Crepant resolution of $\mathbb{A}^{4}/A_{4}$ in characteristic $2$

It is known that in general, under the permutation action of $A_{4}$, the invariant ring $K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}=K[s_{1},s_{2},s_{3},s_{4},\Delta_{4}]$, where $s_{i}$ are the elementary symmetric polynomials and $\Delta_{4}$ is an $A_{4}$-invariant polynomial of degree 6 which is not symmetric. Over fields of characteristic different from 2, $\Delta_{4}$ is always taken as the Vandermonde polynomial. However, in characteristic 2, one should take $\Delta_{4}=\mathcal{O}_{A_{4}}(x_{1}^{3}x_{2}^{2}x_{3})$ instead, where $\mathcal{O}_{G}$ denotes the orbit sum under the group action (see [campbell2011modular], 4.4).

In characteristic 2, computation shows that $\Delta_{4}^{2}+(s_{1}^{2}s_{4}+s_{1}s_{2}s_{3}+s_{3}^{2})\Delta_{4}$ is a symmetric polynomial of degree 12 with elementary representation $s_{1}^{4}s_{4}^{2}+s_{1}^{3}s_{3}^{3}+s_{1}^{2}s_{2}^{3}s_{4}+s_{2}^{3}s_{3}^{2}+s_{3}^{4}$. Let $R=K[A,B,C,D,E]$ be a graded polynomial ring such that $A,B,C,D,E$ are of degree $1,2,3,4,6$ respectively. Consider the ring homomorphism $\phi:R\to K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}$ determined by $\phi(A)=s_{1},\phi(B)=s_{2},\phi(C)=s_{3},\phi(D)=s_{4},\phi(E)=\Delta_{4}$. Then $\phi$ is a surjective homomorphism between two graded rings, and $f:=E^{2}+(A^{2}D+ABC+C^{2})E+A^{4}D^{2}+A^{3}C^{3}+A^{2}B^{3}D+B^{3}C^{2}+C^{4}\in\mathrm{Ker}\phi$.

We want to show that $R/(f)=R/\mathrm{Ker}\phi\cong K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}$. For a graded ring $S$, we consider its Hilbert series $\mathcal{H}(S,\lambda)=\sum_{d\geq 0}\lambda^{d}\dim_{K}S_{d}$. Since $R$ is generated by elements of degree $1,2,3,4,6$ and $f$ is of degree $12$, we have

$\displaystyle\mathcal{H}(R/(f),\lambda)=\frac{1-\lambda^{12}}{(1-\lambda)(1-\lambda^{2})(1-\lambda^{3})(1-\lambda^{4})(1-\lambda^{6})}.$

On the other hand, since $A_{4}$ has a permutation action on $\mathbb{A}^{4}$, the Hilbert series of invariant ring does not change by characteristic of the field, and it is possible to compute the Hilbert series by Molien’s theorem over $\mathbb{C}$ (see [campbell2011modular], 3.7 and 4.5). Therefore,

		$\displaystyle\mathcal{H}(K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}},\lambda)$
	$\displaystyle=$	$\displaystyle\frac{1}{12}(\frac{1}{(1-\lambda)^{4}}+\frac{3}{(1-\lambda)^{2}(1+\lambda)^{2}}+\frac{8}{(1-\lambda)(1-\lambda^{3})})$
	$\displaystyle=$	$\displaystyle\frac{1+\lambda^{6}}{(1-\lambda)(1-\lambda^{2})(1-\lambda^{3})(1-\lambda^{4})}$
	$\displaystyle=$	$\displaystyle\frac{1-\lambda^{12}}{(1-\lambda)(1-\lambda^{2})(1-\lambda^{3})(1-\lambda^{4})(1-\lambda^{6})}.$

Since $R/(f)$ and $K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}$ share the same Hilbert series, the induced surjective homomorphism $\widetilde{\phi}:R/(f)\to K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}$, which is surjective on each degree between two $K$-linear spaces with the same dimension, should also be isomorphic on each degree. Therefore $R/(f)\cong K[x_{1},x_{2},x_{3},x_{4}]^{A_{4}}$. ∎

