1.7 Dimensions

Basic Definitions and Properties of Dimension

Definition

Let $X$ be a variety. Define $\dim X=\mathrm{tr}.{\mathrm{deg}}_{k}k(X)=\mathrm{tr}.{\mathrm{deg}}_k\mathrm{Frac}(R)$, where $\Gamma (U)=R$ for some affine open set $U\subseteq X$.

Remark.

• For nonempty open set $U\subseteq X$, $k(X)=k(U)$ as ${\underline{o}}_X(W)\simeq {\underline{o}}_X(W\cap U)$ and $W\cap U\ne \varnothing$ for any open set $W$. Therefore, $\dim X=\dim U$.
• If $U$ is affine, $\Gamma (U)=R$ and $k(U)=\mathrm{Frac}(R)$. See here.

an equivalent definition

It has an equivalent geometric definition, via chains of irreducible closed subsets.

$$\dim X=\max \{n\mid Z_0⊊Z_1⊊\cdots ⊊Z_n\}$$

where each $Z_i$ is an irreducible closed subset of $X$. The equivalence is shown in ^jp5j6v.

Proposition

Since $k$ is algebraically closed, the following are equivalent:

• $\dim X=0$
• $k(X)=k$
• $X$ is a point.

\begin{proof} i)←>ii) Since $\dim X=\mathrm{tr}.{\mathrm{deg}}_{k}k(X)=0$, we know $k(X)=k$. Conversely, if $k(X)=k$, then $\mathrm{tr}.{\mathrm{deg}}_{k}k(X)=0$ and so $\dim X=0$.

ii)→iii) Assume that $k(X)=k$. Since $X$ is a variety and a prevariety, it has an affine open subset $U\subseteq X$. Then we have $k(X)=k(U)=\mathrm{Frac}(R)=k$ by remark and so $U$ is a point. Since $X$ is irreducible and $U=\{p\}$ is open in $X$, we know $X=U=\{p\}$.

iii)→ii) Since $X$ is a point, $X\simeq {\mathbb{A}}^0=\mathrm{Spec}(k)$ and $\Gamma (X,{\underline{o}}_X)=k$. Then $k(X)=\mathrm{Frac}(k)=k$. \end{proof}

Lemma

Let $R$ be an integral domain over $k,P\subset R$ a prime. Then $\mathrm{tr}.{\mathrm{deg}}_{k}R⩾\mathrm{tr}.{\mathrm{deg}}_{k}R/p$, with equality only if $P=\{0\}$ or both sides are $\infty$.

\begin{proof} Let $\mathrm{tr}{\mathrm{deg}}_{k}R=n<\infty$, and let $P\ne \{0\}$. If the statement does not hold. Then there exist $x_1,\cdots ,x_n\in R$ such that ${\overline{x}}_1,\cdots ,{\overline{x}}_n\in R/p$ are algebraically independent. Let $q\in P\setminus \{0\}$. Then $q,x_1,\cdots ,x_n$ are algebraically dependent, that is, there exists a polynomial $T(y,x_1,\cdots ,x_n)\ne 0$ such that $T(q,x_1,\cdots ,x_n)=0$. We can assume that $y\nmid T(y,x_1,\cdots ,x_n)$. (If $T=y^n{T}_1\cdots T_{\ell }$, then $T(q,x_1,\cdots ,x_n)=q^n{T}_1\cdots T_{\ell }$ with $q^n\ne 0$. So we can replace $T$ by $T_i$ for some $i$.)

Note that

$$0=T(\overline{q},{\overline{x}}_1,\cdots ,{\overline{x}}_n)=T(0,{\overline{x}}_1,\cdots ,{\overline{x}}_n)$$

and $T(0,x_1,\cdots ,x_n)\ne 0$. As $T(0,{\overline{x}}_1,\cdots ,{\overline{x}}_n)=0$, they are algebraically dependent, which is a contradiction. \end{proof}

Proposition

Let $Y$ be a proper closed subvariety of $X$. Then $\dim Y<\dim X$.

\begin{proof} For $y_0\in Y$, define $y_0\in U\subseteq X$ for affine open set $U$. Since $Y\cap U\ne \varnothing$, $\dim X=\dim U$ and $\dim Y=\dim Y\cap U$. Note that $\Gamma (U)=R$ and $\Gamma (Y\cap U)=R/p$ with a prime ideal $p\subseteq R$, then $Y\ne X$ yields $p\ne \{0\}$ and $\dim Y<\dim X$ by ^xsfnsk. \end{proof}

Krull's Principal ideal theorem (PIT)

Define

• $R$ a finitely generated integral domain over $k$
• $f\in R$, $f\ne 0$ and $f\notin R^{\ast }$
• $p$ is a minimal prime ideal containing $f$.

then $\mathrm{tr}.{\mathrm{deg}}_{k}R/p=\mathrm{tr}.{\mathrm{deg}}_{k}R-1$.

