= Solution
\b[Two successive point <blowups of a smooth algebraic surface> suffice.] Let $X_2\to X_1=\mathbb A^2$ be the <blowup of the affine plane at the origin>. In its chart $x=uy$, with coordinates $(y,u)$, the total-transform equation is
$$
x^2-y^5=y^2(u^2-y^3).
$$
Removing the exceptional factor gives the <strict transform>
$$
C_2:\ u^2-y^3=0.
$$
Above the original origin it has just one point, $(y,u)=(0,0)$, which is still singular. The other chart is $y=vx$, where the <strict transform> has equation $1-v^5x^3=0$ and does not meet the exceptional divisor $x=0$. Thus there are no other points above the origin to resolve.
Blow up the remaining point to obtain $X_3\to X_2$. In the chart $u=vy$, with coordinates $(y,v)$, the total transform of $C_2$ is
$$
u^2-y^3=y^2(v^2-y),
$$
and hence its <strict transform> is
$$
\boxed{C_3:\ y=v^2.}
$$
The derivative of $v^2-y$ with respect to $y$ is $-1$, so this is nonsingular, even in characteristics two or five. In the other chart $y=wu$, the <strict transform> has equation $1-w^3u=0$ and does not meet the exceptional divisor $u=0$. Therefore the only point of $C_3$ mapping to the original origin is the smooth point $(y,v)=(0,0)$.
The required sequence is
$$
\boxed{X_3=\operatorname{Bl}_{(0,0)}X_2\longrightarrow X_2=\operatorname{Bl}_0\mathbb A^2\longrightarrow X_1=\mathbb A^2.}
$$
The local parameter $v$ there gives $y=v^2$ and $x=v^5$, also verifying the resolved branch directly. This is the <resolution of the (2,5) cusp by two blowups>. Its tangency to an exceptional divisor does not affect the requested nonsingularity of the <strict transform>; making the whole total transform have normal crossings is a stronger task.
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