Denote the blow-up of $\mathbb{A}^{n}$ along $C$ again by $f:U\to\mathbb{A}^{n}$. Then by abusing notations of exceptional divisors, we have

	$\displaystyle K_{U}$	$\displaystyle=f^{*}K_{\mathbb{A}^{n}}+2E,$
	$\displaystyle f^{*}X$	$\displaystyle=\widetilde{X}+2E.$

Taking them together, and applying the adjunction formula, we obtain

$\displaystyle K_{\widetilde{X}}=(K_{U}+\widetilde{X})|_{\widetilde{X}}=(f^{*}K_{\mathbb{A}^{n}}+f^{*}X)|_{\widetilde{X}}=f^{*}(K_{\mathbb{A}^{n}}+X)|_{\widetilde{X}}=f^{*}K_{X}.$

Therefore, $f$ is crepant. ∎

If $b\neq 0$, then we can take $\alpha\in K$ such that $a\alpha^{2}+b\alpha+c=0$. Therefore

$$h(x,y)=bXY+F,$$

where $X=x+\alpha y+\frac{e+d\alpha}{b}$, $Y=y+\frac{a}{b}(x+\alpha y)+\frac{1}{b}(d+\frac{a}{b}(e+d\alpha))$, $F=f+\frac{1}{b}(e+d\alpha)(d+\frac{a}{b}(e+d\alpha))=f+\frac{1}{b}(de+\frac{ae^{2}}{b}+\frac{cd^{2}}{b})$. Hence $C$ is non-degenerate if and only if $F\neq 0$; when $F=0$, $C$ becomes two intersecting lines.

If $b=0$, we may assume that $a\neq 0$ without loss of generality. Then

$$h(x,y)=aX^{2}+dX+Ey+f,$$

where $X=x+\sqrt{\frac{c}{a}}y$, $E=e+d\sqrt{\frac{c}{a}}$. Here if $E=0$, then $C$ degenerates as two lines; in particular, the two lines become a double line if $d$ is furthermore $0$. If $E\neq 0$, then

$$h(x,y)=aX_{1}^{2}+EY,$$

where $X_{1}=X+\sqrt{\frac{f}{a}}$, and $Y=y+\frac{d}{E}X$. Thus $C$ is non-degenerate under this assumption. ∎

To show that $\pi$ is a crepant resolution, it suffices to check that each blow-up is along a smooth locus of codimension 3 in the whole space, and that the hypersurface has multiplicity $2$ along the center of blow-up (such that each blow-up is a crepant morphism by Lemma 2.2), and that $R$ is smooth. What is more, for each blow-up, we can use Proposition 2.3 to check the structure of exceptional divisor.

Step 1: $\pi_{1}:U\to M$. Take projective coordinates $(u_{0}:u_{1}:u_{2})=(A:C:E)$. Then

	$\displaystyle U=V($	$\displaystyle u_{2}^{2}+(Du_{0}^{2}+Bu_{0}u_{1}+u_{1}^{2})E+A^{2}D^{2}u_{0}^{2}+AC^{3}u_{0}^{2}+B^{3}Du_{0}^{2}$
		$\displaystyle+B^{3}u_{1}^{2}+C^{2}u_{1}^{2},(u_{0}:u_{1}:u_{2})=(A:C:E)).$

Here $E_{1}=V(u_{2}^{2}+B^{3}Du_{0}^{2}+B^{3}u_{1}^{2}).$ By the base change given by the Frobenius $\mathrm{Spec}(K[\sqrt{B},\sqrt{D}])\to\mathrm{Spec}(K[B,D])$, we have $u_{2}^{2}+B^{3}Du_{0}^{2}+B^{3}u_{1}^{2}=(u_{2}+\sqrt{B^{3}D}u_{0}+\sqrt{B^{3}}u_{1})^{2}$, such that $E_{1}$ can be viewed as a trivial $\mathbb{P}^{1}$-bundle over $\mathbb{A}^{2}$ after the base change. Since the Frobenius is a universal homeomorphism, this base change does not change Euler numbers. By abuse of notation, we write $E_{1}\cong\mathbb{A}^{2}\times\mathbb{P}^{1}$ under the necessary base change. In the defining equation of $U$, $u_{0}=u_{1}=0$ implies $u_{2}=0$, which gives a contradiction, so $U$ is covered by its affine pieces determined by $u_{0}\neq 0$ and $u_{1}\neq 0$ respectively. We denote them by $U_{0}=U\cap\{u_{0}\neq 0\}$ and $U_{1}=U\cap\{u_{1}\neq 0\}$. For these 2 affine pieces, we have

