Understanding Varieties

Varieties are a basic structure in algebraic geometry. They were the central objects of study before Grothendieck reinvented the entire theory in his treatise Éléments de géométrie algébrique by introducing schemes. In the following we will introduce varieties and define algebraic curves.

We will restrict our discussion to varieties over the real or complex numbers. As such, let k be \mathbb{R} or \mathbb{C}.

Affine Varieties

We begin by discussing varieties over affine space. Let \mathbb{A}^{n} = \mathbb{A}^{n}(k) = \{(x_{1},\ldots,x_{n}) \mid x_{i} \in k\} be affine n-space over k (this is just k^{n} as a set but we’re “forgetting” the vector space structure). We will denote points of \mathbb{A}^{n} by P. Let k[X] = k[X_{1},\ldots,X_{n}] denote the polnomial ring in n variables over k. If I \in k[X] is an ideal, we call the subset

V_{I} = \{P \in \mathbb{A}^{n} \mid f(P) = 0 \,\, \forall \,\, f \in I\}

of \mathbb{A}^{n} an affine algebraic set. By construction V_{I} is the common zero-set of I. In other words, V_{I} is the largest set for which all functions f \in I vanish on. It is best to illustrate this with a few examples.

  • Affine n-space \mathbb{A}^{n} is an algebraic set if I is the zero ideal.
  • Let I = (X^{2}+Y^{2}-1) \subset \mathbb{R}[X,Y] be the ideal generated by X^{2}+Y^{2}-1. Since elements of I are multiples of X^{2}+Y^{2}-1, any P \in V_{I}, satisfies X^{2}+Y^{2}-1 = 0 or equivalently X^{2}+Y^{2} = 1. Thus V_{I} is the unit circle.
  • Let I = ((X-Y)(Y-X^{2})) \subset \mathbb{R}[X,Y] be the ideal generated by (X-Y)(Y-X^{2}). Similar to the previous example, any P \in V_{I} satisfies X = Y or Y = X^{2}. Therefore V_{I} is the union of the line X = Y and the parabola Y = X^{2}.

The takeaway from these examples is that algebraic sets, geometrically, are curves (and in higher dimensions surfaces, hypersurfaces, etc.). We say that an affine algebric set is an affine algebraic variety if the ideal I corresponding to V_{I} is prime. In the above examples, the first two are algebraic varieties while the third is not. Notice that the algebraic set in the last example turned out to be a union of two graphs. This holds more generally; affine algebraic varieties are irreducible in the sense that they cannot be written as nontrivial unions of smaller algebraic sets.

In general, affine space is not “nice enough” to work with. For example, elliptic curves are certian varieties with a group structure and affine space does not allow for a natural choice of identity element in the group. To get around this, we will work with varities over projective space which resolves this issue by introducing specified points “at infinity”. In the case of elliptic curves, there is a single point “at infity” which will be a natural choice for the identity element.

Projective Varieties

We define projective n-space over k, denoted by \mathbb{P}^{n} = \mathbb{P}^{n}(k), to be the set of all equivalence classes of (n+1)-tuples (x_{0},\ldots,x_{n}) with at least one x_{i} \neq 0 where (x_{0},\ldots,x_{n}) \sim (y_{0},\ldots,y_{n}) if there exists a \lambda \in k such that x_{i} = \lambda y_{i} for all i. We denote the class of (x_{0},\ldots,x_{n}) by [x_{0}:\ldots:x_{n}] and points by P. Since any line in affine n-space is of the form

\{(tx_{0},\ldots,tx_{n}) \in \mathbb{A}^{n+1} \mid t \in k\}

and all of these points are identified in \mathbb{P}^{n} (in fact this is exactly the class [x_{0}:\ldots:x_{n}]), one can think of \mathbb{P}^{n} as the set of lines in \mathbb{A}^{n+1}. We call the set

H = \{[x_{0}:\ldots:x_{n}] \in \mathbb{P}^{n} \mid x_{0} = 0\}

the points at infinity of \mathbb{P}^{n}. To see why this is the case, notice that the map

\phi:\mathbb{P}^{n}-H \to \mathbb{A}^n \qquad [x_{0}:\ldots:x_{n}] \mapsto (x_{1}/x_{0},\ldots,x_{n}/x_{0})

identifies \mathbb{P}^{n}-H and \mathbb{A}^{n}. Moreover the map

\psi:H \to \mathbb{P}^{n-1} \qquad [x_{0}:\ldots:x_{n}] \mapsto [x_{1}:\ldots:x_{n}]

identifies H and \mathbb{P}^{n-1}. So, \mathbb{P}^{n} can also be thought of as a copy of \mathbb{A}^{n} and the points at infinity H (or \mathbb{P}^{n-1}). Moreover, we could of required any x_{i} = 0 to get different copies of H and \mathbb{A}^{n} inisde \mathbb{P}^{n}.

  • Notice \mathbb{P}^{0} always consists of a single point. Thus \mathbb{P}^{1}(\mathbb{C}) is a plane (the complex plane) with a point at infinity while \mathbb{P}^{1}(\mathbb{R}) is a line (the real line) with a point at infinity.

Index the variables of k[X] by X_{0},\ldots,X_{n}. We say that a polynomial f \in k[X] is homogenous of degree d if

f(\lambda X_{0},\ldots,\lambda X_{n}) = \lambda^{d}f(X_{0},\ldots,X_{n})

for all \lambda \in k. We say an ideal I \in k[X] is homogenous if it is generated by homogenous polynomials. Let I_{h} denote the subset of I consisting of homogenous polynomials. If P \in \mathbb{P}^{n}, it makes sense to ask if f(P) = 0 since this is independent of the representative of [x_{0}:\ldots:x_{n}]. To each homogenous ideal I, we call the subset

V_{I} = \{P \in \mathbb{P}^{n} \mid f(P) = 0 \,\, \forall \,\, f \in I_{h}\}

of \mathbb{P}^{n} a projective algebraic set. Similar to the affine case, V_{I} is the common zero-set of I_{h}.

