This post was machine-translated from Korean by Kimi CLI. The translation may contain errors. The Korean original is the source of truth.
In geometry, dimension is one of the most basic invariants. In algebraic geometry as well, dimension is equally important, and there are several equivalent ways to define it. In this post we examine various ways to define the dimension of a variety.
Dimension as a Topological Space
An algebraic variety is already a topological space, so using [Topology] §Dimension, ⁋Definition 10, we can define the dimension of \(X\) as the supremum of the lengths of strictly descending chains of irreducible closed subsets.
Example 1 In \(\mathbb{A}^1\), the closed sets are only \(\mathbb{A}^1\) itself and finite sets. Thus the longest chain is \(\mathbb{A}^1 \supsetneq \{p\} \supsetneq \emptyset\), which is a chain of length \(2\), so \(\mathbb{A}^1\) is \(1\)-dimensional according to this definition.
This definition has the advantage of defining dimension from a purely topological point of view. However, in practice it is not very efficient for computation because we need to know all chains of irreducible closed subsets.
Dimension of an Affine Variety
Meanwhile, we already know that the relationship between an algebraic variety and the functions defined on it is very close. Then, it would not be so surprising that the algebraic structure of functions on an algebraic variety contains information about the dimension. To approach from this perspective, it is best to look at the case of an affine variety, where its coordinate ring \(\mathbb{K}[X]\) is neatly given.
Proposition 2 The dimension of an affine variety \(X\) is equal to the Krull dimension of its coordinate ring \(\mathbb{K}[X]\). ([Commutative Algebra] §Dimension, ⁋Definition 1)
Proof
From §Affine Varieties, ⁋Proposition 12, there is a one-to-one correspondence between irreducible closed subsets of an affine variety and prime ideals of \(\mathbb{K}[X]\).
Corollary 3 \(\dim \mathbb{A}^n = n\).
Meanwhile, for any prime ideal \(\mathfrak{p}\subset \mathbb{K}[\x_1,\ldots, \x_n]\), we know that the following equation
\[\dim \mathbb{K}[\x_1,\ldots, \x_n]/\mathfrak{p}+\codim \mathfrak{p}=\dim \mathbb{K}[\x_1,\ldots, \x_n]=n\tag{$\ast$}\]holds. ([Commutative Algebra] §Regular Local Rings, ⁋Proposition 4) Here, the codimension of \(\mathfrak{p}\) is defined in [Commutative Algebra] §Dimension, ⁋Definition 2, as the supremum of the lengths of chains of prime ideals contained in \(\mathfrak{p}\), and geometrically it is the supremum of the lengths of chains of closed subvarieties of \(\mathbb{A}^n\) containing \(X=Z(\mathfrak{p})\). Geometrically, since we know that \(\dim \mathbb{K}[\x_1,\ldots, \x_n]/\mathfrak{p}\) is the dimension of \(Z(\mathfrak{p})\), we can give ($\ast$) a geometric meaning through this.
Dimension of a Projective Variety
The problem arises when we move to projective varieties. Recall that global functions on \(\mathbb{P}^n\) were only constant functions. In this situation, to define the dimension of a projective variety, we can take an affine chart. That is, given \(X\subset \mathbb{P}^n\), we choose an affine open chart \(U_i\) of \(\mathbb{P}^n\) and consider the dimension of \(X_i=X\cap U_i\) as an affine variety. However, for this definition we would need to show that the dimension of any open subset equals that of the original variety, so we cannot adopt this as a definition yet. Instead, we use the affine cone \(C(X)\) of \(X\).
For a projective variety \(X\subseteq \mathbb{P}^n\), the affine cone \(C(X)\subseteq \mathbb{A}^{n+1}\) is the affine variety in \(\mathbb{A}^{n+1}\) defined by the homogeneous ideal defining \(X\), viewed as an ideal in \(\mathbb{K}[\x_0,\ldots, \x_n]\). That is, for the homogeneous ideal \(I(X)\) defining \(X\), if we define the ring \(S(X)\) as
\[S(X)=\mathbb{K}[\x_0,\ldots, \x_n]/I(X)\]this becomes the coordinate ring of the affine cone. The following result is key for computing the dimension of a projective variety.
Proposition 4 For a projective variety \(X \subseteq \mathbb{P}^n\), \(\dim X = \dim S(X) - 1\).
This can be shown by computation in graded rings. In particular, \(\dim C(X) = \dim X + 1\), and from this we obtain the following.
Proposition 5 \(\dim \mathbb{P}^n = n\).
Proof
The cone of \(\mathbb{P}^n\) is \(\mathbb{A}^{n+1}\) and \(\dim \mathbb{A}^{n+1} = n+1\), so \(\dim \mathbb{P}^n = (n+1) - 1 = n\).
Dimension of a Hypersurface
A hypersurface is a variety defined as the zero set of a single polynomial. Intuitively, adding one equation is the same as imposing one constraint, so it should reduce the dimension by one.
Proposition 6 For an irreducible polynomial \(f \in \mathbb{K}[\x_1, \ldots, \x_n]\), the irreducible hypersurface \(Z(f) \subset \mathbb{A}^n\) has dimension \(n - 1\).
