Algebraic Geometry

Remark 1. Note that we only define codimension when \(Y\) is irreducible. When \(Y\) is reducible, sometimes one defines \(\operatorname{codim}_X(Y)\) to be the minimum value of \(\operatorname{codim}_X(Y')\) where \(Y'\) ranges over the irreducible components of \(Y\text{.}\)

Remark 1. Topologize \(\mathbb{N}\) by declaring a subset to be closed if and only if it is of the form \(\{0, 1, \dots, n\}\) for some \(-1 \leq n \leq \infty\text{.}\) Then \(\mathbb{N}\) is a noetherian space but has infinite dimension. There exist noetherian commutative rings \(R\) such that \(\operatorname{dim}\operatorname{Spec}(R) = \infty\text{.}\) (This cannot occur for local rings nor for rings that are finitely generated as \(k\)-algebras.)

Example 1. Let \(X = V(xy,xz)\text{,}\) the union of the \(x\)-axis and the \(yz\)-plane. The dimension of \(X\) is two. It is clearly at least two, since we have the chain with \(Y_0\) being any point on the \(yz\)-plane, \(Y_1\) being any line in this plane and containing this point, and \(Y_2\) being the whole plane. It is harder to see that it is at most two. This is an example of a variety that fails to be “equi-dimensional” — the variety has two irreducible components, but they have different dimensions.

Let \(Y = V(x) \subseteq X\text{,}\) the \(yz\)-plane. Then \(\operatorname{codim}_X(Y) = 0\) since the only closed subsets strictly between \(Y\) and \(X\) are given by \(Y\) together with a finite set of points on the line.

Let \(Z = V(y,z)\text{,}\) the \(x\)-axis. Then \(\operatorname{codim}_X(Z)\) is also \(0\) by similar reasoning. Since \(\operatorname{dim}(Z) = 1\text{,}\) this is a bit counter-intuitive.

Let \(W = V(x,y,z)\text{.}\) When \(\operatorname{codim}_X(W) = 2\text{.}\) It’s at least two since we have the chain \(W \subsetneq V(x,y) \subsetneq Y \subseteq X\text{;}\) it is less obvious, but true, that it is exactly two.

According to Remark [cross-reference to target(s) "rem131" missing or not unique], what is \(\operatorname{codim}_{\mathbb{A}^3}(X)\text{?}\)

The following result follows from the fact that there is an order-reversing bijections between primes ideals of \(A(X)\) and irreducible closed subsets of \(X\text{:}\)

Proposition 1. The dimension of an affine variety \(X\) is equal to the Krull dimension of its coordinate ring \(A(X)\text{.}\) The codimension of an irreducible closed subset \(Y\) of \(X\) is the height of the corresponding prime ideal of \(A(X)\text{.}\)

Remark 1. Using the definition in Remark [cross-reference to target(s) "rem131" missing or not unique], the codimension of an arbitrary closed subset \(Y\) is the height of the corresponding radical ideal.

Theorem 1 (Properties of dimension). Assume \(k\) is algebraically closed and \(X\) and \(Y\) are irreducible affine \(k\)-varieties

All of these results are best proven in a class on commutative algebra, and so I’ll just skip their proofs.

Example 1. If \(f\) is any non-zero, non-contant polynomial in \(k[x_1, \dots, x_n]\text{,}\) then \(\operatorname{dim}V(f) = n - 1\text{.}\)

Exercise 1. What goes wrong in this Theorem if we don’t assume \(X\) and \(Y\) are irreducible?