  • Projective n-space \mathbb{P}^{n} is a projective algebraic set if I is the zero ideal.
  • A line in \mathbb{P}^{2} is the algebraic set corresponding to a homogenous ideal (aX+bY+cZ) \subset k[X,Y,Z].

A projective algebraic variety is a projective algebraic set where the ideal I is prime. Similar to the affine case, projective algebraic varieties are irreducible in the sense that they cannot be written as nontrivial unions of smaller projective algebraic sets.

Coordinate Rings and Dimension

To give both affine and projective varities the notion of dimension, we need to introduce coordinate rings. Apart from letting us define the dimension of a variety coordinate rings also play a critial role in the general theory.

The defintion is the same in both the affine and projective setting. So, let V be a variety (either affine or projetive) with prime ideal I \subset k[X]. The coordinate ring of V is the quotient ring k[V] defined by

k[V] = k[X]/I.

This defintion should make sense because the elements (which we also call polynomials) of k[V] are defined on V up to a polynomial which vanishes on V. In other words, an element of k[V] is an equivalence class of polynomials where elements of a class agree when their domain is restricted to V. Since k is a field, k[X] is an integral domain and moreover k[V] is an integral domain because I is prime. Therefore, we define the function field k(V) of V to be the field of fractions of k[V].

If R is a commutative ring, then we say that a chain of prime ideals of the form

P_{0} \subsetneq P _{1} \subsetneq \cdots \subsetneq P_{n}

has length n (so the length is the number of strict inclusions). We define the Krull dimension of R to be the supremum of chains of prime ideals of R. In particular, the Krull dimension need not be finite. If V is an affine variety, then the dimension of V, denoted dim(V) is defined to be the Krull dimension of k[V] (this ring is commutative since k[X] is commutative and quotients respect commutativity). The Krull dimension is, in general, not easy to calculate. But we do have a nice theorem:

Theorem: If V is an affine variety, then the dimension of V is equal to the trancendence degree of k(V) over k.

In particular, this tells is that the dimension of V is finite. This also makes computing the dimension of an affine variety significantly easier.

  • The dimension of \mathbb{A}^{n} is n because k[\mathbb{A}^{n}] = k[X]/I = k[X] so that k(\mathbb{A}^{n}) = k(X_{1},\ldots,X_{n}), and this clearly has trancendence degree n over k.
  • If the ideal corresponding to an affine variety V \subset \mathbb{A}^{n} is generated by a single nonconstant polynomial equation, then the dimension of the variety is n-1. This is because f relates any of the X_{i} by a polynomial (hence algebraic) relation in the other variables.

So why didn’t we just define the dimension of an affine variety to be the trancendence degree of k(V) over k? It’s because the Krull dimension definition is intuitive. Standard ring theory tells us that prime ideals of k[V] are in bijective correspondence with prime ideals of k[X] containing I. If J \supseteq I is prime then V_{J} \subseteq V_{I} as varieties (we are being verbose here by writing V_{I} for V). In this case we say V_{J} is an affine subvariety of V_{I} (there is an obvious analgous defintion for projective subvariety). So, if

P_{0} \subsetneq P_{1} \subsetneq \cdots \subsetneq P_{n}

is maximal chain (we can say maximal since the dimension of V is finite) of prime ideals in k[V], then there is a correponding chain of varieties

V_{P_{0}} \supsetneq V_{P_{1}} \supsetneq \cdots \supsetneq V_{P_{n}}.

This correspondence is bijective because the previous correspondence is. So geometrically, the only way to make a variety smaller is by reducing its dimension (because then the corresponding chain of prime ideals decreases in length). To define the notion of dimension for projective varities we need a proposition.

Proposition: Let V \subset \mathbb{P}^{n} be a projective variety. Then for any P \in V there exists a copy of \mathbb{A}^{n} inside \mathbb{P}^{n} such that P \in V \cap \mathbb{A}^{n} and V \cap \mathbb{A}^{n} is an affine variety.

With this proposition, we define the dimension of a projective variety V, denoted dim(V), to be the dimension of V \cap \mathbb{A}^{n} as an affine variety. This is well-defined as it can be shown that the dimension of V \cap \mathbb{A}^{n} is independent of the copy of \mathbb{A}^{n}. The idea behind the definition of dimension for projective varieties is that the points at infinity shouldn’t be accounted for because they are a dimension lower than the affine part (recall \mathbb{P}^{n} can be viewed as \mathbb{A}^{n} together with H \cong \mathbb{P}^{n-1}).

The Defintion of Algebriac Curves

An algebraic curve (or simply a curve) is just a projective variety of dimension one. If f \in \mathbb{C}[X,Y] is a homogenous polynomial and V is the projective variety with corresponding ideal (f), then it’s not too hard to show V is an algebraic curve. To compute the dimension, choose a copy of \mathbb{A}^{1} inside \mathbb{P}^{1}. Then show f|_{V \cap \mathbb{A}^{1}} generates the ideal corresponding to V \cap \mathbb{A}^{1} and use it to prove \mathbb{C}[V \cap \mathbb{A}^{1}] is of transcendence degree one over \mathbb{C}.


The Arithmetic of Elliptic Curves – Joseph Silverman

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