	$\displaystyle U_{0}\cong V($	$\displaystyle u_{2}^{2}+(D+Bu_{1}+u_{1}^{2})Au_{2}+A^{2}D^{2}+A^{4}u_{1}^{3}+B^{3}D+B^{3}u_{1}^{2}+A^{2}u_{1}^{4}),$
	$\displaystyle\mathrm{Sing}(U_{0})$	$\displaystyle\cap E_{1}=V(A,B,u_{2}).$
	$\displaystyle U_{1}\cong V($	$\displaystyle u_{2}^{2}+(Du_{0}^{2}+Bu_{0}+1)Cu_{2}+C^{2}D^{2}u_{0}^{4}+C^{4}u_{0}^{3}+B^{3}Du_{0}^{2}+B^{3}+C^{2}),$
	$\displaystyle\mathrm{Sing}(U_{1})$	$\displaystyle\cap E_{1}=V(C,B,u_{2}).$

And we also obtain $E_{1}\setminus\mathrm{Sing}(U)\cong\mathbb{A}^{2}\times\mathbb{P}^{1}\setminus\mathbb{A}^{1}\times\mathbb{P}^{1}$ by gluing its affine pieces together.

Step 2: $\pi_{2}:V\to U$. From step 1, it suffices to consider blow-ups of $U_{0}$ along $V(A,B,u_{2})$ and $U_{1}$ along $V(C,B,u_{2})$, and then glue them together to obtain $V$. Denote the 2 blow-ups by $\widetilde{U_{0}}$ and $\widetilde{U_{1}}$. Then

	$\displaystyle\widetilde{U_{0}}=V($	$\displaystyle v_{2}^{2}+(D+Bu_{1}+u_{1}^{2})v_{0}v_{2}+D^{2}v_{0}^{2}+A^{2}u_{1}^{3}v_{0}^{2}+BDv_{1}^{2}+Bu_{1}^{2}v_{1}^{2}$
		$\displaystyle+u_{1}^{4}v_{0}^{2},(v_{0}:v_{1}:v_{2})=(A:B:u_{2})),$
	$\displaystyle\widetilde{U_{1}}=V($	$\displaystyle v_{2}^{2}+(Du_{0}^{2}+Bu_{0}+1)v_{0}v_{2}+D^{2}u_{0}^{4}v_{0}^{2}+C^{2}u_{0}^{3}v_{0}^{2}+BDu_{0}^{2}v_{1}^{2}+Bv_{1}^{2}$
		$\displaystyle+v_{0}^{2},(v_{0}:v_{1}:v_{2})=(C:B:u_{2})).$

Here $E_{2}=V(v_{2}^{2}+(Du_{0}^{2}+u_{1}^{2})v_{0}v_{2}+(Du_{0}^{2}+u_{1}^{2})^{2}v_{0}^{2})$ gives two intersecting projective lines if $Du_{0}^{2}+u_{1}^{2}\neq 0$ or a double line $\{v_{2}^{2}=0\}$ if $Du_{0}^{2}+u_{1}^{2}=0$. Similarly to what is done in step 1, we may write

$$E_{2}\cong(\mathbb{A}^{1}\times\mathbb{P}^{1}\setminus\mathbb{A}^{1})\times(\mathbb{P}^{1}\vee\mathbb{P}^{1})\cup\mathbb{A}^{1}\times\mathbb{P}^{1}.$$

