{\pi (w)} Y$ is covered by three affine pieces

$$ \begin{align*} X^{\prime}_1 &= \operatorname{\mathrm{Spec}} k[x, y_1, v_1, z] / (y_1 + x^{m-2} v_1^{m}, x v_1 - z^p), & y_1 &= y/x, & v_1 &= z^p/x, \\ X^{\prime}_2 &= \operatorname{\mathrm{Spec}} k[x_2, y, v_2, z] / (x_2 + y^{m-2} v_2^{m}, y v_2 - z^p), & x_2 &= x/y, & v_2 &= z^p/y, \\ X^{\prime}_3 &= \operatorname{\mathrm{Spec}} k[x_3, y_3, z] / (x_3 y_3 + z^{(m-2)p}), & x_3 &= x/z^p, & y_3 &= y/z^p. \end{align*} $$

One observes that $\operatorname {\mathrm {Sing}}(X')$ consists of two RDPs of type $A_{p-1}$ at the origins of $X^{\prime }_1$ and $X^{\prime }_2$ and, if $m \geq 3$ , one RDP of type $A_{(m-2)p-1}$ at the origin of $X^{\prime }_3$ . Repeating this, we observe that $X \times _Y \tilde {Y}$ has $m A_{p-1}$ .

Now, we show the classification. We say that F has a monomial if the coefficient of that monomial is nonzero. We also write $F = \sum _{h,i,j} a_{hij} x^h y^i z^j$ .

First, assume $p> 2$ . We may assume that the degree $2$ part $F_2$ is either $xy$ or $x^2$ . Assume $F_2 = xy$ . We may assume that F has no $xz^{j}$ and $yz^{j}$ . F must have $z^{j}$ , $j = mp$ , and then it is $A_{mp-1}$ . Then, by replacing x with $x + a_{0ij} y^{i-1} z^j$ and y with $y + a_{h0j} x^{h-1} z^j$ , and so on, we may assume that F has no $y^i z^j$ with $i> 0$ and no $x^h z^j$ with $h> 0$ . Thus, $F = u_1 xy + u_2 z^{mp}$ for some units $u_1, u_2$ , and then by replacing $x,y,F$ by suitable multiples, we obtain $F = xy + z^{mp}$ .

Assume $p> 3$ and $F_2 = x^2$ . We may assume that the degree $3$ part $F_3$ is $y^3$ . If $p \geq 7$ , it cannot be an RDP. If $p = 5$ , then F must have $z^5$ , and then it is $E_8^0$ . We have $F = u_1 x^2 + u_2 y^3 + u_3 z^5$ , and then by replacing $x,y,F$ by suitable multiples, we obtain $F = x^2 + y^3 + z^5$ . (For example, we let $F = u_3 F'$ , $x = (u_3 u_1^{-1})^{1/2} x'$ , $y = (u_3 u_2^{-1})^{1/3} y'$ . Note that we can take nth roots of units provided $p \nmid n$ .)

Assume $p = 3$ and $F_2 = x^2$ . We may assume $F_3 = y^3$ or $F_3 = z^3$ . If $F_3 = z^3$ , then F must have $y^4$ or $y^5$ , and then it is $E_6^0$ or $E_8^0$ . We may assume $a_{130} = a_{140} = 0$ by replacing x with $x + (1/2)(a_{130} y^3 + a_{140} y^4)$ , and then we transform F as above. If $F_3 = y^3$ , then F must have $yz^3$ and then it is $E_7^0$ . We eliminate $a_{1ij}$ as above, then we have $F = u_1 x^2 + u_2 y^3 + u_3 yz^3 + z^6 g(z^3)$ for some power series $g \in k[[z^3]]$ . We may assume $u_i \equiv 1 \pmod {\mathfrak {m}}$ . We eliminate g by replacing y with $y + z^3 g$ , and then we transform F as above.

Now, consider $p = 2$ . We may assume $F_2$ is one of $xy + z^2$ (if irreducible), $xy$ (if reducible but not a square), $z^2$ , or $x^2$ (square, of a linear factor containing z or not). If $F_2 = xy + z^2$ or $F_2 = xy$ , then as above, it is $A_{mp-1}$ and F becomes $xy + z^{mp}$ .

Assume $p = 2$ and $F_2 = x^2$ . If $F_3$ has $yz^2$ , then F must have $xy^m$ and then it is $D_{2m+1}^0$ . We obtain $F = u_1 x^2 + u_2 y z^2 + u_3 x y^m + z^4 g(z) + f(y) + y^{2m} g'(y)$ , where $f(y) = f_0(y)^2 + y f_1(y)^2$ is a polynomial of degree $< 2m$ , $g(z) \in k[[z]]$ , and $g'(y) \in k[[y]]$ . We may assume $u_i \equiv 1 \pmod {\mathfrak {m}}$ . We eliminate f by replacing x with $x + f_0(y)$ and z with $z + f_1(y)$ and so on. Then we eliminate g and $g'$ by replacing y and x suitably, and take multiples by units as above.

