Projective Varieties

Definition of Projective Space

Now we define projective varieties, another important class of algebraic varieties. We begin with the following.

Definition 1 We define the projective \(n\)-space \(\mathbb{P}^n_{\mathbb{K}}\) over a field \(\mathbb{K}\) as follows. As a set,

\[\mathbb{P}^n = (\mathbb{K}^{n+1} \setminus \{0\}) / \sim\]

where the equivalence relation \(\sim\) is given by

\[(x_0, \ldots, x_n) \sim (y_0, \ldots, y_n) \iff \text{$x_i = \lambda y_i$ for some $\lambda \in \mathbb{K}^\times$, for all $i$}\]

When there is no risk of confusion, we write \(\mathbb{P}^n\).

The equivalence class \([(x_0, \ldots, x_n)]\) is usually denoted by \([x_0 : \cdots : x_n]\), and these are called homogeneous coordinates. We call \(x_0, \ldots, x_n\) the coordinates, and at least one of them must not be \(0\). The key point of homogeneous coordinates is that the coordinates determine only a ratio. That is, for all \(\lambda\in \mathbb{K}^\times\), \([x_0 : \cdots : x_n] = [\lambda x_0 : \cdots : \lambda x_n]\) holds.

Homogeneous Polynomials and Projective Space

Now, as in the affine case, we must endow \(\mathbb{P}^n\) with a topological structure. In projective space as well, we will similarly define closed sets as zero sets of polynomials, but what must be noted is that since \(\mathbb{P}^n\) is defined as a quotient set, in general a polynomial does not define a function on \(\mathbb{P}^n\). That is, for an arbitrary \(F \in \mathbb{K}[x_0, \ldots, x_n]\), although \([x_0 : \cdots : x_n] = [\lambda x_0 : \cdots : \lambda x_n]\), in general

\[F(x_0, \ldots, x_n)\neq F(\lambda x_0, \ldots, \lambda x_n)\]

and if \(\mathbb{K}\) is infinite, the only polynomials for which evaluation at every point of \(\mathbb{P}^n\) is well-defined for an arbitrary representative are constant polynomials. However, if we are interested only in the zero set defined by a polynomial, this problem is resolved. For a homogeneous polynomial \(F\) of degree \(d\), since

\[F(\lambda x_0, \ldots, \lambda x_n) = \lambda^d F(x_0, \ldots, x_n)\]

we have

\[F(\lambda x_0, \ldots, \lambda x_n) = 0 \iff F(x_0, \ldots, x_n) = 0\]

Therefore, the zero set of a homogeneous polynomial is well-defined on projective space.

Definition 2 A polynomial \(F \in \mathbb{K}[\x_0, \ldots, \x_n]\) is said to be homogeneous of degree \(d\) if for all \(\lambda \in \mathbb{K}\),

\[F(\lambda \x_0, \ldots, \lambda \x_n) = \lambda^d F(\x_0, \ldots, \x_n)\]

holds.

Although we have stated the definition in a complicated way, if \(\mathbb{K}\) is infinite, this essentially means that when the polynomial is expressed as a sum of monomials, every monomial has degree \(d\). Then we can define the following.

Definition 3 For homogeneous polynomials \(F_1, \ldots, F_k \in \mathbb{K}[\x_0, \ldots, \x_n]\), we define the projective algebraic set \(Z(F_1, \ldots, F_k)\) by

\[Z(F_1, \ldots, F_k) = \{[x_0 : \cdots : x_n] \in \mathbb{P}^n \mid F_1(x) = \cdots = F_k(x) = 0\}\]

Among projective algebraic sets, those that cannot be expressed as a union of finitely many strictly smaller projective algebraic sets are called projective varieties.

As explained above, since each \(F_i\) is homogeneous, one can verify that this is well-defined.

Meanwhile, we recall that when dealing with affine varieties, it sufficed to consider only ideals instead of arbitrary subsets of \(\mathbb{K}[\x_1,\ldots, \x_n]\). In the projective case as well, this philosophy is similar, but by further adding the assumption of homogeneity, we can define the concept of a homogeneous ideal.