For $\widetilde{U_{1}}$, computation shows that $\mathrm{Sing}(\widetilde{U_{1}})\cap E_{2}\subseteq V\cap\{u_{0}\neq 0\}\subseteq\widetilde{U_{0}}$, thus we only need to consider $\widetilde{U_{0}}$. Denote by $V_{0}$ and $V_{1}$ the affine pieces of $\widetilde{U_{0}}$ determined by $v_{0}\neq 0$ and $v_{1}\neq 0$ respectively, and then $\widetilde{U_{0}}=V_{0}\cup V_{1}$. Again like what is done in step 1, we obtain

	$\displaystyle V_{0}\cong V($	$\displaystyle v_{2}^{2}+(D+Au_{1}v_{1}+u_{1}^{2})v_{2}+D^{2}+A^{2}u_{1}^{3}+ADv_{1}^{3}+Au_{1}^{2}v_{1}^{3}+u_{1}^{4}),$
	$\displaystyle\mathrm{Sing}(V_{0})$	$\displaystyle\cap E_{2}=V(A,v_{2},D+u_{1}^{2}).$
	$\displaystyle V_{1}\cong V($	$\displaystyle v_{2}^{2}+(D+Bu_{1}+u_{1}^{2})v_{0}v_{2}+D^{2}v_{0}^{2}+B^{2}u_{1}^{3}v_{0}^{4}+BD+Bu_{1}^{2}+u_{1}^{4}v_{0}^{2}),$
	$\displaystyle\mathrm{Sing}(V_{1})$	$\displaystyle\cap E_{2}=V(B,v_{2},D+u_{1}^{2}).$

It follows that the part $\mathbb{A}^{1}\times\mathbb{P}^{1}$ in $E_{2}$ is exactly singular part, thus $E_{2}\setminus\mathrm{Sing}(V)\cong(\mathbb{A}^{1}\times\mathbb{P}^{1}\setminus\mathbb{A}^{1})\times(\mathbb{P}^{1}\vee\mathbb{P}^{1}).$

Step 3: $\pi_{3}:W\to V$. With similar notations as above, taking $(w_{0}:w_{1}:w_{2})$ as projective coordinates for $(A:v_{2}:D+u_{1}^{2})$ in $\widetilde{V_{0}}$ and $(B:v_{2}:D+u_{1}^{2})$ in $\widetilde{V_{1}}$, we have

	$\displaystyle\widetilde{V_{0}}=V($	$\displaystyle w_{1}^{2}+(w_{2}+u_{1}v_{1}w_{0})w_{1}+w_{2}^{2}+v_{1}^{3}w_{0}w_{2}+u_{1}^{3}w_{0}^{2},$
		$\displaystyle(w_{0}:w_{1}:w_{2})=(A:v_{2}:D+u_{1}^{2})),$
	$\displaystyle\widetilde{V_{1}}=V($	$\displaystyle w_{1}^{2}+(w_{2}+u_{1}w_{0})v_{0}w_{1}+v_{0}^{2}w_{2}^{2}+u_{1}^{3}v_{0}^{4}w_{0}^{2}+w_{0}w_{2},$
		$\displaystyle(w_{0}:w_{1}:w_{2})=(B:v_{2}:D+u_{1}^{2})).$

Consider the conics with coordinates $(w_{0}:w_{1}:w_{2})$ in $E_{3}$. Computation shows that they are degenerate as 2 different projective lines exactly when $v_{0}\neq 0$ and $u_{1}v_{0}^{2}=v_{1}^{2}$. Therefore,

$$E_{3}\cong(\mathbb{A}^{1}\times\mathbb{P}^{1}\setminus\mathbb{A}^{1})\times\mathbb{P}^{1}\cup\mathbb{A}^{1}\times(\mathbb{P}^{1}\vee\mathbb{P}^{1}).$$

For $\widetilde{V_{1}}$, again by computation we obtain $\mathrm{Sing}(\widetilde{V_{1}})\cap E_{3}\subseteq W\cap\{v_{0}\neq 0\}\subseteq\widetilde{V_{0}}$, so it suffices to consider $\widetilde{V_{0}}$. Using similar notations as step 1 and step 2, we consider the affine cover $\widetilde{V_{0}}=W_{0}\cup W_{2}$ as follows.