Theorem

Let $X$ be an variety, and let $U\subseteq X$ be an open set. Take $0\ne g\in \Gamma (U,{\underline{o}}_X)$ and $Z$ is an irreducible component of $\{z\in U:g(x)=0\}\ne \varnothing$. Then $\dim Z=\dim X-1$.

\begin{proof} There exists open affine $U_0\subseteq X$ such that $U_0\cap Z\ne \varnothing$. Define $\Gamma (U_0)=R$ with $f=g{|}_{U_0}\in R$. Then $Z_0=U_0\cap Z$ is an irreducible component of $\{z\in U_0:f(z)=0\}$. As $\dim Z=\dim (U_0\cap Z)$ by remark, it suffices to show $\dim (U_0\cap Z)=\dim U_0-1$.

Firstly, note that $\dim U_0=\mathrm{tr}.{\mathrm{deg}}_k\mathrm{Frac}(R)$. On the other hand, recall that for a closed set $V(f)$ which has irreducible components $V(f)=Z_1^{\prime }\cup \cdots \cup Z_{\ell }^{\prime }$, then each $Z_k^{\prime }$ corresponds a minimal prime ideals $p_k=I(Z_k^{\prime })$ containing $(f)$. So $Z_0=V(p)$ where $p$ is the minimal prime ideal containing $f$, and we have $\Gamma (Z_0)\simeq R/p$ and $k(Z_0)=\mathrm{Frac}(R/p)$. Then by ^a5bed9 there is

$$\dim Z_0=\mathrm{tr}.{\mathrm{deg}}_k\mathrm{Frac}(R/p)=\mathrm{tr}.{\mathrm{deg}}_k\mathrm{Frac}(R)-1=\dim U_0-1.$$

Now we finish the proof. \end{proof}

Finite Morphism

Motivation for Finite Morphisms

Suppose we have a morphism of varieties $f:X\to Y$. Naturally, we start asking:

• How are the dimensions of $X$ and $Y$ related?
• Can the morphism “control” functions on $X$ via the structure of $Y$?
• Is there a way to mimic finite field extensions through geometric maps?

These questions motivate the study of finite morphisms.

finite morphism

Let $f:X\to Y$ be a morphism between affine varieties, and let $f^{\ast }:S\to R$ with $R=\Gamma (X)$ and $S=\Gamma (Y)$. Then $f$ is a finite morphism if $R$ is a finitely generated $f^{\ast }(S)$-module. Note that $R$ is a finitely generated $f^{\ast }(S)$-module iff $R$ is a finitely generated $S$-module iff $R$ is integral dependent on $f^{\ast }(S)$, that is, for any $r\in R$, there exists a monic polynomial $q(x)\in f^{\ast }(S)[x]$ such that $q(r)=0$.

Remark. Recall that $R$ is a finitely generated $S$-module $⟺$ $R$ is integral over $S$ is proved here.

Proposition

Let $f:X\to Y$ be a finite morphism.

• $f$ is a closed map, i.e., $Z\subseteq X$ closed yields $f(Z)\subseteq Y$ closed;
• for any $y\in Y$, $f^{-1}(y)$ is a finite set (quasi-finite);
• $f$ is surjective iff $f^{\ast }:\Gamma (Y)\to \Gamma (X)$ is injective.

\begin{proof} Define $m_{y_0}\subseteq S=\Gamma (Y)$ and $m_{x_0}\subseteq R=\Gamma (X)$, where $m_{x_0}=I(\{x_0\})$. We claim that if $f(x_0)=y_0$, then $m_{y_0}=(f^{\ast }{)}^{-1}(m_{x_0})$. For any $x\in m_{y_0}$, one have $(f^{\ast }(s))(x_0)=(s\circ f)(x_0)=s(y_0)=0$ and so $f^{\ast }(s)\in m_{x_0}$. It deduces that $m_{y_0}\subseteq (f^{\ast }{)}^{-1}(m_{x_0})$. By $m_{y_0}$ maximal, there is $m_{y_0}=(f^{\ast }{)}^{-1}(m_{x_0})$.