Definition 4 An ideal \(\mathfrak{a} \subseteq \mathbb{K}[\x_0, \ldots, \x_n]\) is said to be homogeneous if \(\mathfrak{a}\) is generated by homogeneous polynomials.

For a homogeneous ideal \(\mathfrak{a}\), if we define its zero set \(Z(\mathfrak{a})\) as the set of points where all homogeneous polynomials in \(\mathfrak{a}\) vanish, then as in the affine case, we can define the Zariski topology. For this, we need the following proposition.

Proposition 5 For homogeneous ideals \(\mathfrak{a}, \mathfrak{b}, \mathfrak{a}_i \subseteq \mathbb{K}[\x_0, \ldots, \x_n]\), the following hold:

1. \(Z(0) = \mathbb{P}^n\), \(Z(1) = \emptyset\),
2. \(\bigcap_iZ(\mathfrak{a}_i) = Z\left(\sum_i \mathfrak{a}_i\right)\),
3. \(Z(\mathfrak{a}) \cup Z(\mathfrak{b}) = Z(\mathfrak{a} \cap \mathfrak{b}) = Z(\mathfrak{a}\mathfrak{b})\).

Proof

The only difference from §Affine Varieties, ⁋Proposition 4 is that the polynomials considered here are all homogeneous, but the logic of the proof itself is identical, so we omit the proof.

As in the affine case, this shows that there exists a topology on projective space \(\mathbb{P}^n\) whose closed sets are the projective algebraic sets, and using this, we can endow each projective variety with the subspace topology. Similarly, we call this topology the Zariski topology. (We first examined the Zariski topology in the affine case in §Affine Varieties.)

Projective Nullstellensatz

Definition 6 The homogeneous ideal of a subset \(X \subseteq \mathbb{P}^n\) is defined to be the ideal generated by

\[\{F \in \mathbb{K}[\x_0, \ldots, \x_n] \mid F \text{ is homogeneous and } F(x) = 0 \text{ for all } x \in X\}\]

and is denoted by \(I(X)\).

Theorem 7 (Projective Nullstellensatz) Let \(\mathbb{K}\) be an algebraically closed field and let \(\mathfrak{a} \subseteq \mathbb{K}[\x_0, \ldots, \x_n]\) be a homogeneous ideal. Then

1. \(Z(\mathfrak{a}) = \emptyset \iff \sqrt{\mathfrak{a}} \supseteq (\x_0, \ldots, \x_n)\),
2. \(I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}\) (if \(Z(\mathfrak{a}) \ne \emptyset\)).

The difference from the affine case is that \(Z(\mathfrak{a}) = \emptyset\) does not mean \(\mathfrak{a} = (1)\), but rather that \(\sqrt{\mathfrak{a}}\) contains the irrelevant ideal \((\x_0, \ldots, \x_n)\). This is because \((\x_0, \ldots, \x_n)\) corresponds to the origin of \(\mathbb{K}^{n+1}\), and the origin is excluded from the definition of projective space.

Standard Affine Cover

Projective space \(\mathbb{P}^n\) can be covered by \(n+1\) affine spaces. This is one of the most important ways to understand projective space.

Definition 8 For \(i = 0, 1, \ldots, n\), we define the \(i\)-th standard open set \(U_i\) by

\[U_i = \{[x_0 : \cdots : x_n] \in \mathbb{P}^n \mid x_i \ne 0\}\]

Endow each \(U_i\) with the subspace topology inherited from \(\mathbb{P}^n\). Then the following holds.

Proposition 9 Each \(U_i\) is homeomorphic to affine space \(\mathbb{A}^n\) (under the subspace topology).

Proof

For notational convenience, we prove the case \(i=0\). In the case of \(U_0\), let us define the map \(\varphi_0: U_0 \rightarrow \mathbb{A}^n\) by

\[\varphi_0([x_0 : x_1 : \cdots : x_n]) = \left(\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}\right)\]

The inverse map \(\psi_0: \mathbb{A}^n \rightarrow U_0\) is

\[\psi_0(a_1, \ldots, a_n) = [1 : a_1 : \cdots : a_n]\]

That these are mutual inverses is obvious from the definition. We must now show that both \(\varphi_0\) and \(\psi_0\) are continuous.