	$\displaystyle W_{0}\cong V(w_{1}^{2}+w_{1}w_{2}+u_{1}v_{1}w_{1}+w_{2}^{2}+v_{1}^{3}w_{2}+u_{1}^{3}),$
	$\displaystyle\mathrm{Sing}(W_{0})=\{w_{1}=w_{2}=v_{1}^{3},u_{1}=v_{1}^{2}\}.$
	$\displaystyle W_{2}\cong V(w_{1}^{2}+(1+u_{1}v_{1}w_{0})w_{1}+1+v_{1}^{3}w_{0}+u_{1}^{3}w_{0}^{2}),$
	$\displaystyle\mathrm{Sing}(W_{2})\subseteq\mathrm{Sing}(W)\cap\{w_{0}\neq 0\}\subseteq\mathrm{Sing}(W_{0}).$

Therefore $\mathrm{Sing}(W)=\mathrm{Sing}(W_{0})$ is exactly the strict transform of $P_{2}$ under these 3 blow-ups above, and we only need to consider $W_{0}$ for the next step. As for $E_{3}\setminus\mathrm{Sing}(W)$, since singularities in $E_{3}$ only appear at each crossing point of projective lines, we obtain

		$\displaystyle E_{3}\setminus\mathrm{Sing}(W)$
	$\displaystyle\cong$	$\displaystyle(\mathbb{A}^{1}\times\mathbb{P}^{1}\setminus\mathbb{A}^{1})\times\mathbb{P}^{1}\cup\mathbb{A}^{1}\times(\mathbb{P}^{1}\vee\mathbb{P}^{1}\setminus\{1point\}).$

Step 4: $\pi_{4}:R\to W$. Let $w_{1}^{\prime}=w_{1}+v_{1}^{3},w_{2}^{\prime}=w_{2}+v_{1}^{3},u_{1}^{\prime}=u_{1}+v_{1}^{2}$, and use the projective coordinates $(r_{0}:r_{1}:r_{2})=(w_{1}^{\prime}:w_{2}^{\prime}:u_{1}^{\prime})$. Then

	$\displaystyle\widetilde{W_{0}}=V($	$\displaystyle r_{0}^{2}+r_{1}^{2}+r_{0}r_{1}+u_{1}^{\prime}r_{2}^{2}+v_{1}r_{0}r_{2}+v_{1}^{2}r_{2}^{2},$
		$\displaystyle(r_{0}:r_{1}:r_{2})=(w_{1}^{\prime}:w_{2}^{\prime}:u_{1}^{\prime})).$

For the exceptional divisor, $E_{4}=V(r_{0}^{2}+r_{0}(r_{1}+v_{1}r_{2})+(r_{1}+v_{1}r_{2})^{2})\cong\mathbb{A}^{2}\times(\mathbb{P}^{1}\vee\mathbb{P}^{1})$(note that $A$ is a hidden affine coordinate in the defining equation of $\widetilde{W_{0}}$). For the smoothness of $\widetilde{W_{0}}$, by considering its affine pieces $\widetilde{W_{0}}=R_{0}\cup R_{2}$, we obtain

	$\displaystyle R_{0}\cong V(1+r_{1}^{2}+r_{1}+w_{1}^{\prime}r_{2}^{3}+v_{1}r_{2}+v_{1}^{2}r_{2}^{2}),$
	$\displaystyle R_{2}\cong V(r_{0}^{2}+r_{1}^{2}+r_{0}r_{1}+u_{1}^{\prime}+v_{1}r_{0}+v^{2}).$

And it is easy to check that both $R_{0}$ and $R_{2}$ are smooth.

Above all, we know that $R$ is smooth and $\pi$ is composed of crepant blow-up morphisms, hence $\pi:R\to M$ is a crepant resolution of $M$. ∎