Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In section I.6 of Algebraic Geometry, Hartshorne establishes a that every curve is birationally equivalent to a nonsingular projective curve. To do this, he defines for any given curve $C$ with function field $K$ an abstract nonsingular curve $C_K$ whose points are just DVR's of $K$ with an appropriate topology. It can be shown that $C_K$ is isomorphic to a nonsingular projective curve.

Later on in ex II.3.8, Hartshorne defines the normalization of a scheme, which in the case of curves will be a nonsingular curve that is birational to the original curve.

In general, do these two constructions give the same curve? They give birational curves by II.6.12, but I would like to say that they are actually the same. If this is not true, can we at least say that the normalization has an open immersion into $C_K$? I have a hunch that the latter statement should be true since the points of a nonsingular curve just correspond to DVR's of $K.$

Thanks ahead of time.

share|cite|improve this question

Sorry for resurrecting such an old post, but Section §1.6 of Hartshorne annoys me in the sense that the concept of "abstract nonsingular curve" is not used very much (if at all) in algebraic geometry. We provide an alternative version of that section (specifically, Lemma 1.6.5 through Theorem 1.6.9) that avoids use of said object.

We start with a replacement of Proposition 1.6.8.

Proposition. Let $\phi: X \to Y$ be a rational map from a normal curve $X$ to a projective variety $Y$. Then $\phi$ extends to a (unique) morphism $X \to Y$.

Proof. We may assume that $\phi$ is dominant (replace $Y$ with $\overline{\phi(X)}$), that $Y \subseteq \mathbb{P}^n$, and that $Y$ is not contained in any of the coordinate hyperplanes. Let $U$ be an open subset of $X$ on which $\phi$ is represented by a morphism, and let $P \in X \setminus U$. Since $X$ is normal, $\mathcal{O}_{P,X}$ is a DVR; let $v$ be the associated discrete valuation on $K(X)$. We have that $1 = \phi^*(x_0/x_0), \phi^*(x_1/x_0), \dots, \phi^*(x_n/x_0)$ are nonzero elements of $K(X)$; pick $i$ such that $v(\phi^*(x_i/x_0))$ is minimal. Then, for all $j$,$$v\left(\phi^*\left({{x_j}\over{x_i}}\right)\right) = v\left(\phi^*\left({{x_j}\over{x_0}}\right)\right) - v\left(\phi^*\left({{x_i}\over{x_0}}\right)\right) \ge 0,$$so $\phi^*(x_j/x_i)$ lies in $\mathcal{O}_{P, X}$ for all $j$. Let $V$ be an open neighborhood of $P$ in $X$ such that $\phi^*(x_j/x_i)$ is regular on $V$ for all $j$. Letting $U_i \subseteq \mathbb{P}^n$ denote the open affine $\{x_i \neq 0\}$, we have that the image of the composition$$A(Y \cap U_i) \hookrightarrow K(Y) \overset{\phi^*}{\to} K(X)$$is contained in $\mathcal{O}(V)$, so $\phi^*$ determines a morphism $\phi': V \to Y$ that induces the same map $K(Y) \to K(X)$ as $\phi$. Therefore, $\phi'$ and $\phi$ are the same rational map, and therefore $V$ is contained in the maximal domain of $\phi$. In this way, $\phi$ can be extended to a morphism $X \to Y$. $$\tag*{$\square$}$$

Lemma. Let $W$ be a variety. Then its normalization map $\pi: \widetilde{W} \to W$ is surjective.

Proof. Let $P \in W$. By the lying-over theorem (Atiyah-MacDonald Theorem 5.10 or Eisenbud Proposition 4.15), since $A(\widetilde{W})$ is integral over $A(W)$, it contains a prime ideal whose restriction to $A(W)$ is the maximal (hence prime) ideal $I(P)$. Replacing this prime with a maximal ideal containing it, if necessary we have a maximal ideal $\mathfrak{a}$ for which $\mathfrak{a} \cap A(W) = I(P)$. This maximal ideal corresponds to a point lying over $P$. $$\tag*{$\square$}$$

Theorem. Every curve is birational to a nonsingular projective curve. Moreover, if $C$ is a curve and $\phi: C \to \widetilde{C}$ is a birational map to a nonsingular projective curve $\widetilde{C}$, then $\phi$ is represented by a morphism on the (open) set of nonsingular points of $C$.

