Eventually, we wish to show that every compact, connected Riemann surface 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 in the polynomial ring , and define the set . We say that is an algebraic variety if is a prime ideal. In this case, the quotient ring is an integral domain. We define as the global sections of or the coordinate ring of .

Proposition.

Let be an algebraic variety. Then the map which sends a point to the maximal ideal is a bijection.

*Proof:* Recall that an ideal is maximal if and only if the quotient is a field. It is clear that the map is well-defined and injective. Let be any maximal ideal of . Fix a surjection , and denote as the image of . It is easy to check that for .

Often, we abuse notation and write . If we denote as the quotient field of the integral domain , we define the dimension of as the transcendence degree of over .

### Non-singular Algebraic Curves

We say that is an algebraic curve if . Here is the main result for today’s lecture.

Theorem.

Consider a prime ideal in the polynomial ring , and define the algebraic variety . Assume that is an algebraic curve, that is, that . Then the following are equivalent:

- (i) For each , the matrix

yields an exact sequence:

That is, the Jacobian matrix has rank while the tangent space has dimension as a complex vector space.

- (ii) The Zariski cotangent space has dimension as a complex vector space for each maximal ideal .

- (iii) For each , denote as the localization of at the maximal ideal . Then is a principal ideal.

- (iv) For each , the localization is a discrete valuation ring, that is there exists a surjective map such that the following properties hold.

Multiplicative:.

Ultrametric Inequality:.

Non-Degenerate:if and only if , and if and only if .

- (v) For each , the localization is integrally closed.

- (vi) is a Dedekind Domain.

If any of these equivalent statements holds true, we say that is a non-singular algebraic curve. As is defined by the polynomial equations for , the tangent space at a point is the complex vector space

The Russian mathematician Oscar Zariski (1899 — 1986) realized that is the dual to the quotient for the maximal ideal , 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) (ii). We have a perfect (i.e., bilinear and nondegenerate) pairing defined by . Hence .

(ii) (iii). As is a maximal ideal, Nakayama’s Lemma states that we can find where . Consider the injective map defined by . Clearly this is surjective if and only if is principal. Recall now that .

(iii) (iv). Say that as a principal ideal. In order to show that is a discrete valuation ring, it suffices to show that any nonzero is in the form for some and , because then we have a discrete valuation which sends . Consider the radical of the ideal generated by :

As has a unique nonzero prime ideal, we must have . But then there is largest nonnegative integer such that yet . Hence but .

(iv) (v). Say that is a discrete valuation ring. Since is the quotient field of , we see that . Say that is a root of a polynomial equation for some . Assume by way of contradiction that . Then , so that , and hence is an element of . Upon dividing by we have the relation . This contradiction shows that is indeed integrally closed.

(v) (iii). Say that is integrally closed. We must construct an element such that . Fix a nonzero . By considering the radical and noting that is a finitely generated -module, we see that there exists some such that yet . Choose such that , and let be an element in . Consider the module ; we will show equality. As , we have , so that is not integral over . Then cannot be a finitely generated -module, we have . As there is an element of which is not in , we must have equality: . Hence as desired.

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

### General Remarks

An algebraic curve is non-singular if and only if the coordinate ring is a Dedekind domain. Conversely, given *any* Dedekind Domain , we can define a “non-singular algebraic curve” by setting — although it is more common to write . For example, is a non-singular curve. This is a strange concept because there is no defining equation for in general!

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