If $F_3$ has no $yz^2$ , then F must have $y^3$ and $xz^2$ and then it is $E_6^0$ . We obtain $F = u_1 x^2 + u_2 y^3 + u_3 x z^2 + a y^2 z^2 + z^4 g$ , $g = g_0(z^2) + y g_1(z^2) + y^2 g_2(z^2)$ . We may assume $u_i \equiv 1 \pmod {\mathfrak {m}}$ . We eliminate a and g by replacing y and x suitably, and then we transform F as above.

Assume $p = 2$ and $F_2 = z^2$ . Let $\overline {F_3} = (F_3 \bmod (z)) \in k[[x,y]]$ . If $\overline {F_3}$ has three distinct roots, then we may assume $\overline {F_3} = x^3 + y^3$ and then it is $D_4^0$ . We can transform F to $z^2 + x^3 + y^3$ as above, and then to $z^2 + x^2 y + x y^2$ . If $\overline {F_3}$ has two distinct roots, then we may assume $\overline {F_3} = x^2y$ and F must have $xy^m$ and then it is $D_{2m}^0$ . We obtain $F = u_1 z^2 + u_2 x^2 y + u_3 x y^m + g(x) + f(y) + y^{2m-1} g'(y)$ , where $f(y) = f_0(y)^2 + y f_1(y)^2$ is a polynomial of degree $< 2m-1$ and $g \in k[[x]]$ and $g' \in k[[y]]$ . We argue as in the case of $D_{2m+1}^0$ . If $\overline {F_3}$ has one (triple) root, then we may assume $\overline {F_3} = x^3$ and F must have $xy^3$ or $y^5$ and then it is $E_7^0$ or $E_8^0$ . We transform F as above.

Proof. (1) By Lemma 4.5(1), we have $D = a x \cdot (\partial / \partial x) - a y \cdot (\partial / \partial y)$ with $a \in \mathbb F_p$ for some coordinates $x,y$ , and $a \neq 0$ since D is nontrivial. Hence, w is an isolated fixed point of D. We observe that $a,-a$ are the eigenvalues of the action on the cotangent space. We have $\hat {\mathcal {O}}_{X,w}^D = k[[x^p, xy, y^p]]$ , and it is an RDP of type $A_{p-1}$ .

(2) By Remark 4.3 and assertion (1), w is an isolated fixed point. By Proposition 2.8, D uniquely extends to $D'$ on $X'$ . Let $\omega $ be a generator of $H^0(\operatorname {\mathrm {Spec}} \mathcal {O}_{X,w} \setminus \{w\}, \Omega ^2)$ with $D(\omega ) = 0$ . Since w is an RDP, $X'$ is again an RDP surface, and it follows from Proposition 2.3 that $\omega $ extends to $\omega '$ on $(X')^{\mathrm {sm}}$ , which generates $H^0(\operatorname {\mathrm {Spec}} \mathcal {O}_{X',w'} \setminus \{w'\}, \Omega ^2)$ at any closed point $w' \in X'$ above w, and that $D'(\omega ') = 0$ . Hence, $D'$ is symplectic at every fixed point above w. Since as above such fixed points are isolated, $Y'$ is smooth outside finitely many isolated points. Applying Proposition 2.10 to $\omega $ on $X \setminus \{w\}$ and $\omega '$ on $X' \setminus (\operatorname {\mathrm {Sing}}(X') \cup \operatorname {\mathrm {Fix}}(D'))$ , we obtain $2$ -forms $\psi $ on $Y \setminus \{\pi (w)\}$ and $\psi '$ on $Y' \setminus \pi ((\operatorname {\mathrm {Sing}}(X') \cup \operatorname {\mathrm {Fix}}(D')))$ , which are nonvanishing. Comparing $\psi $ and $\psi '$ , we observe that g is crepant.

Proof of Theorem 4.6(2)

By Remark 4.3 and Lemma 4.8(1), w is an isolated fixed point. By shrinking X, we may assume that D has no fixed point except w.

We construct a finite sequence $(X_j,D_j)_{0 \leq j \leq n}$ ( $n \geq 0$ ) of RDP surfaces $X_j$ and derivations $D_j$ on $X_j$ of multiplicative type that is symplectic at each fixed point. Let $(X_0,D_0) = (X,D)$ . If $X_j$ has no fixed RDP, then we terminate the sequence at $n = j$ . If $X_j$ has at least one fixed RDP, let $X_{j+1}$ be the blowup of $X_j$ at the fixed RDPs and $D_{j+1}$ the extension of $D_j$ to $X_{j+1}$ . Since any RDP becomes smooth after a finite number of blowups at RDPs, this sequence terminates at some $n \geq 0$ . By Lemma 4.8(2), $D_{j+1}$ on $X_{j+1}$ is symplectic at each fixed point, and $(X_{j+1})^{D_{j+1}} \to (X_j)^{D_j}$ is crepant. By Theorem 4.6(1) and Lemma 4.8(1), $Y_n = (X_n)^{D_n}$ has canonical singularity (i.e., has no singularity other than RDPs), and since $Y_n \to Y = X^D$ is crepant, also Y has canonical singularity. If $n> 0$ , then $\pi (w)$ is not a smooth point since $Y_n \to Y$ is a crepant morphism non-isomorphic at that point, and if $n = 0$ , then $\pi (w)$ is not a smooth point by Lemma 4.8(1). Hence, in either case, $\pi (w)$ is an RDP.

