Lemma 33.34.1. 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

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

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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