Corollary 1. The dimension of \(\mathbb{A}^n_k\) is \(n\text{.}\)

Proof. We have \(\mathbb{A}^n_k = \mathbb{A}^1_k \times \cdots \times \mathbb{A}^1_k\) and \(\operatorname{dim}(\mathbb{A}^1_k) = 1\text{.}\) The latter holds since \(k[x]\) is a PID. ◻

A special case of the Catenary Property is:

Corollary 1. For every point \(P\) of an irreducible variety \(X\text{,}\) we have \(\operatorname{codim}_X(P) = \operatorname{dim}(X)\text{.}\)

The previous result says that the dimension of an irreducible variety is that same near every point. In algebra terms, it says that if \(R\) is an integral domain and is a finitely generated \(k\)-algebra, then \(\operatorname{dim}R_{\mathfrak m}\) is the same for all maximal ideals \({\mathfrak m}\text{.}\)

Corollary 1. Assume \(X\) is a (not necessarily irreducible) affine variety \(X\) and let \(f \in A(X)\text{.}\) Let \(X = X_1 \cup \cdots \cup X_m\) be the decomposition of \(X\) into its irreducible components and assume they are ordered so that \(\operatorname{dim}(X_i) = \operatorname{dim}(X)\) for \(1 \leq i \leq k\) and \(\operatorname{dim}(X_i) < \operatorname{dim}(X)\) for \(i > k\text{.}\)

For any \(i\) such that \(\emptyset \subsetneq V_X(f) \cap X_i \subsetneq X_i\) the Theorem applies (using that \(V_X(f) \cap X_i = V_{X_i}(f)\)) to give that \(V_X(f) \cap X_i\) is equi-dimensional of dimension equal to \(\operatorname{dim}(X_i) -1\text{.}\) If \(1 \leq i \leq k\text{,}\) then by assumption we have \(\operatorname{dim}(X_i) = \operatorname{dim}(X)\) and either (a) \(V_X(f) \cap X_i = \emptyset\) or (b) \(\emptyset \subsetneq V_X(f) \cap X_i \subsetneq X_i\text{,}\) and moreover condition (b) holds for at least one such \(i\text{.}\) It follows that \(\max \{ \operatorname{dim}V_X(f) \cap X_i \mid 1 \leq i \leq k\} =\operatorname{dim}(X) - 1\text{.}\) If \(i > k\text{,}\) then \(\operatorname{dim}(V_X(f) \cap X_i) \leq \operatorname{dim}(X_i) \leq \operatorname{dim}(X) - 1\text{.}\) The first assertion thus follows from [cross-reference to target(s) "E22" missing or not unique].

If \(k = m\text{,}\) then for each \(i\) we have that \(V_X(f) \cap X_i\) is either empty or is equi-dimensional of dimension \(\operatorname{dim}(X) -1\text{,}\) and it is non-empty for at least one \(i\text{.}\) The second assertion follows. ◻

Example 1. Take \(X = V(xy, xz)\text{.}\) Let \(f = x-1\text{.}\) Then \(\operatorname{dim}V_X(f) = 0 = \operatorname{dim}(X) - 2\text{.}\) Why doesn’t this contradict the Corollary? Let \(g = x\text{.}\) Then \(\operatorname{dim}V_X(g) = 2 = \operatorname{dim}(X)\text{.}\) Why doesn’t this contradict the Corollary?

Let \(h = y-z\text{.}\) Then the Corollary does apply, and we have \(V_X(h) = 1 = \operatorname{dim}(X) -1\text{.}\)

Example 1. Note that if \(X = V_{{\mathbb{R}}}(z - x^2 - y^2)\) and \(f = z\text{,}\) then \(X \cap V_{{\mathbb{R}}}(f) = \{(0,0,0)\}\text{.}\) Why does this not contradict Krull’s Theorem?

Example 1. Say \(f, g \in k[x_1, \dots, x_n]\) are non-constant polynomials with \(n \geq 2\) and assume that \(f\) and \(g\) have no common irreducible factors and that \(V(f,g) \ne \emptyset\text{.}\) Then \(\operatorname{dim}V(f,g) = n - 2\text{.}\)

To see this, recall the irreducible components of \(V(f)\) are \(V(p_i)\) where \(f = \prod_i p_i^{e_i}\text{,}\) and they are each of dimension \(n-1\text{.}\) The fact that \(p_i\) is not a factor of \(g\) gives that \(g\) is not identically \(0\) on \(V(p_i)\text{,}\) for all \(i\text{.}\) So, the Corollary applies.