Proof of Proposition 4.9

We consider tuples $(a,b,c,F)$ with $a,b,c \in \mathbb F_p$ , not all $0$ , and $F \in k[[x,y,z]]$ such that $F = 0$ defines an RDP and only monomials of weight $a+b+c$ ( $\in \mathbb F_p$ ) appear in F, where $x,y,z$ have respective weights $a,b,c$ . By Lemma 4.5(2), it suffices to consider $k[[x,y,z]]/(F)$ of this form. We show that there exist a $\mu _p$ -equivariant isomorphism $k[[x,y,z]]/(F) \cong k[[x',y',z']]/(F')$ with $F'$ in Table 4 and $\operatorname {\mathrm {wt}}(x',y',z') = (1,-1,0)$ up to $\operatorname {\mathrm {Aut}}(\mu _p) = \mathbb F_p^*$ (which amounts to replacing $(a,b,c)$ with $(ta,tb,tc)$ for some $t \in \mathbb F_p^*$ ).

We write $F = \sum _{h,i,j} a_{hij} x^h y^i z^j$ , and we say that a polynomial or a formal power series has a monomial if its coefficient is nonzero.

First, assume the degree $2$ part $F_2$ is a non-square. Then we may assume that $F_2$ contains a non-square monomial, say $xy$ . (Indeed, if this is not the case, then $p \neq 2$ and $F_2$ contains at least two square monomials, say $x^2$ and $y^2$ , then x and y has the same weight, and then after a linear coordinate change, we may assume $F_2$ contains $xy$ .) Then we have $c = 0$ . If $a + b \not \in \{0,a,b\}$ , then $F \in (x,y)^2$ , which implies that $F = 0$ is not an RDP. If $a + b = a \neq 0$ , then $F \in (x)$ , again not an RDP. The same if $a + b = b \neq 0$ . So we have $a + b = 0$ and hence $F \in k[[x^p,xy,y^p,z]]$ . Since F cannot belong to $(x,y)$ , there exists an integer m such that F has the monomial $z^{m}$ . Let m be the smallest such integer. We have $F = u_1 z^m + u_2 xy + g_1(x^p) + g_2(y^p)$ for some units $u_1,u_2 \in k[[x^p, x y, y^p, z]]^*$ and power series $g_1, g_2$ . We may assume $u_1,u_2 \equiv 1 \pmod {\mathfrak {m}}$ . We eliminate $g_1, g_2$ by replacing x with $x + g_2/y$ and y with $y + g_1/x$ (and repeating this), and we obtain $F = u_1 z^m + u_2 xy$ . By replacing $x,y,z,F$ with suitable multiples, we obtain $F = z^m + xy$ .

Next, assume $p \geq 3$ and $F_2$ is square. We may assume $F_2 = z^2$ . We may assume $F_3 \not \equiv 0 \pmod {z}$ . If $F_3$ has $x^2y$ , then by $2c = 2a + b = a + b + c$ , we have $b = 0$ and $a = c$ , and hence $F \in (x,z)^2$ , which is absurd. Hence, we may assume $F_3$ has $y^3$ . By $2c = 3b = a+b+c$ , we have $(a,b,c) = (a,2a,3a)$ . If F does not have $x^3$ , then $F \in (z^2, x^3z, xyz, x^6, x^4y, x^2y^2, y^3)$ , and $F = 0$ cannot define an RDP. Hence, F has $x^3$ , hence $p = 3$ , and then $F \in k[[x^3,xy,y^3,z]]$ . We may assume that F does not have $xyz$ . To define an RDP, F must have one of $x^2y^2, x^4y, xy^4$ .

If it has $x^2y^2$ , then it is $E_6^1$ . We can eliminate $x^h y^i z$ , and we have

$$ \begin{align*} F &= z^2 u_1 + x^3 + y^3 + x^2 y^2 u_2 + \sum_{(h,i,j) \in S_1} a_{hij} x^h y^i z^j + \sum_{(h,i) \in S_2} b_{hi} x^h y^i + \sum_{(h,i) \in S_3} c_{hi} x^h y^i, \\ &\qquad S_1 = \{(4,1,0), (1,4,0)\}, \qquad S_2 = \{(6,0), (7,1)\}, \qquad S_3 = \{(0,6), (1,7)\}, \end{align*} $$

where $a_{hij} \in k$ , $b_{hi} \in k[[x^3]]$ , $c_{hi} \in k[[y^3]]$ , and $u_1, u_2 \in k[[x^3, x y, y^3, z]]^*$ . By replacing x with $x + t y^2$ and y with $y + t' x^2$ , we eliminate $a_{410}$ and $a_{140}$ . Then, by replacing F with $(1 + x^3 b_{60} + x^4 y b_{71} + y^3 c_{06} + x y^4 c_{17}) F$ , we eliminate all $b_{hi}$ and $c_{hi}$ . Finally, we replace $x,y,z,F$ with suitable multiples and achieve $u_1 = u_2 = 1$ . (For example, we let $F = u_2^{-3} F'$ , $x = u_2^{-1} x'$ , $y = u_2^{-1} y'$ , and $z = (u_1 u_2^3)^{-1/2} z'$ .)