First, to show the continuity of \(\varphi_0\), consider a closed set \(Z(f)\) in \(\mathbb{A}^n\). Then

\[\varphi_0^{-1}(Z(f)) = \left\{[x_0 : \cdots : x_n] \in U_0 \mid f\left(\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}\right) = 0\right\}\]

Now if \(f\) is a polynomial of degree \(d\), then

\[F(\x_0,\ldots, \x_n)=\x_0^d f(\x_1/\x_0, \ldots, \x_n/\x_0)\]

is a homogeneous polynomial, and \(\varphi_0^{-1}(Z(f)) = Z(F) \cap U_0\). This is a closed set in the subspace topology of \(U_0\).

Now let us show the continuity of its inverse \(\psi_0\). Consider a closed set \(Z(F) \cap U_0\) in \(U_0\), where \(F\) is a homogeneous polynomial of degree \(d\). Then

\[\psi_0^{-1}(Z(F) \cap U_0) = \{(x_1, \ldots, x_n) \in \mathbb{A}^n \mid F(1, x_1, \ldots, x_n) = 0\}\]

Since \(F(1, \x_1, \ldots, \x_n)\) is a polynomial in \(\mathbb{K}[\x_1, \ldots, \x_n]\), \(\psi_0^{-1}(Z(F) \cap U_0)\) is a closed set in \(\mathbb{A}^n\).

Therefore, since \(\varphi_0\) and \(\psi_0\) are mutual inverses and both are continuous, \(\varphi_0\) is a homeomorphism.

That is, intuitively, we can think of \(U_i\) as the points where the coordinate \(x_i\) is not infinity. Also, \(\mathbb{P}^n = U_0 \cup \cdots \cup U_n\), and by the proposition above, each \(U_i \cong \mathbb{A}^n\). Since the key point in the proof of the proposition above was the following proposition, let us separate it out.

Proposition 10 For a projective variety \(X \subseteq \mathbb{P}^n\) and a standard open set \(U_i\), if \(X \cap U_i\) is nonempty, then it is an affine variety in \(U_i \cong \mathbb{A}^n\).

Proof

For notational convenience, we prove the case \(i=0\). Let \(X = Z(F_1, \ldots, F_k)\), and suppose that each \(F_j\) is homogeneous of degree \(d_j\). Now if we set

\[f_j(\x_1, \ldots, \x_n) = F_j(1, \x_1, \ldots, \x_n)\]

then by the homogeneity of \(F_j\), for any point with \(x_0 \ne 0\),

\[F_j(x_0, x_1, \ldots, x_n) = x_0^{d_j} f_j\left(\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}\right)\]

holds. Therefore, that \([x_0 : \cdots : x_n] \in U_0\) belongs to the zero set of all \(F_j\) is equivalent to the condition that, under the homeomorphism \(\varphi_0\) of Proposition 9, the point in \(\mathbb{A}^n\) corresponding to this point, \((x_1/x_0, \ldots, x_n/x_0)\), belongs to the zero set of all \(f_j\), and from this we obtain \(\varphi_0(X \cap U_0) = Z(f_1, \ldots, f_k) \subseteq \mathbb{A}^n\).

It remains to show irreducibility. Since \(U_0\) is an open set of \(\mathbb{P}^n\), \(X \cap U_0\) is an open subset of \(X\), and it is nonempty by assumption. Since a nonempty open subset of an irreducible space is irreducible, \(X \cap U_0\) is irreducible, and since a homeomorphism preserves irreducibility, \(Z(f_1, \ldots, f_k)\) is also irreducible. That is, \(X \cap U_0\) is an affine variety.

Example 11 To interpret the above proposition geometrically, let \(\mathbb{K}=\mathbb{R}\), and consider in \(\mathbb{P}^2\) the conic \(X = Z(\x_0^2 + \x_1^2 - \x_2^2)\).

