Commutative Algebra 62

Irreducible Subsets of Projective Space

Throughout this article, k is an algebraically closed field.

We wish to consider irreducible closed subsets of \mathbb P^n_k. For that we need the following preliminary result.

Lemma 1.

Let A be a graded ring; a proper homogeneous ideal \mathfrak p \subsetneq A is prime if and only if:

a, b \in A \text{ homogeneous}, ab \in \mathfrak p \implies a \in \mathfrak p \text{ or } b\in\mathfrak p.


Suppose \mathfrak p is not prime so there exists a, b\in A-\mathfrak p such that ab\in \mathfrak p. Since a\not\in \mathfrak p, among all homogeneous components of a pick a_d of maximum degree such that a_d \not\in\mathfrak p; similarly pick b_e fo b so b_e \not\in\mathfrak p.

The degree-(d+e) component of ab is congruent to a_d b_e mod \mathfrak p. Since \mathfrak p is homogeneous we have a_d b_e \in \mathfrak p, and by the given condition this means a_d \in \mathfrak p or b_e \in \mathfrak p, a contradiction. ♦

Exercise A

Prove that a proper homogeneous ideal \mathfrak q of a graded ring A is primary if and only if

a, b \in A \text{ homogeneous }, ab\in\mathfrak q \implies a \in \mathfrak q \text{ or } b\in r(\mathfrak q).

Proposition 1.

Suppose the closed subset V\subseteq \mathbb P^n_k corresponds to the homogeneous radical ideal \mathfrak a \subseteq B, \mathfrak a \ne B_+, where B = k[T_0, \ldots, T_n].

Then V is irreducible if and only if \mathfrak a is prime.


(⇒) Suppose V is irreducible; let \mathfrak a = I_0(V). If f, g \in B - I_0(V) are homogeneous, then C := V_0(\mathfrak a + fB) and D := V_0(\mathfrak a + gB) are closed subsets of \mathbb P^n_k properly contained in V. Since V is irreducible the following shows fg\not\in\mathfrak a:

V \supsetneq C \cup D = V_0(\mathfrak a + fB) \cup V_0(\mathfrak a + gB) = V_0((\mathfrak a + fB)(\mathfrak a + gB)) \supseteq V_0(\mathfrak a + fgB).

(⇐) Let V = V_0(\mathfrak p) where \mathfrak p is prime. Let C, D\subseteq V be closed subsets with union V. Now write C = V_0(\mathfrak a) and D = V_0(\mathfrak b) for homogeneous radical ideals \mathfrak a and \mathfrak b. Then V = C\cup D = V_0(\mathfrak a \cap \mathfrak b). Since \mathfrak a\cap \mathfrak b is a homogeneous radical ideal, \mathfrak p = \mathfrak a \cap \mathfrak b. By exercise B here, \mathfrak a = \mathfrak p or \mathfrak b = \mathfrak p. ♦

Corollary 1.

Let V\subseteq \mathbb P^n_k be a non-empty closed subset. Then V is irreducible if and only if \mathrm{cone}(V) is irreducible.


By proposition 1, V is irreducible if and only if I_0(V) is prime. But I_0(V) = I(\mathrm{cone}(V)) (from an exercise here) so the result follows. ♦


Quasi-projective Varieties

Recall that a projective variety is a closed subset of some \mathbb P^n_k.


A quasi-projective variety is an open subset of a projective variety.

This merely defines it as a set; we need a geometric structure on it.

First, let F_0, \ldots, F_m \in k[T_0, \ldots, T_n] be homogeneous polynomials of the same degree. If \mathbf v \in \mathbb P^n is such that not all F_i(\mathbf v) = 0, then we can define a function on an open subset U of \mathbb P^n containing \mathbf v as follows:

U \longrightarrow \mathbb P^m,\quad \mathbf v' \mapsto ( F_0(\mathbf v') : F_1(\mathbf v') : \ldots : F_m(\mathbf v') ).

The map is well-defined: indeed if F_i(\mathbf v) \ne 0 we can find an open neighbourhood U of \mathbf v such that 0 \not\in F_i(U). Also, if we replace projective coordinates (t_0 : \ldots : t_n) with (\lambda t_0 : \ldots : \lambda t_n), then each F_i(\lambda t_0, \ldots, \lambda t_n) = \lambda^d F_i(t_0, \ldots, t_n) where d = \deg F_i so

\begin{aligned} (F_0(\lambda t_0, \ldots, \lambda t_n) : \ldots : F_m(\lambda t_0, \ldots, \lambda t_n)) &= (\lambda^d F_0(t_0, \ldots, t_n) : \ldots: \lambda^d F_m(t_0, \ldots, t_n)) \\ &= (F_0(t_0, \ldots, t_n) : \ldots : F_m(t_0, \ldots, t_n)).\end{aligned}

We write (F_0 : \ldots : F_m) : U\to \mathbb P^m for the resulting function.