If it does not have $x^2 y^2$ but has $x^4 y$ or $x y^4$ , then it is $E_8^1$ . By replacing F with a unit multiple, we may assume that it has $x^4 y$ and does not have $x y^4$ . We can eliminate $x^h y^i z$ , and we have

$$ \begin{align*} &F = z^2 u_1 + x^3 + y^3 + x^4 y u_2 + \sum_{(h,i,j) \in S_1} a_{hij} x^h y^i z^j + \sum_{(h,i) \in S_2} b_{hi} x^h y^i + \sum_{(h,i) \in S_3} c_{hi} x^h y^i, \\ S_1 &= \{(6,0,0), (3,3,0), (0,6,0)\}, \qquad S_2 = \{(9,0)\}, \qquad S_3 = \{(0,9), (1,7), (2,5), (3,6)\}, \end{align*} $$

with $a_{hij}$ , $b_{hi}$ , $c_{hi}$ , and $u_1,u_2$ as in the previous case. By replacing y with $y + t x^2$ , we eliminate $a_{600}$ . By replacing x with $x + t' y^2$ and F with $(1 + t" y^3) F$ , we eliminate $a_{330}$ and $a_{060}$ . By replacing F with $(1 + x^2 y^2 c_{25} + x y^4 c_{17} + y^6 c_{09}) F$ , then with $(1 + x^6 b_{90} + x^3 y^3 c_{36}) F$ , we eliminate all $b_{hi}$ and $c_{hi}$ . We replace $x,y,z,F$ with suitable multiples and achieve $u_1 = u_2 = 1$ .

Hereafter, assume $p = 2$ and that $F_2$ is a square. If $\sqrt {F_2}$ is homogeneous, then we may assume $F_2 = x^2$ , we may assume $F_3$ has $y^2 z$ or $z^3$ , and then we have $(a,b,c) = (1,1,0)$ , and hence $F \in k[[x^2,xy,y^2,z]]$ . If $\sqrt {F_2}$ is not homogeneous, then we may assume $F_2 = x^2 + z^2$ and $a = 1$ and $c = 0$ , and then again we have $(a,b,c) = (1,1,0)$ , and hence $F \in k[[x^2,xy,y^2,z]]$ , and $F_3$ has $x y z$ , $y^2 z$ , $z^3$ , or $x^2 z$ .

Assume ( $F_2$ is $x^2$ or $x^2 + z^2$ and) $F_3$ contains $y^2 z$ . Furthermore, since $F \notin (x,y)^2$ , we see that F has $z^l$ ( $l \geq 2$ ). We have $F \equiv x^2 + y^2 z + z^l \pmod {(x^4, x^3 y, x^2 y^2, x y^3, y^4, x^2 z, x y z, y^2 z^2, z^{l+1})}$ . Write $F = F_0(x^2, y^2, z) + xy F_1(x^2, y^2, z)$ . Then there exist unique $f, g \in k[[z]]$ such that $F_0 \in (x^2 - f(z)^2, y^2 - g(z)^2)$ , and they satisfy $l = \min \{2 \operatorname {\mathrm {ord}}_z(f), 2 \operatorname {\mathrm {ord}}_z(g) + 1\}$ . If l is even, then, by replacing y with $y - x g/f$ , we may assume $g = 0$ . If l is odd, then, by replacing x with $x - y f/g$ , we may assume $f = 0$ . We eliminate $a_{hij}$ with $h \geq 2$ , $(h,i,j) \neq (2,0,0)$ , by replacing F with $(1 + a_{hij} x^{h-2} y^i z^j) F$ , and $a_{hij}$ with $i \geq 2$ , $(h,i,j) \neq (0,2,0), (0,2,1)$ , by replacing z with $z + a_{hij} x^h y^{i-2} z^j$ . We obtain $F = x^2 + y^2 z + z^l u(z) + x y e(z)$ , where $e(z) \in k[[z]]$ and $u(z) \in k[[z]]^*$ . We have $e(z) \neq 0$ , since if $e(z) = 0$ , then $F = F_0 \in ((x - f(z))^2, (y - g(z))^2)$ , which is absurd. Write $e(z) = z^k v(z)$ , $k \geq 1$ , and $v(z) \in k[[z]]^*$ . It is $D_{2k+l}^{\lfloor l/2 \rfloor }$ . If l is even, then, since $F_0 \in (x^2 - f(z)^2, y^2 - g(z)^2)$ and $g = 0$ , we have $z^l u(z) = f(z)^2$ and hence $u(z)$ is a square, and then by replacing x with $u(z)^{1/2} x$ and by replacing F with a unit multiple, we obtain $F = x^2 + y^2 z u'(z) + z^l + x y e(z)$ for some $u'(z) \in k[[z]]^*$ . Similarly, if l is odd, then, since $f = 0$ , we have $z^l u(z) = z g(z)^2$ and hence $u(z)$ is a square, and then (by replacing y) we obtain $F = x^2 u'(z) + y^2 z + z^l + x y e(z)$ . By replacing $x,y,z,F$ with unit multiples, we can achieve $u' = v = 1$ .

