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
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
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