# Blowing up a singular point on a curve reduces its singular multiplicity by at least one

Let $X$ be the affine plane curve given by $y^2=x^3$, and $O=(0,0)$. Then $X$ has a double singularity at $O$, since its tangent space at $O$ is the doubled $x$-axis. How do we see that, if $\widetilde{X}$ is the blow-up of $X$ at $O$, then $O$ is a nodal point of $\widetilde{X}$, i.e. the tangent space of $\widetilde{X}$ at $O$ consists of two distinct tangent lines?

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One can compute this quite easily in local charts following e.g. Lecture 20 in Harris' Book "Algebraic Geometry". Recall that $\tilde X$ can be computed by blowing up the affine plane first and then taking the proper transform of the curve $X$. So: The blow-up of the affine plane is given by the points $z_1 W_2=z_2 W_1$ in $\mathbb A^2\times \mathbb P^1$ with coordinates $z_1, z_2$ on $\mathbb A^2$ and $W_1,W_2$ on $\mathbb P^1$. Taking Euclidean coordinates $w_1=W_1/W_2$ on $U_2=\{W_2\neq0\}$ yields an isomorphism from $U_2$ to $\mathbb A^2$ with coordinates $(z_2,w_1)$. We have $z_1^2-z_2^3=z_2^2w_1^2-z_2^3$ on $U_2$ and thus the proper transform $\tilde X$ is defined by the polynomial $w_1^2-z_2$. Hence it is smooth!