## 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 $X$.

Proposition.
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.

Theorem.
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:

• (i) For each $P \in X$, the $m \times n$ matrix
$\text{Jac}_P(X) = \left[ \begin{matrix} \dfrac {\partial f_1}{\partial z_1}(P) & \dfrac {\partial f_1}{\partial z_2}(P) & \cdots & \dfrac {\partial f_1}{\partial z_n}(P) \\[10pt] \dfrac {\partial f_2}{\partial z_1}(P) & \dfrac {\partial f_2}{\partial z_2}(P) & \cdots & \dfrac {\partial f_2}{\partial z_n}(P) \\[10pt] \vdots & \vdots & \ddots & \vdots \\[5pt] \dfrac {\partial f_m}{\partial z_1}(P) & \dfrac {\partial f_m}{\partial z_2}(P) & \cdots & \dfrac {\partial f_m}{\partial z_n}(P) \end{matrix} \right]$
yields an exact sequence:
$\{ 0 \} \longrightarrow T_P(X) \longrightarrow \mathbb A^n(\mathbb C)\overset{\text{Jac}_P(X)}{\longrightarrow} \mathbb A^m(\mathbb C) \longrightarrow \{ 0 \}.$
That is, the Jacobian matrix $\text{Jac}_P(X)$ has rank $m$ while the tangent space has dimension $\text{dim}_{\mathbb C} \bigl( T_P(X) \bigr) = \text{dim}(X)$ as a complex vector space.

• (ii) The Zariski cotangent space has dimension $\text{dim}_{\mathbb C} \bigl( \mathfrak m / {\mathfrak m}^2 \bigr) = \text{dim}(X)$ as a complex vector space for each maximal ideal $\mathfrak m \in \text{mSpec} \, \mathcal O(X)$.

• (iii) For each $P \in X$, denote $\mathcal O_P$ as the localization of $\mathcal O(X)$ at the maximal ideal $\mathfrak m_P$. Then $\mathfrak m_P \, \mathcal O_P$ is a principal ideal.

• (iv) For each $P \in X$, the localization $\mathcal O_P$ is a discrete valuation ring, that is there exists a surjective map $\text{ord}_P: K \twoheadrightarrow \mathbb Z \cup \{ \infty \}$ such that the following properties hold.

• Multiplicative: $\text{ord}_P(x \, y) = \text{ord}_P(x) + \text{ord}_P(y)$.

• Ultrametric Inequality: $\text{ord}_P(x + y) \geq \min \bigl \{ \text{ord}_P(x), \, \text{ord}_P(y) \bigr \}$.

• Non-Degenerate: $\text{ord}_P(x) = \infty$ if and only if $x = 0$, and $\text{ord}_P(x) = 0$ if and only if $x \in \mathcal O(X)^\times$.

• (v) For each $P \in X$, the localization $\mathcal O_P$ is integrally closed.

• (vi) $\mathcal O(X)$ is a Dedekind Domain.

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!