Assume $F_2 = x^2$ and $F_3$ has $z^3$ but no $y^2z$ . To define an RDP, F must have $y^4$ and must have $xyz$ or $xy^3$ . If F has $xyz$ , then it is $E_7^3$ . We have

$$ \begin{align*} &\qquad \quad F = x^2 + y^4 + z^3 u_1+ x y z u_2 + \sum_{(h,i,j) \in S_1} a_{hij} x^h y^i z^j \\ &\qquad \quad + \sum_{(h,i) \in S_2} b_{hi} x^h y^i + \sum_{(h,i) \in S_3} c_{hi} x^h y^i + \sum_{(h,i) \in S_4} d_{hi} x^h y^i, \\ S_1 &= \{(3,1,0), (1,3,0), (1,5,0), (0,2,2), (2,0,1), (2,0,2), (0,4,1), (0,4,2)\}, \\ &\quad S_2 = \{(4,0), (5,1)\}, \qquad S_3 = \{(0,6), (1,7)\}, \qquad S_4 = \{(2,2)\}, \end{align*} $$

where $a_{hij} \in k$ , $b_{hi} = \sum _{j = 0}^{2} b_{hij} z^j$ with $b_{hij} \in k[[x^2]]$ , $c_{hi} = \sum _{j = 0}^{2} c_{hij} z^j$ with $c_{hij} \in k[[y^2]]$ , $d_{hi} \in k[[x^2, x y, y^2]]$ , and $u_1,u_2 \in k[[x^2, x y, y^2, z]]^*$ . We may assume $u_1, u_2 \equiv 1 \pmod {\mathfrak {m}}$ . We replace z with $z + a_{310} x^2 + a_{130} y^2 + a_{150} y^4$ , x with $x + t y z$ (which eliminates $a_{022}$ ), y with $y + a_{201} x + a_{202} x z$ , and x with $x + a_{041} y^3 + a_{042} y^3 z$ , and thus eliminate all $a_{hij}$ ( $(h,i,j) \in S_1$ ). We replace F with $(1 + x^2 b_{40} + y^2 c_{06}) F$ , F with $(1 + x^3 y b_{51} + x y^3 c_{17}) F$ , and z with $z + d_{22} x y$ , and thus eliminate all $b_{hi}$ , $c_{hi}$ , and $d_{hi}$ . We replace $x,y,z,F$ with suitable multiples and achieve $u_1= u_2 = 1$ .

Next, if F does not have $xyz$ but has $x y^3$ , then it is $E_8^3$ . We have

$$ \begin{align*} &F = x^2 + y^4 + z^3 u_1 + x y^3 u_2 + \sum_{(h,i,j) \in S_1} a_{hij} x^h y^i z^j + \sum_{(h,i) \in S_2} b_{hi} x^h y^i + \sum_{(h,i) \in S_3} c_{hi} x^h y^i, \\ &\quad S_1 = \{(2,0,1), (2,0,2), (0,4,1), (0,2,2), (1,1,2), (0,6,0), (0,4,2), (0,6,1), (0,6,2), \\ &\qquad (2,2,0), (2,2,1), (2,2,2)\}, \qquad S_2 = \{(4,0), (3,1), (4,2))\}, \qquad S_3 = \{(0,8)\}, \end{align*} $$

with $a_{hij}$ , $b_{hi}$ , $c_{hi}$ , and $u_1,u_2$ as in the previous case. We may assume $u_1, u_2 \equiv 1 \pmod {\mathfrak {m}}$ . We replace F with $(1 + a_{201} z + a_{202} z^2) F$ , x with $x + t y z$ and z with $z + t' y^2$ (which eliminates $a_{041}$ and $a_{022}$ ), z with $z + a_{112} x y$ , x with $x + t" y^3$ (which eliminates $a_{060}$ ), x with $x + a_{042} y z^2 + a_{061} y^3 z$ , x with $x + a_{062} y^3 z^2$ , and y with $y + (a_{220} + a_{221} z + a_{222} z^2) x$ , and thus eliminate all $a_{hij}$ ( $(h,i,j) \in S_1$ ). We replace F with $(1 + x^2 b_{40} + x y b_{31} + x^2 y^2 b_{42} + y^4 c_{08}) F$ , and thus eliminate all $b_{hi}$ and $c_{hi}$ . We replace $x,y,z,F$ with suitable multiples and achieve $u_1 = u_2 = 1$ .