Proof. Let $C$ be a curve. We may assume that it is quasi-projective. It is birational to its projective closure, so we may assume that $C$ is projective. Let $\{U_1, \dots, U_m\}$ be a covering of $C$ by open affines. For each $i$, let $\pi_i: V_i \to U_i$ be the normalization (Exercise 1.3.17). This map is birational (since $A(U_i) \subseteq A(V_i) \subseteq K(U_i)$), so there are nonempty open subsets $U_i' \subseteq U_i$ and $V_i' \subseteq V_i$ such that $\pi_i$ induces an isomorphism $V_i' \overset{\cong}{\to} U_i'$.

Each $V_i$ is affine, say $V$ is closed in $\mathbb{A}^{n_i}$ for all $i$. Let $Y_i$ be its projective closure in $P^{n_i}$. We have a rational map$$\theta: C\,\, -\!\! \rightarrow Y_1 \times \dots \times Y_m \subseteq \mathbb{P}^{n_1} \times \dots \times \mathbb{P}^{n_m},$$represented by a morphism on the set $U := \bigcap U_i'$. Let $Y$ be the closure of the graph. Then $Y$ is a projective curve, and we have morphisms $\phi: Y \to C$ and $\psi_i : Y \to Y_i$ for all $i$. As noted earlier, $\phi$ is birational, and is an isomorphism over $U$. In particular, $Y$ is birational to $C$.

In addition, $Y \cap (U \times \prod \pi_i^{-1} (U))$ is the (closed) graph of the morphism$$(\pi_1^{-1}, \dots, \pi_m^{-1}) : U \to \prod Y_i,$$so this nonempty open subset of $Y$ is isomorphic to $U$ and to $\pi_i^{-1}(U)$ for all $i$; hence all $\psi_i$ are also birational. Moreover, we have commutative diagrams

                           enter image description here

in which the maps on the left induce the maps on the right.

It remains only to show that $Y$ is nonsingular.

Pick $P \in Y$, and choose $i$ such that $\phi(P) \in U_i$. Let $W$ be an open affine neighborhood of $P$ in $\phi^{-1}(U_i)$, and let $\alpha: \widetilde{W} \to W$ be its normalization. By the lemma, there is a point $\tilde{P} \in \widetilde{W}$ such that $\alpha(\widetilde{P}) = P$. We have a commutative diagram

                                                                              enter image description here

in which $\beta$ exists because $\widetilde{W}$ is normal, $\phi \circ \alpha$ is dominant, and $\pi_i$ is the normalization. However, $\beta$ coincides with $\psi_i \circ \alpha$ on some nonempty open set, since they are both associated to the same map of function fields. Therefore, $\beta = \psi_i \circ \alpha$, so in particular, $\psi_i(P) = \psi_i(\alpha(\tilde{P})) = \beta(\tilde{P})$ lies in $V_i$

Therefore, the local ring $\mathcal{O}_{P, Y}$ dominates $\mathcal{O}_{\psi_i(P), V_i}$. But the latter is normal, hence a DVR (by Theorem 6.2A), hence a valuation ring. By maximality of valuation rings with respect to dominance (Theorem 6.1A), $\mathcal{O}_{P, Y}$ equals $\mathcal{O}_{\psi_i(P), V_i}$; in particular it is regular (by 6.2A again).

This implies the first assertion of the theorem. The second assertion follows immediately from the proposition.$$\tag*{$\square$}$$

share|cite|improve this answer
Very nice post: +1. – Georges Elencwajg Nov 23 '15 at 10:22

If $k$ is an algebraically closed field, then the procedure described in Hartshorne's text establishes an (anti)equivalence of categories between:

  1. The category whose objects are finitely generated field extensions of $k$ of transcendence degree 1, and whose morphisms are $k$-algebra homomorphisms.
  2. The category whose objects are nonsingular projective $k$-curves, and whose morphisms are morphisms of $k$-schemes.

In particular, if you have a nonsingular projective $k$-curve, the corresponding function field is unique up to isomorphism, and two such curves are isomorphic if and only if they are birational.

If you start with a curve that is not projective, then its normalization admits an open immersion to a projective curve, and that curve is unique up to isomorphism.

share|cite|improve this answer
So in particular we know that the abstract singular curve and the normalization are birational so isomorphic. Thanks! That does it. – Lalit Jain Jul 13 '11 at 19:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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