Suppose $X$ is a complex algebraic projective surface with one dimensional singular locus. For example consider the hypersurface $z^5=t^2(tx^2+y^3+t^3)$, whose singular locus is along the double line $t^2=0$ in $\mathbb{P}^2$ with a special singularity above the point of intersection of $t=0$ and the elliptic curve $tx^2+y^3+t^3=0$. How can one calculate the normalization of such a surface? I have a feeling that in the example above (thought of as fifth cover of $\mathbb{P^2}$), the normalized surface has singularity over the point of intersection of the branch loci and locally it looks like $z^5=t(tx^2+y^3+t^3)$. However I am unable to back it up by any concrete reasoning. Can someone point out the appropriate ways to go about normalizing such surfaces? It is intuitively clear to me that the normalization of the above example is not a hypersurface, however a precise description of the normalization near its isolated singular points is what I am after.

  • 2
    $\begingroup$ I thought I would point out that Macaulay2 can normalize things. $\endgroup$ Dec 24, 2010 at 16:17

1 Answer 1


Let $f = t x^2 + y^3 + t^3$ be the equation of the elliptic curve, so that your hypersurface has equation $z^5 = t^2 f$.

When you have an equation between monomials like this, there are always some elements of the normalization that are easy to write down. In this case, one has the element $w := z^3/t$. Note that $w^2 = z f$ (which shows that $w$ lie in the normalization) and $w^5 = t f^3$.

If I'm not miscalculating, this shows that by adjoining the element $w$, you make your surface non-singular everywhere where $f \neq 0$, and it was already non-zero already in the region $t \neq 0$, so adjoining $w$ is enough to cut down the singular locus to the intersection of $f = 0$ and $t= 0$.

Unfortunately, I'm not practiced enough at these kinds of computations to go further.

  • $\begingroup$ Thanks Emerton, this indeed shows that the singularity of the normalization lies on the intersection of the branch curves. $\endgroup$
    – user5893
    Dec 25, 2010 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.