This conic is the cone in \(\mathbb{A}^3\) given by \(\x_0^2 + \x_1^2 = \x_2^2\), expressed in homogeneous coordinates. Then how \(X\) looks in the standard open sets can be seen from Proposition 10. That is, to see in \(U_i\) what \(X\) looks like, we simply replace \(\x_i\) with \(1\) and regard the remaining \(n\) variables as coordinates on \(\mathbb{A}^n\). In particular, we then obtain the following:

1. In \(U_0, U_1\), \(X\) is the hyperbolas \(1+y^2-z^2=0\) and \(x^2+1-z^2=0\).
2. In \(U_2\), \(X\) is the circle \(x^2+y^2=1\).

This happens because in \(\mathbb{A}^3\) the equation \(\x_0^2 + \x_1^2 = \x_2^2\) is a cone, and its traces cut by the planes \(\x_0=1, \x_1=1, \x_2=1\) become hyperbolas and a circle.

On the other hand, we can also interpret this directly in \(\mathbb{P}^2\). To this end, let us obtain \(\mathbb{P}^2\) as follows. For points with \(\x_2\neq 0\), we radially project onto the upper hemisphere satisfying \(\x_2>0\), and for points with \(\x_2=0\), we identify antipodal points. Through this, \(\mathbb{P}^2\) can be thought of as the “line at infinity” \(\mathbb{P}^1\) combined with the plane \(\mathbb{A}^2\) corresponding to the surface of the upper hemisphere. Then the given cone first becomes a circle contained in the upper hemisphere via the first radial projection, and from this we know that \(X\) appears as a circle in \(\mathbb{P}^2\).

Of course, by radially projecting points satisfying \(\x_0\neq 0\) onto the upper hemisphere with \(\x_0>0\) and taking points with \(\x_0=0\) as \(\mathbb{P}^1\), one could initially have constructed \(\mathbb{P}^2\) as well. In this process, two semicircles would be drawn on the upper hemisphere, but the boundary points of these two semicircles would be treated as the same in the process of identifying points with \(\x_0=0\), so in this picture too \(X\) becomes a circle.

Now from this perspective, looking in \(U_i\) at \(X\) corresponds to removing from \(\mathbb{P}^2\) the line at infinity \(\x_i=0\). If in \(U_2\) we view \(X\), then as examined above, \(X\) does not meet the line at infinity \(\x_2=0\), so removing this line leaves a complete circle. However, if for example we remove the line at infinity \(\x_1=0\), then \(X\) meets this line at two points, and so it can be understood that removing these two points from the circle \(X\) and unfolding it yields a hyperbola.

Affine Cone

The preceding example shows how to view a curve on projective space in each affine open chart, but one may still find it somewhat unintuitive. Another way to understand a projective variety as a geometric object in affine space is to consider an affine cone.

Definition 12 For a projective variety \(X \subseteq \mathbb{P}^n\), we define its affine cone \(C(X) \subseteq \mathbb{A}^{n+1}\) as follows:

\[C(X) = \{(x_0, \ldots, x_n) \in \mathbb{A}^{n+1} \setminus \{0\} \mid [x_0 : \cdots : x_n] \in X\} \cup \{0\}\]

That is, \(C(X)\) is the union of the points that, when all points of \(X\) are expressed in homogeneous coordinates, appear in \(\mathbb{A}^{n+1}\), and the origin.

Example 13 For the conic \(X = Z(\x_0^2 + \x_1^2 - \x_2^2) \subseteq \mathbb{P}^2\) from Example 11, the affine cone \(C(X)\) is the cone in \(\mathbb{A}^3\) given by \(\x_0^2 + \x_1^2 = \x_2^2\).

Then the following holds, and the proofs are not difficult either.

Proposition 14 For a projective variety \(X \subseteq \mathbb{P}^n\), the affine cone \(C(X)\) satisfies the following properties:

1. (Homogeneity) \(C(X)\) consists of lines passing through the origin. That is, if \((x_0, \ldots, x_n) \in C(X)\) and \(\lambda \in \mathbb{K}\), then \((\lambda x_0, \ldots, \lambda x_n) \in C(X)\).