i) For any closed $Z\subseteq X$ such that $Z=V(I)$, we aim to show $f(V(I))=V(J)$ with $J=(f^{\ast }{)}^{-1}(I)$. Define $J=(f^{\ast }{)}^{-1}(I)$, then $R$ is a finitely generated $S$-module yields that $R/I$ is a finitely generated $S/J$-module. Thus $R/I$ is integral over $S/J$. By ^6qz19f, every prime ideal of $S/J$ is the restriction of a prime ideal from $R/I$.

For any $y_0\in V(J)$, take the corresponding maximal ideal $m_{y_0}/J$. By Going-Up, there exists prime ideal $Q/I$ of $R/I$ such that $m_{y_0}/J=(Q/I)\cap S/J$. It deduces that $m_{y_0}\subseteq (f^{\ast }{)}^{-1}(Q)\subseteq (f^{\ast }{)}^{-1}(m_{x_0})$, for some maximal ideal $m_{x_0}\supseteq Q\supseteq I$. So $m_{y_0}=(f^{\ast }{)}^{-1}(m_{x_0})$ and then $f(x_0)=y_1(=y_0)$ by the argument above. Hence, for any $y_0\in V(J)$, there exists $x_0\in V(I)$ such that $f(x_0)=y_0$. Therefore, $V(J)\subseteq f(V(I))$.

Conversely, to show $f(V(I))\subseteq V(J)$, it suffices to show for any $x\in V(I)$, we have $f(x)\in V(J)$. Take $x_0\in V(I)$ and $y_0=f(x)$. For any $j\in J$, $f^{\ast }(j)\in I$ and so $f^{\ast }(j)(x)=j(f(x))=j(y_0)=0$. Thus $y\in V(J)$ and so $f(V(I))\subseteq V(J)$. Therefore, $V(J)=f(V(I))$ and $f$ is a closed map.

ii) Since $R$ is a finitely generated $S$-module, we know $R^{\prime }=R/f^{\ast }(m_y)R$ is a finitely generated $S/m_y$-module, where $k=S/m_y$ is a field. It deduces that $R^{\prime }$ is a finite-dimensional $k$-algebra and so it is Artinian. By ^steden, $R^{\prime }$ has finitely many maximal ideals. Recall that for any $x\in f^{-1}(y)$, there is $(f^{\ast }{)}^{-1}(m_x)=m_y$. Thus $f^{-1}(y)$ is a finite set.

iii) Assume that $f$ is surjective. If $f^{\ast }(s)=0$, then $s\circ f=0$ on $X$ and so $s(f(X))=s(Y)=0$. Thus $s=0$ and $f^{\ast }$ is injective. Conversely, if $f^{\ast }$ is injective, then by ^j3610u $f(X)$ is dense in $Y$. Since $f$ is a closed map, $Y=\overline{f(X)}=f(X)$ and so $f$ is surjective. \end{proof}

Counterexample. ${\mathbb{A}}^1\setminus \{0\}\hookrightarrow {\mathbb{A}}^1$ satisfies ii) but not i). It is not a finite morphism, because $\Gamma ({\mathbb{A}}^1\setminus \{0\})=k[x,x^{-1}]$ is not a finitely generated $k[x]$-module.

Remark. If $Z\subseteq X$ is a closed subvariety, then $Z\stackrel{j}{\hookrightarrow }X\stackrel{f}{\to }Y$ is still finite, because $\Gamma (Z)=\Gamma (X)/I$ is still finitely generated.

Codimension and Intersection Theory

Definition

Let $X$ be a variety, and let $Z\subseteq X$ be a closed subset. We say $Z$ has pure dimension $r$ if each of its irreducible component of $Z$ is $\dim r$. Define ${\mathrm{codim}}_X{Z}_j=\dim X-\dim Z_j$. In general, $\dim Z=\max \{\dim Z_j:Z=\cup Z_j\text{ is irreducible decomposition}\}$.

Remark. By ^a5bed9, for any $g\in \Gamma (X,{\underline{o}}_X)$ with $V(g)\ne \varnothing$, $V(g)$ has pure $\mathrm{codim}1$.

