Take the tour ×

I need to compute explicitly the normalisation of a singular algebraic curve $C$ which is given by an explicit equation in $\mathbb{A}^2$. This task is mostly reduced to finding the integral closure $\tilde{A}$ of the regular function ring $A := \mathcal{O}_C(C)$ in the function field $k(C)$, and then taking the normalisation $\tilde{C}$ to be the variety whose ring of regular functions is $\tilde{A}$.

One aspect of the problem that bothers me slightly is the case when $A$ has zero divisors and computations in the field of fractions become substantially less intuitive. For instance, if $C$ is given by $x^2 = y^2$, then it is intuitively quite clear that the normalisation is a disjoint union of two lines, but deriving this via explicit construction feels somewhat slippery at a number of points. Thus, I would like to ask the following:

  1. It feels natural that $\pi : \tilde{C} \to C$ should be the normalisation as soon as the following hold:
    • $\tilde{C}$ is a normal variety, $\pi$ is a morphism.
    • $\pi$ is an isomorphism away from the singular points.

Does this criterion hold, or am I missing something here? If it does hold, then a practical way to find the normalisation could be to make an informed guess about what it should be, and then the verification is almost immediate.

(The "official" definition of normalisation requires that $\pi$ should be a finite morphism and it should induce the isomorphism of function fields $ k(C) \to k(\tilde{C})$. )

  1. What are the technical difficulties and traps in using the "standard" procedure of finding the normalisation in the case when $C$ is the zero locus of a reducible polynomial (if any)?

  2. Is it generally true that if $C$ is a sum of irreducible curves $C_i$, whose normalisations are $\tilde{C}_i$, then $\tilde{C}$ is a disjoint sum of $\tilde{C_i}$? (I think it should follow immediately from the definitions, but I might be missing something)

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted
  1. The conditions you listed are not sufficient for $\pi$ to be the normalization map. For example, the inclusion of $C_{\mathrm{reg}}\subseteq C$ satisfies these conditions, but is not the normalization. You really need $\pi$ to be finite (and birational of course).

  2. If $C_1, \dots, C_n$ are the irreducible components of $C$ with normalizations $\tilde{C}_1, \dots, \tilde{C}_n$, then yes, the disjoint union of the $\tilde{C}_i$ is the normalization of $C$. This is because the natural map from this disjoint union to $C$ is finite and birational, and it is itself normal. In particular, if $C$ is defined by a reducible polynomial, just decompose the latter into product of irreducible polynomials and normalize the curves defined by the irreducible factors.

share|improve this answer
Thank you! I had a distorted intuition as to what a finite morphism is until now. By the way, would it be enough to additionally require that $\pi$ is surjective? –  Feanor Jan 26 at 22:42
@Feanor: no because you can take the normalization and remove one point above a singular point. This remains surjective if the singular point is not a cusp. –  user18119 Jan 26 at 22:46
add comment

Your Answer


