Consider $$f(x,y)=x^2-y^5 \in k[x,y].$$ For simplicity assume $\operatorname{char} k=0$. Then $C=(f=0)\subset \mathbb{A}^2$ has a singularity at $(0,0)$. I was told that the singularity at $(0,0)$ needs two blow-ups, however, according to my calculations only one is enough. I am almost certain that something is horribly wrong in what follows, but I cannot quite put my finger on it.

The blow-up is given by $$ \operatorname{Bl}_{(0,0)}\mathbb{A}^2:=\{((a,b)\in \mathbb{A}^2, [l]\in \mathbb{P}^1): (a,b)\in l\}\subset \mathbb{A}^2\times \mathbb{P}^1 $$ If I take the co-ordinates on $\mathbb{P}^1$ to be $X$ and $Y$, then the blow-up is cut out by the equation $$ xY=Xy. $$ Hence, the pre-image of $C$ is cut out by $$ x^2-y^5=0, $$ $$ xY=Xy. $$ Now I want to show that this pre-image is nonsingular. This is done locally on the two standard charts.

If $X\neq 0$ then $X=1$ and we have $$ x^2=y^5 $$ $$ xY=y $$ Call these equations $f$ and $g$. To show that the resulting curve is nonsingular, we need to compute the Jacobian, which is $$ J=\begin{pmatrix} 2x & -5y^4 & 0 \\ Y & 1 & x \end{pmatrix} $$ I claim that the rank of $J$ is 2. Indeed, look at the row rank. Suppose there is $\lambda\in \mathbb{C}^*$ such that $$ \lambda Y=2x, $$ $$ \lambda =-5y^4, $$ $$ \lambda x=0. $$ This at once implies $x=0$, so $y=0$ and $\lambda=0$, so the rows are always linearly independent. There is a similar calculation involving the chart $Y\neq 0$ with the same conclusion.

Where am I wrong?


I do not understand your computations. In the chart $Y \neq 0$ your equations are $$x-Xy=0, \quad x^2-y^5=0.$$ The equations of the strict transform of your curve are obtained by removing the exceptional divisor (counted twice) $y^2=0$, so they are $$x-Xy=0, \quad X^2 - y^3=0.$$ An immediate computation of the rank of the Jacobian matrix now shows that the such a strict transform has a singular point at $(x, \, y, \, X)=(0, \, 0, \,0)$.


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