Assume $F_2 = x^2 + z^2$ and $F_3$ does not have $y^2 z$ and has $xyz$ . F moreover needs $x y^i$ , $y^i z$ , or $y^l$ . Replacing z with $z + xy^{i-1}$ (resp. x with $x + y^{i-1}z$ ), we may assume that there are no $x y^i$ (resp. $y^i z$ ) of low degree. Thus, we have

$$ \begin{align*} &\qquad\qquad\qquad\qquad\quad F \equiv x^2 + z^2 + xyz + y^{2n-2} \\ &\pmod{(x^4, x^3 y, x^2 y^2, x y^{m} (m> n), y^{2n}, x^2 z, x y z^2, y^2 z^2, y^{m} z (m > n), z^3)}, \end{align*} $$

$n \geq 3$ , and this is $D_{2n}^{n-1}[*]$ . We eliminate $x^h y^i$ ( $h ,i \geq 1$ , $(h, i) \neq (1, 1)$ ) by replacing z with $z + a_{hi0} x^{h-1} y^{i-1}$ , and $x^h z^j$ and $y^i z^j$ similarly. We eliminate $x^h$ , $y^i$ , $z^j$ ( $h \geq 3$ , $i \geq 2n-1$ , $j \geq 3$ ) by replacing F with a unit multiple. We obtain $F = x^2 + z^2 + xyz u + y^{2n-2}$ for some $u \in k[[x^2, x y, y^2, z]]^*$ , and we can achieve $u = 1$ .

Assume $F_2 = x^2 + z^2$ and $F_3$ does not have $y^2 z$ nor $xyz$ . By replacing F with $(1 + a_{201} x^2 z)^{-1} F$ , we may assume that F does not have $x^2 z$ . Then F has $z^3$ and F moreover needs $xy^3$ , and then it is $E_7^2$ . We have

$$ \begin{align*} &F = x^2 + z^2 + z^3 u_1 + x y^3 u_2 + \sum_{(h,i,j) \in S_1} a_{hij} x^h y^i z^j + \sum_{(h,i) \in S_2} b_{hi} x^h y^i + \sum_{(h,i) \in S_3} c_{hi} x^h y^i, \\ & \ \ S_1 = \{(0,4,0), (0,4,1), (0,6,0), (0,2,2), (3,1,0), (2,2,0), (1,1,2), (4,0,0), (2,0,2) \}, \\ &\qquad\qquad\qquad\qquad S_2 = \{(4,0), (3,1), (2,2)\}, \qquad S_3 = \{(0,4)\}, \end{align*} $$

with $a_{hij}$ , $b_{hi}$ , $c_{hi}$ , and $u_1,u_2$ as in the case of $E_7^3$ , and moreover $b_{hi} \in (x^2, z)$ and $c_{hi} \in (y^4, y^2 z, z^2)$ . We may assume $u_1, u_2 \equiv 1 \pmod {\mathfrak {m}}$ . We replace z with $z + a_{040}^{1/2} y^2$ , x with $x + a_{041} y z$ , x with $x + t y^3$ (which eliminates $a_{060}$ ), F with $(1 + a_{022} y^2) F$ , y with $y + t' x$ and z with $z + t" x y$ and F with $(1 + t"' x y) F$ (which eliminates $a_{310}$ , $a_{220}$ , and $a_{112}$ ), z with $z + t"" x^2$ and F with $(1 + t""' x^2) F$ (which eliminates $a_{400}$ and $a_{202}$ ), and thus eliminate all $a_{hij}$ ( $(h,i,j) \in S_1$ ). We replace x with $x + y c_{04}$ and F with $(1 + b_{40} x^2 + b_{31} x y + b_{22} y^2) F$ to eliminate all $b_{hi}$ and $c_{hi}$ . We replace $x,y,z,F$ with suitable multiples and achieve $u_1 = u_2 = 1$ .

Proof. Let $\mathcal {O}_{X,w} = B = \bigoplus _{i \in \mathbb Z/p^2\mathbb Z} B_i$ be the corresponding decomposition. Since w is non-fixed by $\mu _p$ , there exists $y \in \mathfrak {m}_{w} \subset B$ with $1 + y \in \bigoplus _{i \equiv 1 \pmod {p}} B_i$ (Lemma 2.9). Then, since $y^p \in \mathfrak {m}_{\pi _1(w)} \subset \mathcal {O}_{X_1, \pi _1(w)}$ satisfies $1 + y^p = (1 + y)^p \in B_p$ , we conclude by Proposition 2.8 that $\pi _1(w)$ is non-fixed by $\mu _{p^2}/\mu _p$ .