Corollary

Let $X$ be a variety. For a maximal irreducible subset $Z⊊X$ (i.e. if $Z\subseteq W\subseteq X$ with irreducible $W$, then $Z=W$), we have $\dim Z=\dim X-1$.

\begin{proof} Since $Z\ne X$, there is $\dim Z<\dim X$ by ^7w5xag and so $I(Z)\ne (0)$. If $X$ is an affine variety, then take any nonzero $g\in I(Z)$ and consider $V_X(g)=\{x\in X:g(x)=0\}=W_1\cup \cdots \cup W_s$. By ^a5bed9, each irreducible component $W_j\subseteq V_X(g)$ has codimension $1$. Since $Z\subseteq V_X(g)$ is irreducible, there exists $j_0$ such that $Z\subseteq W_j$. Additionally, since $Z$ is maximal, one can check $Z=W_j$ and so $\dim Z=\dim X-1$.

If $X$ is not affine, take an affine open set $U\subseteq X$ such that $Z\cap U\ne \varnothing$ and define $Z_U=Z\cap U$. We first prove that $Z_U$ is a maximal irreducible subset of $U$. For any irreducible closed subset $W_U$ such that $Z_U\subseteq W_U⊊U$, notice that ${\overline{W_U}}^X\cap U=W_U$ and ${\overline{W_U}}^X$ is an irreducible closed subset of $X$. Claim that ${\overline{W_U}}^X⊊X$. Otherwise, ${\overline{W_U}}^X=X$ and $W_U=X\cap U=U$, which is impossible. Since $Z\cap U$ is an open set of $Z$, it is dense, i.e., ${\overline{Z\cap U}}^X=Z$. As $Z_U\subseteq W_U$, there is $Z={\overline{Z_U}}^X\subseteq {\overline{W_U}}^X=X$, which deduces that $Z={\overline{W_U}}^X$. Therefore, $Z_U={\overline{W_U}}^X\cap U=W_U$ and by the arbitrary of $W_U$ we know $Z_U$ is maximal.

By homework, $Z_U$ is an irreducible closed subset of $U$ and $Z_U⊊U$ (otherwise $Z\supseteq U$ and $\overline{Z}\supseteq \overline{U}=X$, which is impossible). Take $g\in \Gamma (U,{\underline{o}}_U)$ such that $g{|}_{Z_U}\equiv 0$ and $g$ is not always zero over $U$. Consider the set $V_U(g)=\{x\in U:g(x)=0\}$, which can be written as $V_U(g)=W_1\cup \cdots \cup W_s$ and $V_U(g)\supseteq Z_U$. Since $Z_U$ is irreducible and maximal in $U$, we have $Z=W_j$ for some $j$ and so $\dim Z_U=\dim U-1$. Since $Z_U\subseteq Z$ is an open subset of $Z$ and $U$ is an open subset of $X$, by remark we have $\dim Z=\dim Z_U=\dim X-1$. \end{proof}

Topological characterization of dimension

Suppose

$$\varnothing \ne Z_1⊈Z_2⊈\dots Z_1⊈Z$$

is any maximal chain of closed irreducible subsets of $X$. Then $\dim X=r$.

\begin{proof} Use induction and apply ^3de0e2. \end{proof}

Corollary

Let $X$ be a variety and let $Z$ be a component of $V\left(\left(f_1,\dots ,f_r\right)\right)$, where $f_1,\dots ,f_r\in \Gamma \left(X,{\underline{o}}_X\right)$. Then $\mathrm{codim}Z⩽r$.

\begin{proof} We prove it by induction on $r$. When $r=1$, $Z$ is a component of $V(f_1)$. If $f_1=0$, then $V(f_1)=X$ and $\mathrm{codim}Z=0$. Otherwise, if $f_1\ne 0$, then $V(f_1)$ has codimension $1$ by ^a5bed9 and so for $Z$. Hence ${\mathrm{codim}}_{X}Z⩽1=r$. Assume that the statement holds for $r=k-1$, and assume $Z$ is an irreducible component of $V(f_1,\cdots ,f_k)=V(f_1,\cdots ,f_{k-1})\cap V(f_k)\subseteq V(f_1,\cdots ,f_{k-1})$. Then $Z$ is contained in some irreducible component $Z^{\prime }\subseteq V(f_1,\cdots ,f_{k-1})$ and so $Z$ is an irreducible component of $Z^{\prime }\cap V(f_k)$.