2. (Algebraic structure) If \(X = Z(F_1, \ldots, F_k)\), then \(C(X) = Z(F_1, \ldots, F_k) \subseteq \mathbb{A}^{n+1}\). Here the \(F_i\) are regarded as polynomials on \(\mathbb{A}^{n+1}\).

3. (Correspondence) The correspondence \(X \leftrightarrow C(X)\) gives a one-to-one correspondence between projective varieties and irreducible affine algebraic sets other than \(\{0\}\) consisting of lines passing through the origin.

Through this proposition, by studying the properties of the affine cone \(C(X)\), we can indirectly understand the properties of \(X\).

Morphisms between Projective Varieties

Finally, we define morphisms of projective varieties. Earlier, when defining projective algebraic sets, we confirmed that the zero sets of polynomials do not well-define subsets of projective space; a similar thing occurs when defining morphisms, and the solution once again is homogeneous polynomials.

Definition 15 A function \(\varphi: X \rightarrow Y\) is a morphism between projective varieties \(X \subseteq \mathbb{P}^n\) and \(Y \subseteq \mathbb{P}^m\) if for each point \(x \in X\), there exists an open subset containing \(x\) in \(X\), \(U\), and suitable homogeneous polynomials \(F_0, \ldots, F_m \in \mathbb{K}[\x_0, \ldots, \x_n]\) of the same degree such that for all \(y \in U\),

\[\varphi(y) = [F_0(y) : \cdots : F_m(y)] \in \mathbb{P}^m\]

holds.

If \(F_0, \ldots, F_m\) are all homogeneous polynomials of the same degree \(d\), then since \(F_i(\lambda x) = \lambda^d F_i(x)\),

\[[F_0(\lambda x) : \cdots : F_m(\lambda x)] = [\lambda^d F_0(x) : \cdots : \lambda^d F_m(x)] = [F_0(x) : \cdots : F_m(x)]\]

holds, showing that well-definedness is guaranteed. Meanwhile, as in §Affine Varieties, ⁋Definition 17, when, for a morphism \(\varphi: X \rightarrow Y\), an inverse function \(\psi: Y \rightarrow X\) exists and is also a morphism, we call \(\varphi\) an isomorphism, and two projective varieties for which such a \(\varphi\) exists are said to be isomorphic. The following examples are representative morphisms.

Example 16 First, if we define the Veronese embedding (of degree 2) from \(\mathbb{P}^1\) to \(\mathbb{P}^2\) by

\[[x:y]\mapsto [x^2: xy:y^2]\]

this becomes a morphism between projective spaces. As another example, the Segre embedding from \(\mathbb{P}^1\times \mathbb{P}^1\) to \(\mathbb{P}^3\) is given by the formula

\[([x:y], [u:v])\mapsto [xu: xv: yu: yv]\]

However, since we have not yet defined the product of two projective varieties, we postpone verifying whether this actually defines a morphism until later.

Example 17 Twisted cubic in \(\mathbb{P}^3\)

\[C = \{[1 : t : t^2 : t^3] \mid t \in \mathbb{K}\} \cup \{[0 : 0 : 0 : 1]\}\]

is the common zero locus of the three quadratic polynomials

\[\x_0 \x_2 - \x_1^2, \quad \x_0 \x_3 - \x_1 \x_2, \quad \x_1 \x_3 - \x_2^2\]

and is isomorphic to \(\mathbb{P}^1\). In fact, if we extend the concept of the Veronese embedding examined in Example 16 above to \(d=3\),

\[[x:y]\mapsto [x^3: x^2y: xy^2: y^3]\]

becomes an isomorphism from \(\mathbb{P}^1\) to \(C\).

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References

[Har] J. Harris, Algebraic Geometry: A First Course, Springer, 1992.
[Sha] I. R. Shafarevich, Basic Algebraic Geometry I: Varieties in Projective Space, Springer, 2013.
[Ful] W. Fulton, Algebraic Curves, 2008.