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. $\square$

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