It is mentioned in this question https://mathoverflow.net/questions/148826/do-there-exist-double-points-on-an-algebraic-surface-in-mathbbp-mathbbc that $X=\{x^2+y^3+z^6=0\}\subset \mathbb C^3$ defines an elliptic singularity at the origin. This means that there is a desingularization $$f:\tilde{X}\to X$$ such that the curve over $0$ has the arithmetic genus $p_a(C_0)=1$.

In page 109, exercise 18 of Miles Reid's notes, weighted blowup is suggested to resolve the singularity.

Edited: As @Sasha pointed out, the singularity can also be resolved just by a sequence of ordinary blowups, so let's try it: Blowup the origin gives the system of equations


in $\mathbb C^3\times \mathbb P^2_{[u,v,w]}$. By setting $v=1$ and substitute, we get $y^2(u^2+y+y^4w^6)=0.$ Therefore the proper transform is defined by $u^2+y+y^4w^6=0$ in the affine chart and it is smooth, moreover the exceptional divisor is the rational curve $\{u=0\}\subset \mathbb P^2_{[u,v,w]}$ with multiplicity $2$.

However, I didn't see an elliptic curve appear in the blowup. Is there anything wrong?


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    $\begingroup$ Any singularity can be resolved by a sequence of ordinary blowups. In this case, the sequence is rather long, while the weighted blowup is a shorter way. $\endgroup$ – Sasha Mar 9 at 8:02
  • $\begingroup$ Dear @Sasha: Thanks for your comment! I'm thinking $z$ as a parameter of a family of curves with total space $x^2+y^3+z^6=0$. It seems to me that a single ordinary blowup produces normal crossing on the special fiber (over $0$), but I didn't see an elliptic curve (as computed above). Is there anything wrong? Thanks a lot! $\endgroup$ – AG learner Mar 9 at 16:43
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    $\begingroup$ My computations give the following result. 1) After the first blowup there is an isolated singularity with local equation $x^2 + y^3z + z^4 = 0$. 2) After the second blowup there is a 1-dimensional singularity with affine equation $x^2 + y^2(z + z^4) = 0$ in one of the charts. 3) After the third blowup (this time with center the line $x = y = 0$), the surface is smooth and preimage of the singular locus contains the curve $t^2 + z + z^4 = 0$, which is an elliptic curve. $\endgroup$ – Sasha Mar 9 at 18:59
  • $\begingroup$ Dear @Sasha: Thanks a lot for your help! Finally, I managed to repeat all the computations. I also found that the variety $x^2+y^2(z+z^4)=0$ is not normal, and its normalization coincides with the blowup in the step 3) in your computation. $\endgroup$ – AG learner Mar 10 at 8:31

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