If $f_k{|}_{Z^{\prime }}\equiv 0$, then $Z^{\prime }\subseteq V(f_k)$ and $Z^{\prime }\cap V(f_k)=Z=Z^{\prime }$. Hence ${\mathrm{codim}}_{X}Z={\mathrm{codim}}_X{Z}^{\prime }⩽k-1⩽k$ and we have done. If $f_k{|}_{Z^{\prime }}\not\equiv 0$, then $Z$ is an irreducible component of $Z^{\prime }\cap V(f_k{|}_{Z^{\prime }})$. By ^a5bed9, we have ${\mathrm{codim}}_{Z^{\prime }}Z=1$ and so

$${\mathrm{codim}}_{X}Z=\dim X-\dim Z=\dim X-(\dim Z^{\prime }-1)={\mathrm{codim}}_X{Z}^{\prime }+1.$$

By induction hypothesis, ${\mathrm{codim}}_{X}Z⩽k$. Now we finish the proof. \end{proof}

Example. Define $Z=\{(t^3,t^4,t^5):t\in k\}\subseteq {\mathbb{A}}^3$, then $Z=V(x^4-y^3,x^5-z^3,y^5-z^4)$ and $\mathrm{codim}Z=2$. However, $I(Z)=(x^4-y^3,x^5-z^3,y^5-z^4)$ cannot be generated by $2$ elements (not easy to prove).

Corollary

Let $U$ be an affine variety, and let $Z$ be a closed irreducible subset. Let $r=$ codim $Z$. Then there exist $f_1,\dots ,f_r$ in $R=\Gamma \left(U,{\underline{o}}_U\right)$ such that $Z$ is a component of $V\left(\left(f_1,\dots ,f_r\right)\right)$.

\begin{proof} We prove it by induction on $r$. When $r=0$, there is $Z=U$ and so $\{f_i\}=\varnothing$ is what we desired. When $r=1$, take $f_1\in I(Z)$ and then $Z\subseteq V(f_1)$. Since $Z\subseteq V(f_1)$ is irreducible, there exists an irreducible component $W_j\subseteq V(f_1)$ such that $Z\subseteq W_j$. By ^a5bed9, $\dim Z=\dim W_j=\dim U-1$ and so $Z=W_j$, i.e., $Z$ is a component of $V(f_1)$.

Assume that the statement holds for all $r^{\prime }<k$, that is, if $r^{\prime }=\mathrm{codim}Z<k$, then there exists $f_1,\cdots ,f_{r^{\prime }}$ such that $Z$ is a component of $V(f_1,\cdots ,f_{r^{\prime }})$. Now suppose $\mathrm{codim}Z=k$. Since $Z⊊U$, there exists nonzero $f_1\in I(Z)$ and $Z\subseteq V(f_1)$. Then $Z\subseteq W$ for some irreducible component $W\subseteq V(f_1)$. If $Z=W$, then ${\mathrm{codim}}_{U}Z={\mathrm{codim}}_{U}W=1$ and we have done. If $Z\ne W$, then ${\mathrm{codim}}_{W}Z=\dim W-\dim Z=(\dim U-1)-\dim Z=k-1<k$. Note that $W$ is an affine variety as $U$ is an affine variety, then by induction hypothesis there exist $\overline{g_2},\cdots ,\overline{g_k}\in \Gamma (U)/I(W)$ such that $Z$ is a component of $V_W(\overline{g_2},\cdots ,\overline{g_k})$. Suppose $f_i=g_i$ is a representative of $\overline{g_i}$.

Consider the set $Y_0=V_U\left(f_1,f_2,\dots ,f_k\right)$. We can prove that $Z$ is an irreducible component of $Y_0$; although the proof is nontrivial, we omit it here. This completes the argument. \end{proof}

Proposition

Let $X$ be an affine variety with coordinate ring $R$. Assume $R$ is a UFD. Then every closed subset $Z\subset X$ of pure codimension 1 equals $V((f))$ for some $f\in R$.

\begin{proof} Since $R$ is a UFD, every minimal prime ideal of $R$ is principal. For a minimal prime ideal $P\subset R$ and $f\in P$, $P$ contains one of the prime factors $f^{\prime }$ of $f$. By the UFD property, $\left(f^{\prime }\right)$ is also a prime ideal and since $\left(f^{\prime }\right)\subset P$, we must have $\left(f^{\prime }\right)=P$. Let $Z_1,\dots ,Z_t$ be the components of $Z$. Then $I\left(Z_1\right),\dots ,I\left(Z_t\right)$ are minimal prime ideals. If $I\left(Z_i\right)=\left(f_i\right)$, then

$$Z=V\left(\left(f_1,\dots ,f_t\right)\right)=V(f_1\cdots f_t).$$

Now we finish the proof. \end{proof}

Example. Localization can get better ring (non UFD → UFD for example).

