# Projective Space

Definition.

Let $n\ge 0$. On the set $\mathbb A^{n+1}_k - \{\mathbf 0\}$, we consider the equivalence relation:

$v, w \in \mathbb A^{n+1}_k - \{\mathbf 0\} \implies (v \sim w \iff \exists \lambda \in k-\{0\}, w = \lambda v).$

The projective n-space $\mathbb P^n_k$ is the set of equivalence classes under this relation. An element of $\mathbb P^n_k$ is denoted by $(t_0 : t_1 : \ldots : t_n)$ for any representative $(t_0, \ldots, t_n) \in \mathbb A^{n+1}_k - \{\mathbf 0\}$.

For example $(5 : 2 : 3) = (\frac 1 6 : \frac 1 {15} : \frac 1 {10}) \in \mathbb P^2_{\mathbb C}$.

The above only defines $\mathbb P^n_k$ as a set; we need some geometric structure on it for the concept to be meaningful. Throughout this article we will fix $B = k[T_0, \ldots, T_n]$, which is graded by degree.

Definition.

Suppose $F\in B$ is homogeneous and $\mathbf v = (t_0 : t_1 : \ldots : t_n) \in \mathbb P^n_k$. We write $F(\mathbf v) = 0$ if  $F(t_0, \ldots, t_n) = 0$.

Note that the value $F(\mathbf v)$ is in general not well-defined. Indeed if we have two representatives $(t_0 : \ldots : t_n) = (\lambda t_0 : \ldots :\lambda t_n)$ for $\mathbf v$, then

$F(\lambda t_0, \ldots, \lambda t_n) = \lambda^{\deg f} F(t_0, \ldots, t_n).$

Despite this, $F(\lambda t_0, \ldots, \lambda t_n) = 0$ if and only if $F(t_0, \ldots, t_n) = 0$ so the definition is sensible. As in the case of the affine n-space, we will define a correspondence between ideals of B and subsets of $\mathbb P^n_k$.

Definition.

Let $\mathfrak a\subseteq B$ be a graded ideal. Write

$V_0(\mathfrak a) := \{ \mathbf v \in \mathbb P^n : F(\mathbf v) = 0 \text{ for all homogeneous } F \in \mathfrak a\}$.

For a finite sequence of homogeneous polynomials $F_1, \ldots, F_m$ we also write $V_0(F_1, \ldots, F_m)$ for $V_0(\mathfrak a)$ where $\mathfrak a = (F_1, \ldots, F_m)$.

Also let $D_0(F) := \mathbb P^n_k - V_0(F)$.

Exercise A

Prove the following, for any graded ideals $\mathfrak a, \mathfrak b\subseteq B$ and collection of graded ideals $(\mathfrak a_i)$ of B.

• $\mathfrak a \subseteq \mathfrak b \implies V(\mathfrak a)\supseteq V(\mathfrak b)$.
• If $(F_i)$ is a set of homogeneous generators of $\mathfrak a$, then

$V_0(\mathfrak a) = \{\mathbf v \in \mathbb P^n_k : F_i(\mathbf v) = 0 \text{ for all } i\}$.

• $\cap_i V_0(\mathfrak a_i) = V_0(\sum_i \mathfrak a_i)$.
• $V_0(\mathfrak a) \cup V_0(\mathfrak b) = V_0(\mathfrak a \cap \mathfrak b) = V_0(\mathfrak {ab})$.

In the other direction, we define:

Definition.

Let $V\subseteq \mathbb P^n$ be any subset. Then $I_0(V)$ denotes the (graded) ideal of B generated by:

$\{F \in B \text{ homogeneous } : F(\mathbf v) = 0 \text{ for all } \mathbf v \in V\}$.

In summary, we defined the following maps.

# Zariski Topology of Projective Space

We wish to define the Zariski topology on $\mathbb P^n_k$; for that let us take subsets of $\mathbb P^n$ which can be identified with the affine space $\mathbb A^n$. Fix $0\le i \le n$; let

$U_i = \{ (t_0 : t_1 : \ldots : t_n ) \in \mathbb P^n_k : t_i \ne 0\}.$

Note that for $(t_0 : t_1 : \ldots : t_n) \in U_i$ the same point can be represented by $(\frac{t_0}{t_i} : \frac{t_1}{t_i} : \ldots : \frac{t_n}{t_i})$ where the i-th coordinate is 1. This gives a bijection $\varphi_i : U_i \to \mathbb A^n_k$. E.g. for n = 2, we have:

