The Stacks project

33.34 Projective space

Some results on projective space over a field.

slogan

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 $k$.

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B2N. Beware of the difference between the letter 'O' and the digit '0'.