Let V \subseteq \mathbb P^n and W \subseteq \mathbb P^m be quasi-projective varieties and \phi : V\to W be a function.

We say \phi is regular at \mathbf v \in V if there is an open neighbourhood U of \mathbf v in V such that

\phi|_U = (F_0 : \ldots : F_m) for some homogeneous F_i \in k[T_0, \ldots, T_n] of the same degree.

We say \phi is regular if it is regular at every \mathbf v\in V, in which case we also say \phi : V\to W is a morphism of quasi-projective varieties.

From the above definitions, we obtain the category of all quasi-projective varieties and morphisms between them.

Example 1

First consider the case where V\subseteq \mathbb A^n and W\subseteq \mathbb A^m are closed subsets.

E.g., let V\subseteq \mathbb A^3. A regular map \phi : V\to \mathbb A^1 in the earlier sense can be expressed as a polynomial f(X, Y, Z), e.g. take f = X^3 - Y^2 + 3Z. Via embeddings \mathbb A^3 \hookrightarrow \mathbb P^3 and \mathbb A^1 \hookrightarrow \mathbb P^1 taking (x, y, z) \mapsto (1:x:y:z) and t \mapsto (1:t) respectively, f can be written in terms of homogeneous coordinates as

(T_0 : T_1 : T_2 : T_3) \mapsto (T_0^3 : T_1^3 - T_2^2 T_0 + 3 T_3 T_0^2)

since it is the homogenization of the map (\frac{T_1}{T_0}, \frac{T_2}{T_0}, \frac{T_3}{T_0}) \mapsto (\frac{T_1}{T_0})^3 - (\frac{T_2}{T_0})^2 + 3(\frac{T_3}{T_0}). This generalizes to an arbitrary regular map of closed subsets \phi : (V\subseteq \mathbb A^n) \to (W \subseteq \mathbb A^m).

Conversely we have:

Lemma 2.

Let \phi :V\to W be regular under the new definition. Then there exist polynomials f_1, \ldots, f_m \in k[X_1, \ldots, X_n] which represent \phi.


We will prove this for the case where V is irreducible.

For each of 1\le i\le m, let \pi_i : \mathbb A^m \to \mathbb A^1 be projection onto the i-th coordinate. Then \pi_i \circ \phi : V \to \mathbb A^1 is regular under the new definition, and by proposition 2 here (and its preceding discussion) \pi_i \circ \phi can be represented as a polynomial f_i(X_1, \ldots, X_n). Hence we see that

\phi(\mathbf v) = (f_1 (\mathbf v), \ldots, f_m(\mathbf v)) for polynomials f_1, \ldots, f_m \in k[X_1, \ldots, X_n]. ♦

Example 2

Take the map \phi : \mathbb P^1 \to \mathbb P^3 given by

\phi : (T_0 : T_1) \mapsto (T_0^3 : T_0^2 T_1 : T_0 T_1^2 : T_1^3)

Note that the same set of polynomials (F_0, F_1, F_2, F_3) works globally over the whole of \mathbb P^1.

Example 3

Suppose \mathrm{char} k \ne 2. Let V\subset \mathbb P^2 be the closed subset defined by T_0^2 = T_1^2 + T_2^2. We define a map \phi : V \to \mathbb P^1 as follows

  • Outside the point (1 : 1 : 0), take (T_0 : T_1 : T_2) \mapsto (T_0 - T_1 : T_2).
  • Outside the point (1 : -1 : 0), take (T_0 : T_1 : T_2) \mapsto (T_2 : T_0 + T_1).


The map agrees outside those two points since (T_0 - T_1 : T_2) = (T_2 : T_0 + T_1) due to the equality T_0^2 = T_1^2 + T_2^2.




Consider the category of all quasi-projective k-varieties, with morphisms defined as above. Two such varieties are said to be isomorphic if they are isomorphic in the category.