$\varphi_0 : (x : y : z) \mapsto (\frac y x, \frac z x), \quad \varphi_1 : (x : y : z) \mapsto (\frac x y, \frac z y), \quad \varphi_2 : (x : y : z) \mapsto (\frac x z, \frac y z).$

Note that for any $i\ne j$, the intersection $U_i \cap U_j$ maps to an open subset of $\mathbb A^n_k$ via both $\varphi_i$ and $\varphi_j$. Indeed if i < j then $\varphi_j(U_i\cap U_j) \subset \mathbb A^n_k$ is the set of all $(t_0, \ldots, t_n)$ satisfying $t_i \ne 0$ while $\varphi_i(U_i\cap U_j)\subset \mathbb A^n_k$ is the set of all $(t_0, \ldots, t_n)$ satisfying $t_{j-1} \ne 0$. Hence, the following is well-defined.

Definition.

The Zariski topology on $\mathbb P^n_k$ is defined by specifying every $U_i \subseteq \mathbb P^n$ as an open subset, where $U_i$ obtains the Zariski topology of $\mathbb A^n$ from $\varphi_i: U_i \to \mathbb A^n$.

projective variety is a closed subspace $V\subseteq \mathbb P^n_k$.

First, we have the following preliminary results.

Lemma 1.

For any homogeneous $F \in B$, the set $V_0(F)$ is (Zariski) closed in $\mathbb P^n_k$.

Proof

It suffices to show that $V_0(F) \cap U_i$ is closed in $U_i$ for each $0\le i\le n$. But $V_0(F) \cap U_0 = V(f)$, where $f(X_1, \ldots, X_n) = F(1, X_1, \ldots, X_n) \in k[X_1, \ldots, X_n]$. The same holds for $U_1, \ldots, U_n$. ♦

Definition.

For any $f \in k[X_1, \ldots, X_n]$, let

$F(T_0, \ldots, T_n) := T_0^{\deg f} f(\frac {T_1} {T_0}, \ldots, \frac{T_n}{T_0})$

be its homogenization.

Exercise B

1. Prove that if $f,g\in k[X_1, \ldots, X_n]$, the homogenization of fg is the product of the homogenizations of f and g.

2. Let $F\in k[T_0, \ldots, T_n]$ be the homogenization of a non-constant $f \in k[X_1, \ldots, X_n]$. Then f is irreducible if and only if F is irreducible. [ Hint: you may find lemma 2 here helpful. ]

The Zariski topology on $\mathbb P^n$ is consistent with our earlier notions of closed subsets:

Proposition 1.

A subset $V\subseteq \mathbb P^n_k$ is closed under the Zariski topology if and only if $V = V_0(\mathfrak a)$ for some graded ideal $\mathfrak a \subseteq B$.

Proof

(⇐) Suppose $\mathfrak a = (F_1, \ldots, F_m)$ for some homogeneous $F_i \in B$. Since $V_0(\mathfrak a) = \cap_i V_0(F_i)$, by lemma 1 this is Zariski closed.

(⇒) Let $U= \mathbb P^n_k - V$; it suffices to show that any $\mathbf v \in U$ is contained in $D_0(F)$ for some homogeneous $F \in B$. Now $\mathbf v$ is contained in some $U_i$, say $U_0$ without loss of generality. Hence $\mathbf v \in D(f) \subseteq U \cap U_0$ for some $f \in k[X_1, \ldots, X_n]$. If $\mathbf v = (1, t_1, \ldots, t_n)$, then

$f(t_1, \ldots, t_n) \ne 0 \implies F(\mathbf v) \ne 0 \implies \mathbf v \in D_0(F)$, where F = homogenization of f. ♦

### Example

Let $V\subseteq \mathbb P^2_k$ be the projective variety defined by the homogeneous equation $T_0^2 + T_1^2 = T_2^2$. Then

• $V \cap U_0$ is cut out from $\mathbb A^2_k$ by $1 + y^2 = z^2$;
• $V \cap U_1$ is cut out from $\mathbb A^2_k$ by $x^2 + 1 = z^2$;
• $V \cap U_2$ is cut out from $\mathbb A^2_k$ by $x^2 + y^2 = 1$.

# Cone of Projective Variety

We wish to prove the bijective correspondence between graded radical ideals of B and closed subsets $V\subseteq \mathbb P^n_k$. For that, we can piggyback on existing results for the affine case.

Definition.

Let $C\subseteq \mathbb P^n_k$ be any subset. The cone of C is

$\mathrm{cone}(C) := \{\mathbf v = (x_0, \ldots, x_n) \in \mathbb A^{n+1}_k : \mathbf v = \mathbf 0 \text{ or } (x_0 : \ldots : x_n) \in C\}$.

