The Stacks project

Proof. We claim the lemma holds with $U = \mathbf{P}(V) \setminus X$. For a closed point $p$ of $U$ we indeed obtain a morphism $r_ p|_ X : X \to \mathbf{P}(W_ p)$. This morphism is proper because $X$ is a proper scheme (Morphisms, Lemmas 29.43.5 and 29.41.7). On the other hand, the fibres of $r_ p$ are affine lines as can be seen by a direct calculation. Hence the fibres of $r_ p|X$ are proper and affine, whence finite (Morphisms, Lemma 29.44.11). Finally, a proper morphism with finite fibres is finite (Cohomology of Schemes, Lemma 30.21.1). $\square$

Proof. Proof of (1). Note that if $x, y \in X$ have the same image under $r_ p$ then $p$ is on the line $\overline{xy}$. Consider the finite type scheme

and the morphisms $T \to X$ and $T \to \mathbf{P}(V)$ given by $(y, p) \mapsto y$ and $(y, p) \mapsto p$. Since each fibre of $T \to X$ is a line, we see that the dimension of $T$ is $\dim (X) + 1 < \dim (\mathbf{P}(V))$. Hence $T \to \mathbf{P}(V)$ is not surjective. On the other hand, consider the finite type scheme

Then the dimension of $T'$ is $\dim (X) < \dim (\mathbf{P}(V))$. Thus the morphism $T' \to \mathbf{P}(V)$ is not surjective either. Let $U \subset \mathbf{P}(V) \setminus X$ be nonempty open and disjoint from these images; such a $U$ exists because the images of $T$ and $T'$ in $\mathbf{P}(V)$ are constructible by Morphisms, Lemma 29.22.2. Then for $p \in U$ closed the projection $r_ p|_ X : X \to \mathbf{P}(W_ p)$ is injective on the tangent space at $x$ and $r_ p^{-1}(\{ r_ p(x)\} ) = \{ x\} $. This means that $r_ p$ is unramified at $x$ (Varieties, Lemma 33.16.8), finite by Lemma 43.23.1, and $r_ p^{-1}(\{ r_ p(x)\} ) = \{ x\} $ thus Étale Morphisms, Lemma 41.7.3 applies and there is an open neighbourhood $R$ of $r_ p(x)$ in $\mathbf{P}(W_ p)$ such that $(r_ p|_ X)^{-1}(R) \to R$ is a closed immersion which proves (1).

Proof of (2). In this case we still conclude that the morphism $T' \to \mathbf{P}(V)$ is not surjective. Arguing as above we conclude that for $U \subset \mathbf{P}(V)$ avoiding $X$ and the image of $T'$, the projection $r_ p|_ X : X \to \mathbf{P}(W_ p)$ is étale at $x$ and finite. $\square$

Proof. Set $Y' = Y \setminus Y \cap Z$. Let $y \in Y'$, $z \in Z$ be closed points with $r_ p(y) = r_ p(z)$. Then $p$ is on the line $\overline{yz}$ passing through $y$ and $z$. Consider the finite type scheme

and the morphism $T \to \mathbf{P}(V)$ given by $(y, z, p) \mapsto p$. Observe that $T$ is irreducible and that $\dim (T) = \dim (Y) + \dim (Z) + 1$. Hence the general fibre of $T \to \mathbf{P}(V)$ has dimension at most $\dim (Y) + \dim (Z) + 1 - n$, more precisely, there exists a nonempty open $U \subset \mathbf{P}(V) \setminus (Y \cup Z)$ over which the fibre has dimension at most $\dim (Y) + \dim (Z) + 1 - n$ (Varieties, Lemma 33.20.4). Let $p \in U$ be a closed point and let $F \subset T$ be the fibre of $T \to \mathbf{P}(V)$ over $p$. Then

is the image of $F \to Y$, $(y, z, p) \mapsto y$. Again by Varieties, Lemma 33.20.4 the closure of the image of $F \to Y$ has dimension at most $\dim (Y) + \dim (Z) + 1 - n$. $\square$