Proof
Since \(f\) is irreducible, \((f)\) is a prime ideal, and thus the coordinate ring of \(Z(f)\) is \(\mathbb{K}[\x_1, \ldots, \x_n]/(f)\). Now let us show that the codimension of \((f)\) in \(\mathbb{K}[\x_1, \ldots, \x_n]\) is 1. \((0) \subsetneq (f)\) is a chain of length 1, so \(\codim(f) \ge 1\). On the other hand, in the UFD \(\mathbb{K}[\x_1, \ldots, \x_n]\), every prime ideal of codimension 1 is a principal prime ideal, so no other prime ideal can lie between \((0)\) and \((f)\). Therefore \(\codim(f) = 1\), and
\[\dim \mathbb{K}[\x_1, \ldots, \x_n]/(f) = \dim \mathbb{K}[\x_1, \ldots, \x_n] - \codim(f) = n - 1\]In general, for a regular local ring \(R\) and a prime ideal \(\mathfrak{p}\), we have \(\dim R/\mathfrak{p} = \dim R - \codim(\mathfrak{p})\). ([Commutative Algebra] §Regular Local Rings, ⁋Proposition 4)
Dimension via Function Fields
Another way to define dimension is by using the function field. The function field \(\mathbb{K}(X)\) carries information about the generic point of the variety, and it is also a birational invariant. The following proposition is also deduced from an algebraic fact.
Proposition 7 The dimension of a variety \(X\) equals the transcendence degree of its function field \(\mathbb{K}(X)\) over \(\mathbb{K}\).
Example 8 The following are examples of computing dimension via the function field.
- \(\mathbb{K}(\mathbb{A}^n) = \mathbb{K}(x_1, \ldots, x_n)\), and \(x_1, \ldots, x_n\) are algebraically independent over \(\mathbb{K}\), so \(\dim \mathbb{A}^n = n\).
- \(\mathbb{K}(V(\y - \x^2)) = \mathbb{K}(x)\), and \(x\) is algebraically independent over \(\mathbb{K}\), so \(\dim V(\y - \x^2) = 1\). This agrees with the intuition that a parabola is a curve.
- \(\mathbb{K}(\mathbb{P}^n) = \mathbb{K}(\x_1/\x_0, \ldots, \x_n/\x_0)\), and \(\x_1/\x_0, \ldots, \x_n/\x_0\) are algebraically independent over \(\mathbb{K}\), so \(\dim \mathbb{P}^n = n\). This reflects the fact that projective space is birationally equivalent to affine space.
Basic Properties of Dimension
The most basic property of dimension is that a proper subset has smaller dimension. This is geometrically obvious.
Proposition 9 For a closed subvariety \(Y \subsetneq X\) of \(X\), we have \(\dim Y < \dim X\).
Proof
Consider a maximal chain of closed subsets of \(Y\)
\[Y = Y_0 \supsetneq Y_1 \supsetneq \cdots \supsetneq Y_n \neq \emptyset\]Then
\[X \supsetneq Y = Y_0 \supsetneq Y_1 \supsetneq \cdots \supsetneq Y_n\]is a chain in \(X\) of length \(n+1\).
This can be thought of as a generalization of Proposition 6. Now let us examine the relationship between regular maps and dimension.
Proposition 10 For two varieties \(X, Y\) and a regular map \(\varphi: X \to Y\), the following hold.
- \(\dim \varphi(X) \le \dim X\).
- If \(\varphi\) is dominant, then \(\dim Y \le \dim X\). (§Rational Maps, ⁋Definition 8)
Proof
-
For a chain of closed subsets of \(\varphi(X)\),
\[Z_0 \supsetneq Z_1 \supsetneq \cdots \supsetneq Z_n\]their preimages
\[\varphi^{-1}(Z_0) \supsetneq \varphi^{-1}(Z_1) \supsetneq \cdots \supsetneq \varphi^{-1}(Z_n)\]form a chain of closed subsets in \(X\).
-
If \(\varphi\) is dominant, the pullback \(\varphi^\ast: \mathbb{K}(Y)\rightarrow \mathbb{K}(X)\) is injective, and thus we obtain the desired result from Proposition 7.
The first result supports our intuition that a geometric map cannot increase dimension in general. The second result roughly shows that if \(\varphi\) is surjective (up to birational equivalence), then the dimension of the target cannot exceed that of the domain.
Definition 11 A regular map \(\varphi: X \to Y\) between irreducible varieties is called finite if for every affine open \(U \subseteq Y\), the preimage \(\varphi^{-1}(U)\) is affine and \(\mathbb{K}[\varphi^{-1}(U)]\) is a finitely generated module over \(\mathbb{K}[U]\).
It can be shown that a finite morphism has finite fibers. Then the following is obvious.
Proposition 12 For two varieties \(X, Y\) and a finite map \(\varphi: X \to Y\), we have \(\dim X = \dim Y\).
Proof
If \(\varphi\) is finite, then at the coordinate ring level \(\mathbb{K}[X]\) is finitely generated as a \(\mathbb{K}[Y]\)-module. Hence \(\mathbb{K}(X)\) is a finite-degree extension of \(\mathbb{K}(Y)\), and the transcendence degrees are equal. Thus \(\dim X = \dim Y\).
Example 13 A \(k\)-dimensional linear subspace \(L\) of \(\mathbb{A}^n\) satisfies \(\dim L = k\). This is because \(L \cong \mathbb{A}^k\). Likewise, a \(k\)-dimensional linear subspace \(L\) of \(\mathbb{P}^n\) satisfies \(\dim L = k\).
Example 14 For two varieties \(X, Y \subseteq \mathbb{A}^n\), in general
\[\dim(X \cap Y) \ge \dim X + \dim Y - n\]This is called the dimension inequality. The reason this is an inequality is that, for example, in extreme situations such as \(X=Y\), the desired equality may not hold. When equality holds, we call it a proper intersection.
References
[Har] J. Harris, Algebraic Geometry: A First Course, Springer, 1992.
[Hart] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics, Springer, 1977.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Varieties in Projective Space, Springer, 2013.
[AM] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Algebra, Addison-Wesley, 1969.
댓글남기기