Proof. We first show that there exist elements $x,y,z \in \mathfrak {m}$ generating $\mathfrak {m}$ and satisfying $\operatorname {\mathrm {wt}}(x,y,1+z) = (a,b,c)$ with $c = n/p^{e-f} = p^f r$ . Since the stabilizer of the $\mu _{n}$ -action is $\mu _{p^f r} \subsetneq \mu _{n }$ , there exist $i \in \mathbb Z/n\mathbb Z$ and a homogeneous element t of weight i such that $t \notin \mathfrak {m}$ and $\gcd \{i, n\} = p^f r$ . We may assume $t \equiv 1 \pmod {\mathfrak {m}}$ , and then $z' := -1 + t \in \mathfrak {m}$ satisfies $\operatorname {\mathrm {wt}}(1 + z') = i$ . Take an integer q such that $q i = p^f r$ (in $\mathbb Z/n\mathbb Z$ ), and then $1 + z := (1 + z')^q$ satisfies $z \in \mathfrak {m}$ and $\operatorname {\mathrm {wt}}(1 + z) = p^f r$ . Now, take $x^{\prime (1)}, x^{\prime (2)}, x^{\prime (3)}$ generating $\mathfrak {m}$ . We may assume each $x^{\prime (h)}$ is homogeneous with respect to $\mu _{p^f r}$ , that is, there exists $i^{(1)}, i^{(2)}, i^{(3)} \in \mathbb Z/n\mathbb Z$ such that $x^{\prime (h)} = \sum _{j = 0}^{p^{e-f}-1} x^{\prime (h,j)}$ with $\operatorname {\mathrm {wt}}(x^{\prime (h,j)}) = i^{(h)} + j p^f r$ . Let $x^{(h)} := \sum _{j = 0}^{p^{e-f}-1} (1+z)^{-j} x^{\prime (h,j)}$ . Then $\operatorname {\mathrm {wt}}(x^{(h)}) = i^{(h)}$ and, since $x^{(h)} \equiv x^{\prime (h)} \pmod {(z)}$ , the elements $x^{(1)}, x^{(2)}, x^{(3)}, z$ generate $\mathfrak {m}$ . We can omit one of these four elements, which cannot be z since $\mathfrak {m}$ is not generated by homogeneous elements.

In this proof, by a monomial, we mean a polynomial of the form $x^i y^j z^{p^{e-f} l} (1+z)^m$ with $0 \leq m < p^{e-f}$ . Any polynomial (resp. formal power series) is uniquely expressed as a finite (resp. possibly infinite) sum of monomials with k-coefficients, and we say that a polynomial or a formal power series has a monomial if its coefficient is nonzero. Expressions such as $F = x^2 + y^3 (1+z) + \cdots $ will indicate that F has these monomials. However, when we say of degree m part $F_m$ of F, this is understood with respect to the usual monomials $x^i y^j z^l$ .

Assume the degree $2$ part $F_2$ of F is either irreducible or the product of two distinct homogeneous linear factors. Then we may assume $F = xy (1+z)^i + \cdots $ , and F must have $z^{m'} (1+z)^j$ , we may assume $i = j = 0$ by replacing y and F, and it is $A_{m'-1}$ with $m' = p^{e-f} m$ .

Assume $F_2$ is the product of two distinct nonhomogeneous linear factors. Then we have $p \neq 2$ and we may assume $F = x^2 (1+z)^i + y^2 (1+z)^j + z^{m'} (1+z)^k + \cdots $ and $b \equiv a + r/2 \pmod {r}$ . We may assume $k = 0$ by replacing F with $(1+z)^{-k} F$ and $i = j = 0$ by replacing x and y with $x (1+z)^{i(p^e+1)/2}$ and $y (1+z)^{j(p^e+1)/2}$ . We have $2a = 2b = 0$ and then we have $f = 0$ and $r = 2$ . (Otherwise, $a,b,c$ cannot generate $\mathbb Z/p^e r \mathbb Z$ .) It is $A_{m'-1}$ with $m' = p^e m$ , and we may assume $(a,b) = (0,n/2)$ .

Assume $p \geq 5$ and $F_2$ is a square. We may assume $F_2 = x^2$ , $F_3 = y^3$ , $F = x^2 + y^3 + \cdots $ , and then F must have $z^5 (1+z)^i$ and we have $p = 5$ , it is $E_8^0$ , and we may assume $i = 0$ . By $2a = 3b = 0$ , we have $n \mid 30$ and we may assume $a = 15$ and $b = 10$ .

Assume $p = 3$ and $F_2$ is a square. We may assume $F = x^2 + \cdots $ . We may assume $F_3 \bmod (x)$ is either $y^3$ , $z^3$ , or $y^3 + z^3$ . If $F = x^2 + y^3 + \cdots $ , then F must have $y z^3$ and it is $E_7^0$ . If $F = x^2 + z^3 + \cdots $ , then F must have $y^4$ or $y^5$ and it is $E_6^0$ or $E_8^0$ . If $F = x^2 + y^3 + z^3 (1+z)^i + \cdots $ , then we may assume $i = 1$ (if $i = 0$ , then by replacing z with $y (1 + z)^l + z $ , we reduce this case to the previous case) and then it is $E_6^0$ .

Assume $p = 2$ and $F_2$ is a square. We may assume $F_2$ is $x^2$ , $z^2$ , $x^2 + z^2$ , $x^2 + y^2$ , or $x^2 + y^2 + z^2$ .

