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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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up vote 4 down vote accepted

Blowing up a cuspidal plane curve actually yields a nonsingular curve. So the multiplicity is actually reduced by more than one. This is e.g. Exercise 19.4.C in Vakil's notes "Foundations of Algebraic Geometry".

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!

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