• Define $X=V(xy-z^2)\subseteq {\mathbb{A}}^3$ with $\Gamma (X)=R=k[x,y,z]/(xy-z^2)$, and $W=\{x=0\}\cap X=\{(0,y,0):y\in {\mathbb{A}}^1\}$.
  • $I(W)=(x,z)$ is a prime ideal, which is not a principal ideal.
• Consider its localization $X_y=\{(x,y,z)\in X:y\ne 0\}=X\setminus V(y)$.
  • $\Gamma (X_y)=R_y=k[y,y^{-1},z/y]$ is a UFD, because $k[y,z/y]$ is a UFD and $R_y$ is a localization of $k[y,z/y]$ and so also a UFD.
  • $(I(W){)}_y=(z/y)$ is generated by $f=z/y$, which is consistent with ^yof8h3.

Proposition

$\dim X\times Y=\dim X+\dim Y$

\begin{proof} Suppose $U\subseteq X$ and $V\subseteq Y$ are affine open sets. Then $\dim X\times Y=\dim U\times V$ where $\Gamma (U)=R$, $\Gamma (V)=S$. Because $\Gamma (U\times V)=R{\otimes }_{k}S$, we have

$$\dim U\times V=\mathrm{tr}{\mathrm{deg}}_{k}R{\otimes }_{k}S\stackrel{\Delta }{=}\mathrm{tr}\mathrm{deg}R+\mathrm{tr}\mathrm{deg}S=n+m.$$

It suffices to show $\Delta$: note that $R\supseteq k[x_1,\cdots ,x_n]$ and $S\supseteq k[y_1,\cdots ,y_m]$ are algebraic by ^d13b03, then $R\otimes S\supseteq k[x_1,\cdots ,x_n,y_1,\cdots ,y_m]$ are algebraic. \end{proof}

Dimension in Projective Varieties

Theorem

Let $X\subset {\mathbb{P}}_n(k)$ be a projective variety, and let $I(X)\subset k\left[X_0,\dots ,X_n\right]$ be its ideal. If $f\in k\left[X_0,\dots ,X_n\right]$ is homogeneous and not a constant and $f\notin I$, then $X\cap V((f))$ is non-empty and of pure codimension $1$ in $X$, unless $\dim X=0$.

Remark. $X\cap V((f))\ne \varnothing$ does not always hold in affine case. For example, $X=\{x=0\}\subseteq {\mathbb{A}}^3$ and $V(f)=\{x=1\}\subseteq {\mathbb{A}}^3$ do not have intersection.

\begin{proof} It is enough to show $X\cap V((f))\ne \varnothing$ for all non-constant $f\notin I(X)$, because remains part has proved by PIT.

Suppose $\dim X⩾1$. Define $C(X)=V^{\ast }(I(X))\subseteq {\mathbb{A}}^{n+1}$. Since $I(X)$ is homogeneous, there is $0\in C(X)$ and so $0\in Z^{\ast }=C(X)\cap V^{\ast }(f)$. Note that $\dim C(X)=\dim X+1⩾2$ and $f$ is not always $0$ on $C(X)$ as $f\notin I(X)$. By ^jp5j6v, every irreducible component of $C(X)\cap V^{\ast }(f)$ has dimension $\dim C(X)-1=\dim X⩾1$. It deduces that $Z^{\ast }$ has a nonzero point $(a_0,\cdots ,a_n)$, which corresponds to $[a_0:\cdots :a_n]$ in ${\mathbb{P}}^n(k)$. Then $[a_0:\cdots :a_n]\in X\cap V(f)$ and so the set is nonempty. \end{proof}

Corollary

Take homogeneous $f_1,f_2\in k[x_0,\cdots ,x_n]\setminus k$, where $n⩾2$. If $\dim V(f_1)⩾1$, then $V(f_1)\cap V(f_2)\ne \varnothing$ in ${\mathbb{P}}^n$.