Note

For any non-empty collection of subsets $C_i \subseteq \mathbb P^n_k$ we have

$\bigcup_i \mathrm{cone}(C_i) = \mathrm{cone}(\cup_i C_i), \quad \bigcap_i \mathrm{cone}(C_i) = \mathrm{cone}(\cap_i C_i).$

Also, we have:

Lemma 2.

If $\mathfrak a \subsetneq B$ is a proper graded ideal then $\mathrm{cone}(V_0(\mathfrak a)) = V(\mathfrak a)$.

In particular, by proposition 1, the cone of a closed $V \subseteq \mathbb P^n_k$ is closed in $\mathbb A^{n+1}_k$.

Proof

Note: if $F\in B$ is non-constant homogeneous, then $\mathrm{cone} V_0(F) = V(F) \subseteq \mathbb A^{n+1}_k$. Now pick a set of homogeneous generators $F_i$ for $\mathfrak a$; each $F_i$ is non-constant so

$\mathrm{cone}(V_0(\mathfrak a)) = \mathrm{cone}(\cap_i V_0(F_i)) = \bigcap_i \mathrm{cone} V_0(F_i) = \bigcap_i V(F_i) = V(\mathfrak a)$.

This completes the proof. ♦

Furthermore we have:

Lemma 3.

A non-empty closed subset $W\subseteq \mathbb A^{n+1}$ is of the form $\mathrm{cone}(V)$ for some closed $V\subseteq \mathbb P^n$ if and only if

$\mathbf v \in W, \lambda \in k \implies \lambda \mathbf v \in W$.

When that happens, we call W a closed cone in $\mathbb A^{n+1}$.

Proof

(⇒) is obvious; for (⇐) clearly W = cone(V) for some subset $C\subseteq \mathbb P^n$. Let $\mathfrak a = I(W)$ so that $V(\mathfrak a) = W$. It remains to show $\mathfrak a$ is graded, for we would get $W = \mathrm{cone}(V_0(\mathfrak a))$ by lemma 2.

Indeed if $f\in \mathfrak a$, write $f = f_0 + \ldots + f_d$ as a sum of homogeneous components. Then for any $\mathbf v \in W$ and $\lambda \in k$ we have $\lambda \mathbf v \in W$ which gives

$0 = f(\lambda \mathbf v) = f_0 (\mathbf v) + \lambda \cdot f_1(\mathbf v) + \ldots + \lambda^d \cdot f_d(\mathbf v).$

Thus $f_0, \ldots, f_d$ vanish for any $\mathbf v \in W$, i.e. $f_0, \ldots, f_d \in \mathfrak a$. ♦

# Projective Nullstellensatz

Thus we have the following correspondences:

The top-left column is a bijection by lemma 3; the bottom row is a bijection by Nullstellensatz. In the proof of lemma 3, we also showed that for a closed cone $W\subseteq \mathbb A^{n+1}_k$, the ideal $I(W)$ is graded. Conversely, if $\mathfrak a\subsetneq k[T_0, \ldots, T_n]$ is graded, $V(\mathfrak a)$ is the (non-empty) solution set of a collection of graded polynomials; hence it is a closed cone too.

Hence we have a bijection between

• closed subsets $V\subseteq \mathbb P^n_k$, and
• proper homogeneous radical ideals $\mathfrak a \subsetneq B = k[T_0, \ldots, T_n]$.

The correspondence takes $V \mapsto I(\mathrm{cone}(V)) =: \mathfrak a$ and so

$\mathrm{cone}(V) = V(\mathfrak a) = \mathrm{cone} (V_0 (\mathfrak a)) \implies V = V_0(\mathfrak a).$

The last piece of the puzzle is the map $I_0$ which takes closed subsets of $\mathbb P^n_k$ to homogeneous ideals of B. As an easy exercise, show that

$V \subseteq \mathbb P^n_k \text{ closed non-empty } \implies I_0(V) = I(\mathrm{cone}(V))$.

However $I_0(\emptyset) = (1)$ so we modify our bijection to:

Theorem (Projective Nullstellensatz).

There is a bijection between:

• closed subsets $V \subseteq \mathbb P^n_k$;
• homogeneous radical ideals $\mathfrak a \subseteq B$ such that $\mathfrak a \ne B_+$ where $B_+ := (T_0, \ldots, T_n)$ is the irrelevant ideal of B.

Exercise C

Prove that if $\mathfrak a \subseteq B$ is a homogeneous ideal then $V(\mathfrak a)$ is empty if and only if $\mathfrak a$ contains a power of $B_+$.

[ Hint: prove that the radical of 𝔞 is either (1) or B+. ]

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