Lemma 33.34.1.slogan Let k be a field and n \geq 0. Then \mathbf{P}^ n_ k is a smooth projective variety of dimension n over k.
33.34 Projective space
Some results on projective space over a field.
Proof. Omitted. \square
Lemma 33.34.2. Let k be a field and n \geq 0. Let X, Y \subset \mathbf{A}^ n_ k be closed subsets. Assume that X and Y are equidimensional, \dim (X) = r and \dim (Y) = s. Then every irreducible component of X \cap Y has dimension \geq r + s - n.
Proof. Consider the closed subscheme X \times Y \subset \mathbf{A}^{2n}_ k where we use coordinates x_1, \ldots , x_ n, y_1, \ldots , y_ n. Then X \cap Y = X \times Y \cap V(x_1 - y_1, \ldots , x_ n - y_ n). Let t \in X \cap Y \subset X \times Y be a closed point. By Lemma 33.20.5 we have \dim _ t(X \times Y) = \dim (X) + \dim (Y). Thus \dim (\mathcal{O}_{X \times Y, t}) = r + s by Lemma 33.20.3. By Algebra, Lemma 10.60.13 we conclude that
\dim (\mathcal{O}_{X \cap Y, t}) = \dim (\mathcal{O}_{X \times Y, t}/(x_1 - y_1, \ldots , x_ n - y_ n)) \geq r + s - n
This implies the result by Lemma 33.20.3. \square
Lemma 33.34.3. Let k be a field and n \geq 0. Let X, Y \subset \mathbf{P}^ n_ k be nonempty closed subsets. If \dim (X) = r and \dim (Y) = s and r + s \geq n, then X \cap Y is nonempty and \dim (X \cap Y) \geq r + s - n.
Proof. Write \mathbf{A}^ n = \mathop{\mathrm{Spec}}(k[x_0, \ldots , x_ n]) and \mathbf{P}^ n = \text{Proj}(k[T_0, \ldots , T_ n]). Consider the morphism \pi : \mathbf{A}^{n + 1} \setminus \{ 0\} \to \mathbf{P}^ n which sends (x_0, \ldots , x_ n) to the point [x_0 : \ldots : x_ n]. More precisely, it is the morphism associated to the pair (\mathcal{O}_{\mathbf{A}^{n + 1} \setminus \{ 0\} }, (x_0, \ldots , x_ n)), see Constructions, Lemma 27.13.1. Over the standard affine open D_+(T_ i) we get the morphism associated to the ring map
k\left[\frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i}\right] \longrightarrow k\left[T_0, \ldots , T_ n, \frac{1}{T_ i}\right] \cong k\left[\frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i}\right] \left[T_ i, \frac{1}{T_ i}\right]
which is surjective and smooth of relative dimension 1 with irreducible fibres (details omitted). Hence \pi ^{-1}(X) and \pi ^{-1}(Y) are nonempty closed subsets of dimension r + 1 and s + 1. Choose an irreducible component V \subset \pi ^{-1}(X) of dimension r + 1 and an irreducible component W \subset \pi ^{-1}(Y) of dimension s + 1. Observe that this implies V and W contain every fibre of \pi they meet (since \pi has irreducible fibres of dimension 1 and since Lemma 33.20.4 says the fibres of V \to \pi (V) and W \to \pi (W) have dimension \geq 1). Let \overline{V} and \overline{W} be the closure of V and W in \mathbf{A}^{n + 1}. Since 0 \in \mathbf{A}^{n + 1} is in the closure of every fibre of \pi we see that 0 \in \overline{V} \cap \overline{W}. By Lemma 33.34.2 we have \dim (\overline{V} \cap \overline{W}) \geq r + s - n + 1. Arguing as above using Lemma 33.20.4 again, we conclude that \pi (V \cap W) \subset X \cap Y has dimension at least r + s - n as desired. \square
Lemma 33.34.4. Let k be a field. Let Z \subset \mathbf{P}^ n_ k be a closed subscheme which has no embedded points such that every irreducible component of Z has dimension n - 1. Then the ideal I(Z) \subset k[T_0, \ldots , T_ n] corresponding to Z is principal.
Proof. This is a special case of Divisors, Lemma 31.31.3. \square
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