Lecture 14: Friday, September 20, 2013

Eventually, we wish to show that every compact, connected Riemann surface X is a nonsingular algebraic curve. Today, we discuss what is means to be a “nonsingular algebraic curve” using Dedekind Domains.

Coordinate Rings

Consider an ideal I in the polynomial ring \mathbb C[z_1, z_2, \dots, z_n], and define the set X = \left \{ P \in \mathbb A^n(\mathbb C) \, \biggl| \, f(P) = 0 \ \text{for all} \ f \in I \right \}. We say that X is an algebraic variety if I \subseteq \mathbb C[z_1, z_2, \dots, z_n] is a prime ideal. In this case, the quotient ring \mathcal O(X) = \mathbb C[z_1, z_2, \dots, z_n] / I is an integral domain. We define \mathcal O(X) as the global sections of X or the coordinate ring of X.

Let X be an algebraic variety. Then the map X \to \text{mSpec} \, \mathcal O(X) which sends a point P = (a_1, a_2, \dots, a_n) to the maximal ideal \mathfrak m_P = \langle z_1 - a_1, \, z_2 - a_2, \, \dots, \, z_n - a_n \rangle is a bijection.

Proof: Recall that an ideal \mathfrak m \subseteq \mathcal O(X) is maximal if and only if the quotient \mathcal O(X) / \mathfrak m \simeq \mathbb C is a field. It is clear that the map P \mapsto \mathfrak m_P is well-defined and injective. Let \mathfrak m be any maximal ideal of \mathcal O(X). Fix a surjection \mathcal O(X) \twoheadrightarrow \mathcal O(X) / \mathfrak m \simeq \mathbb C, and denote a_i \in \mathbb C as the image of z_i \in \mathcal O(X). It is easy to check that \mathfrak m = \mathfrak m_P for P = (a_1, a_2, \dots, a_n). \square

Often, we abuse notation and write X = \text{Spec} \, \mathcal O(X). If we denote K = \mathbb C(X) as the quotient field of the integral domain \mathcal O(X), we define the dimension of X as the transcendence degree of K over \mathbb C.

Non-singular Algebraic Curves

We say that X is an algebraic curve if \text{dim}(X) = 1. Here is the main result for today’s lecture.

Consider a prime ideal I = \langle f_1, f_2, \dots, f_m \rangle in the polynomial ring \mathbb C[z_1, z_2, \dots, z_n], and define the algebraic variety X = \left \{ P \in \mathbb A^n(\mathbb C) \, \biggl| \, f_i(P) = 0 \ \text{for all} \ 1 \leq i \leq n \right \}. Assume that X is an algebraic curve, that is, that \text{dim}(X) = n - m = 1. Then the following are equivalent:

If any of these equivalent statements holds true, we say that X is a non-singular algebraic curve. As X is defined by the m polynomial equations f_i(z_1, z_2, \dots, z_n) = 0 for 1 \leq i \leq m, the tangent space at a point P = (a_1, a_2, \dots, a_n) is the complex vector space
T_P(X) = \left \{ (b_1, b_2, \dots, b_n) \in \mathbb A^n(\mathbb C) \ \left| \ \displaystyle \sum_{j=1}^n \dfrac {\partial f_i}{\partial z_j}(P) \, b_j = 0 \ \text{for all} \ 1 \leq i \leq m \right. \right \}.
The Russian mathematician Oscar Zariski (1899 — 1986) realized that T_P(X) is the dual to the quotient \mathfrak m_P / {\mathfrak m_P}^2 for the maximal ideal \mathfrak m_P = \langle z_1 - a_1, \, z_2 - a_2, \, \dots, \, z_n - a_n \rangle, so this quotient is often called the Zariski cotangent space.

Proof: This is essentially a a restatement of Proposition 9.2 on pages 94-95 in Atiyah and Macdonald’s Introduction to Commutative Algebra.

(i) \iff (ii). We have a perfect (i.e., bilinear and nondegenerate) pairing \bigl( \mathfrak m_P / \mathfrak m_P^2 \bigr) \times T_P(X) \to \mathbb A^1(\mathbb C) defined by \bigl( f, \, (b_1, b_2, \dots, b_n) \bigr) \mapsto \displaystyle \sum_{i=1}^n \dfrac {\partial f}{\partial z_i}(P) \, b_i. Hence \dim_{\mathbb C} \bigl( \mathfrak m_P / \mathfrak m_P^2 \bigr) = \dim_{\mathbb C} \bigl( T_P(X) \bigr) = n - m = \text{dim}(X).

(ii) iff (iii). As \mathfrak m = \mathfrak m_P is a maximal ideal, Nakayama’s Lemma states that we can find \varpi \in \mathfrak m_P where \varpi \notin {\mathfrak m_P}^2. Consider the injective map \mathcal O(X) / \mathfrak m_P \to \mathfrak m_P / \mathfrak m_P^2 defined by x \mapsto \varpi \, x. Clearly this is surjective if and only if \mathfrak m_P \, \mathcal O_P = \varpi \, \mathcal O_P is principal. Recall now that \dim_{\mathbb C} \bigl( \mathcal O(X) / \mathfrak m_P \bigr) = \dim_{\mathbb C} \bigl( \mathbb C \bigr) = 1.