Assume $F_2 = x^2$ . We may assume $F = x^2 + \cdots $ . Then F must have $y z^2 (1+z)^i$ or $y^3 (1+z)^j$ . If F has $y z^2 (1+z)^i$ , then we may assume $i = 0$ and F must have $x y^m (1+z)^i$ ( $m \geq 2$ ) or $y^m (1+z)$ ( $m \geq 3$ ). In the former case, we may assume $i = 0$ (by replacing x with $x (1+z)^i$ ) and we have $D_{2m+1}^0$ with $(a,b,c) = (2m,2,2m-1)$ , $n \mid 2(2m-1)$ . In the latter case, we have $D_{2m}^0$ with $(a,b,c) = (1,2,2(m-1))$ , $n \mid 4(m-1)$ . Now, assume that F does not have $y z^2 (1+z)^i$ and has $y^3 (1+z)^j$ . We may assume $j = 0$ . Then F must have either $x z^2 (1+z)^i$ or $z^4 (1+z)^i$ . If F has $x z^2 (1+z)^i$ , then we may assume $i = 0$ and then $F = x^2 + y^3 + x z^2 + \cdots $ is $E_6^0$ , and we have $e = 1$ , $f = 0$ , $a = 0$ , and $c = n/2$ . If F does not have $x z^2 (1+z)^i$ and has $z^4 (1+z)^i$ , then we may assume $i = 1$ and then $F = x^2 + y^3 + z^4 (1+z) + \cdots $ is $E_8^0$ , and we have $e - f \leq 2$ , $f = 1$ , $(a,b,c) \equiv (1,2/3,2) \pmod {2^e}$ .

Assume $F_2 = z^2$ . We have $e - f = 1$ . Suppose F has $x^2 y (1 + z)^i$ . We may assume $i = 0$ . We may assume F does not have monomials $z^2 M$ ( $M \neq 1$ ) nor $x^2 M$ ( $M \neq 1, y$ ) of low degree (by replacing F or y with $(1 + M)^{-1} F$ or $y + M$ , respectively). F must have $x y^m$ or $y^m (1+z)$ and then it is $D_{2m}^0$ or $D_{2m+1}^0$ . Now, suppose F does not have $x^2 y (1 + z)^i$ nor $x y^2 (1 + z)^i$ and has $x^3 (1 + z)^i$ . We may assume $i = 0$ . Then F must have $y^3 (1 + z)^j$ ( $D_4^0$ ), $x y^3 (1 + z)^j$ ( $E_7^0$ ), or $y^5 (1 + z)^j$ ( $E_8^0$ ).

Assume $F_2 = x^2 + z^2$ . We may assume $F = x^2 + z^2 (1+z)^i + \cdots $ . If $i = 0$ , then by replacing z with $z + x$ or $z + x (1+z)$ , we reduce this case to the previous case. Assume $i = 1$ . F cannot have $x^3 (1+z)^i$ nor $x z^2 (1+z)^i$ . (If F has $x^3 (1+z)^i$ , then we have $2a = c = 3a + ic$ and this implies $(2i+1)c = 0$ , contradicting $c = n/p^{e-f}$ . Other cases are similar.) We may assume that F does not have $y z^2 (1+z)^i$ . If F has $y^3 (1+z)^i$ , then F does not have $x y^2 (1+z)^j$ and it is $D_4^0$ . If F does not have $y^3 (1+z)^i$ and has $x^2 y (1+z)^i$ , then F cannot have $x y^2 (1+z)^j$ , and F must have $y^{2m+1} (1+z)^j$ , and it is $D_{4m}^0$ . If F does not have $y^3 (1+z)^i$ nor $x^2 y (1+z)^i$ and has $xy^2 (1+z)^i$ , then it is $D_5^0$ . If F does not have $y^3 (1+z)^i$ nor $xy^2 (1+z)^i$ nor $x^2 y (1+z)^i$ , then F must have $xy^3 (1+z)^i$ or $y^5 (1+z)^i$ , and (we may assume $i = 0$ and) it is $E_7^2$ or $E_8^0$ . (This $E_7^2$ is the only example of $D_n^r$ or $E_n^r$ with $r> 0$ in this proposition.)

Assume $F_2 = x^2 + y^2$ . Write $F = x^2 + y^2 (1+z)^j + \cdots $ . If $j = 0$ , then by replacing x with $x + y (1+z)^k$ , we reduce to the $F_2 = x^2$ case. Suppose $j = 1$ . We may assume F does not have monomials $x^2 M$ ( $M \neq (1+z)^i$ ) nor $y^2 M$ ( $M \neq (1+z)^i$ ) of low degree (by replacing F or z with $(1 + M)^{-1} F$ or $z + M$ , respectively). F must have $x z^{2m} (1+z)^k$ or $y z^{2m} (1+z)^k$ , by symmetry, we may assume F has $x z^{2m} (1+z)^k$ , we may assume $k = 0$ , and then it is $D_{4m}^0$ .