Proof. Consider the image of $B$ under the projection $r_ p : \mathbf{P}(V) \to \mathbf{P}(W_ p)$. Since $\dim (W_ p) = \dim (V) - 1$, we see that $r_ p(B)$ has codimension $\geq 1$ in $\mathbf{P}(W_ p)$. For any $q \in \mathbf{P}(V)$ with $r_ p(q) \not\in r_ p(B)$ we see that the line $\ell = \overline{pq}$ connecting $p$ and $q$ works. $\square$

Proof. Omitted. Hint: This follows from the fact that $\text{GL}(V)$ acts doubly transitive on pairs of linearly independent vectors. $\square$

Proof. Let $V$ be an $n$ dimensional vector space. Translating into geometry the lemma signifies that the variety $C$ of non-invertible linear maps $V \to V$ is irreducible. Let $W$ be a vector space of dimension $n - 1$. By elementary linear algebra, the morphism

has image $C$. Since the source is irreducible, so is the image. $\square$

Proof. Let $B \subset \mathbf{P}$ be the set of “bad” points, i.e., those points $[g]$ that violate either (1) or (2). Note that $[\text{id}_ V] \not\in B$ by assumption. Moreover, $B$ is closed. Hence it suffices to prove that $\dim (B) \leq \dim (\mathbf{P}) - 2$ (Lemma 43.23.4).

First, consider the open $G = \text{PGL}(V) \subset \mathbf{P}$ consisting of points $[g]$ such that $g : V \to V$ is invertible. Since $G$ acts doubly transitively on $\mathbf{P}(V)$ (Lemma 43.23.5) we see that

is a locally trivial fibration over $X \times Y$ with fibre equal to the stabilizer of a point in $G$. Hence $T$ is a variety. Observe that the fibre of $T \to G$ over $[g]$ is $r_ g(X) \cap Y$. The morphism $T \to G$ is surjective, because any translate of $X$ intersects $Y$ (note that by the assumption that $X$ and $Y$ intersect properly and that $X \cap Y \not= \emptyset $ we see that $\dim (X) + \dim (Y) \geq \dim (\mathbf{P}(V))$ and then Varieties, Lemma 33.34.3 implies all translates of $X$ intersect $Y$). Since the dimensions of fibres of a dominant morphism of varieties do not jump in codimension $1$ (Varieties, Lemma 33.20.4) we conclude that $B \cap G$ has codimension $\geq 2$.

Next we look at the complement $Z = \mathbf{P} \setminus G$. This is an irreducible variety because the determinant is an irreducible polynomial (Lemma 43.23.6). Thus it suffices to prove that $B$ does not contain the generic point of $Z$. For a general point $[g] \in Z$ the cokernel $V \to \mathop{\mathrm{Coker}}(g)$ has dimension $1$, hence $U(g)$ is the complement of a point. Since $X \not= \mathbf{P}(V)$ we see that for a general $[g] \in Z$ we have $X \subset U(g)$. Moreover, the morphism $r_ g|_ X : X \to r_ g(X)$ is finite, hence $\dim (r_ g(X)) = \dim (X)$. On the other hand, for such a $g$ the image of $r_ g$ is the closed subspace $H = \mathbf{P}(\mathop{\mathrm{Im}}(g)) \subset \mathbf{P}(V)$ which has codimension $1$. For general point of $Z$ we see that $H \cap Y$ has dimension $1$ less than $Y$ (compare with Varieties, Lemma 33.35.3). Thus we see that we have to show that $r_ g(X)$ and $H \cap Y$ intersect properly in $H$. For a fixed choice of $H$, we can by postcomposing $g$ by an automorphism, move $r_ g(X)$ by an arbitrary automorphism of $H = \mathbf{P}(\mathop{\mathrm{Im}}(g))$. Thus we can argue as above to conclude that the intersection of $H \cap Y$ with $r_ g(X)$ is proper for general $g$ with given $H = \mathbf{P}(\mathop{\mathrm{Im}}(g))$. Some details omitted. $\square$