A quasi-projective variety is said to be

  • projective if it is isomorphic to a closed subset of some \mathbb P^n_k (this generalizes the existing definition of projective varieties);
  • affine if it is isomorphic to an affine k-variety (closed subspace of some \mathbb A^n);
  • quasi-affine if it is isomorphic to an open subset of an affine variety.

Example 4

In example 3 above, we get an isomorphism \phi : V\to \mathbb P^1 since we have the reverse map

\mathbb \psi : \mathbb P^1 \to V, \quad (U_0 : U_1) \mapsto (U_0^2 + U_1^2 : U_1^2 - U_0^2 : 2U_0 U_1).

As an exercise, prove that \phi\circ \psi = 1_{\mathbb P^1} and \psi\circ \phi = 1_V.


The coordinate ring of a quasi-projective variety V is the set

k[V] := \{ f : V\to \mathbb A^1 : f \text{ regular } \},

taken to be a k-algebra via point-wise addition and multiplication:

f, g : V\to\mathbb A^1 \text{ regular } \implies \begin{cases} (f+g) :V \to \mathbb A^1, \ &\mathbf v \mapsto f(\mathbf v) + g(\mathbf v), \\ (fg) : V\to \mathbb A^1, \ &\mathbf v \mapsto f(\mathbf v)g(\mathbf v). \end{cases}


As before, a regular map \phi:V\to W of quasi-projective varieties induces a ring homomorphism \phi^* : k[W] \to k[V]. By lemma 2, when V is affine k[V] agrees with our earlier version (we proved this in the case where V is irreducible).

Example 5

For each g\in GL_{n+1}(k), we have an automorphism 

\phi_g : \mathbb P^n_k \longrightarrow \mathbb P^n_k, \quad (t_0 : \ldots : t_n) \mapsto (\sum_{j=0}^n g_{0j} t_j : \ldots : \sum_{j=0}^n g_{nj} t_j).

Note that \phi_{gh} = \phi_g \circ \phi_h for g, h \in GL_{n+1}(k). Also \phi_g = 1 if and only if g is a scalar multiple of the identity matrix, so we get an injective homomorphism PGL_{n+1}(k) = GL_{n+1}(k)/k^* \hookrightarrow \mathrm{Aut} \mathbb P^n_k. In fact this is an isomorphism of groups.

E.g. when n = 1, we get the Möbius transformations:

\left[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2 k \right] : (t_0 : t_1) \mapsto (at_0 + bt_1 : ct_0 + dt_1).

Example 6

We have an isomorphism between the quasi-affine variety \mathbb A^1 - \{0\} and V = \{(x,y) \in \mathbb A^2 : xy = 1\} via the maps

\mathbb A^1 - \{0\} \to V, \ x \mapsto (x, \frac 1 x), \quad V \to \mathbb A^1 - \{0\}, \ (x, y) \mapsto x.

Hence \mathbb A^1-\{0\} is an affine variety even though it is not closed in \mathbb A^1. From the isomorphism we also have:

k[\mathbb A^1 - \{0\}] \cong k[V] = k[X, Y]/(XY - 1) \implies k[\mathbb A^1 - \{0\}] = k[X, \frac 1 X].

Example 7

Let V = \mathbb A^2 -\{(0, 0)\}. We will show that V is not affine. Indeed consider the injection \phi : V \hookrightarrow \mathbb A^2 which induces

\phi^* : k[X, Y] \cong k[\mathbb A^2] \longrightarrow k[V].

The map is injective since V is dense in \mathbb A^2. Let us show that it is surjective. Suppose f \in k[V] so that f: V \to \mathbb A^1 is regular. Write

V = U \cup U', where U = (\mathbb A^1 - \{0\}) \times \mathbb A^1, \ U' = \mathbb A^1 \times (\mathbb A^1 - \{0\}).

By example 6, we have f|_U \in k[U] \cong k[X, Y, \frac 1 X] and f|_{U'} \in k[U']\cong k[X, Y, \frac 1 Y]. Since U, U', U\cap U' are all dense in V we have injections k[V] \to k[U] \to k[U\cap U'] and k[V] \to k[U'] \to k[U\cap U'] so that f \in k[X, Y, \frac 1 X] \cap k[X, Y, \frac 1 Y]. It is easy to show that this means f\in k[X, Y].

Hence \phi induces an isomorphism of the coordinate rings k[\mathbb A^2] \to k[V]. If V is affine, by proposition 1 here \phi would be an isomorphism of varieties, which is a contradiction since \phi is not surjective.


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