(iii) \implies (iv). Say that \mathfrak m_P \, \mathcal O_P = \varpi \, \mathcal O_P as a principal ideal. In order to show that \mathcal O_P is a discrete valuation ring, it suffices to show that any nonzero x \in \mathcal O_P is in the form x = {\varpi}^m \, y for some m \in \mathbb Z and y \in \mathcal O_P^\times, because then we have a discrete valuation \text{ord}_P: K^\times \twoheadrightarrow \mathbb Z which sends x \mapsto m. Consider the radical of the ideal generated by x:
\sqrt{\langle x \rangle} = \left \{ y \in \mathcal O_P \, \biggl| \, y^n \in x \, \mathcal O_P \ \text{for some nonnegative integer} \ n \right \}.
As \mathcal O_P has a unique nonzero prime ideal, we must have \sqrt{\langle x \rangle} = \mathfrak m_P \, \mathcal O_P. But then there is largest nonnegative integer m such that \varpi^{m-1} \notin x \, \mathcal O_P yet \varpi^m \in x \, \mathcal O_P. Hence y = x/\varpi^m \in \mathcal O_P but y \notin \mathfrak m_P.

(iv) \implies (v). Say that \mathcal O_P is a discrete valuation ring. Since K is the quotient field of \mathcal O(X), we see that \mathcal O(X) \subseteq \mathcal O_P \subseteq K. Say that x \in K is a root of a polynomial equation x^n + a_1 \, x^{n-1} + \cdots + a_n = 0 for some a_i \in \mathcal O_P. Assume by way of contradiction that x \notin \mathcal O_P. Then \text{ord}_P(x) < 0, so that \text{ord}_P(1/x) > 0, and hence y = 1/x is an element of \mathcal O_P. Upon dividing by x^{n-1} we have the relation x = - \bigl( a_1 + a_2 \, y + \cdots + a_n \, y^{n-1} \bigr) \in \mathcal O_P. This contradiction shows that \mathcal O_P is indeed integrally closed.

(v) \implies (iii). Say that \mathcal O_P is integrally closed. We must construct an element \varpi \in \mathcal O_P such that \mathfrak m_P \, \mathcal O_P = \varpi \, \mathcal O_P. Fix a nonzero x \in \mathfrak m_P. By considering the radical \sqrt{\langle x \rangle} and noting that \mathfrak m_P \, \mathcal O_P is a finitely generated \mathcal O_P-module, we see that there exists some m \in \mathbb Z such that \mathfrak m_P^m \, \mathcal O_P \subseteq x \, \mathcal O_P yet \mathfrak m_P^{m-1} \, \mathcal O_P \not \subseteq x \, \mathcal O_P. Choose y \in \mathfrak m_P^{m-1} such that y \notin x \, \mathcal O_P, and let \varpi = x/y be an element in K. Consider the module (1/\varpi) \, \mathfrak m_P \, \mathcal O_P \subseteq \mathcal O_P; we will show equality. As y \notin \mathcal O_P, we have 1/\varpi \notin \mathcal O_P, so that 1/\varpi is not integral over \mathcal O_P. Then (1/\varpi) \, \mathfrak m_P\, \mathcal O_P cannot be a finitely generated \mathcal O_P-module, we have (1/\varpi) \, \mathfrak m_P \, \mathcal O_P \not \subseteq \mathfrak m_P. As there is an element of (1/\varpi) \, \mathfrak m_P\, \mathcal O_P which is not in \mathfrak m_P, we must have equality: (1/\varpi) \, \mathfrak m_P \, \mathcal O_P = \mathcal O_P. Hence \mathfrak m_P \, \mathcal O_P = \varpi \, \mathcal O_P as desired.

(v) \iff (vi). A Dedekind domain is a Noetherian integral domain of dimension 1 that is integrally closed. But the localization \mathcal O_P is integrally closed for each maximal ideal \mathfrak m_P if and only if \mathcal O is integrally closed. (Consult Theorem 5.13 on page 63 of Atiyah and Macdonald’s Introduction to Commutative Algebra.) \square

General Remarks

An algebraic curve X \simeq \text{mSpec} \, \mathcal O(X) is non-singular if and only if the coordinate ring \mathcal O(X) is a Dedekind domain. Conversely, given any Dedekind Domain \mathcal O, we can define a “non-singular algebraic curve” by setting X = \text{mSpec} \, \mathcal O — although it is more common to write X = \text{Spec} \, \mathcal O. For example, \text{Spec} \, \mathbb Z is a non-singular curve. This is a strange concept because there is no defining equation for X in general!


About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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2 Responses to Lecture 14: Friday, September 20, 2013

  1. Pingback: Lecture 15: Monday, September 23, 2013 | Lectures on Dessins d'Enfants

  2. Pingback: MA 59800 Course Syllabus | Lectures on Dessins d'Enfants

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