Let \(k\) be algebraically closed and let \(n\ge0\). A hypersurface defined by a linear polynomial is called a hyperplane.
Show that the following two conditions are equivalent for a projective variety \(Y \subseteq \PP_k^n\).
\(I(Y)\) can be generated by linear polynomials.
- \(Y\) can be written as an intersection of hyperplanes.
In this case \(Y\) is called a linear variety in \(\PP^n\).
If \(Y\) is a linear variety of dimension \(r\) in \(\PP^n\), show that \(I(Y)\) is minimally generated by \(n-r\) linear polynomials.
Let \(Y, Z\) be linear varieties in \(\PP^n\) with \(\dim Y = r\) and \(\dim Z = s\). If \(r + s - n \geq 0\), then \(Y \intersect Z \neq \emptyset\). Furthermore, if \(Y \intersect Z \neq \emptyset\), then \(Y \intersect Z\) is a linear variety of dimension \(\geq r+s-n\).
Hint
Think of \(\AA^{n+1}\) as a vector space over \(k\) and work with its subspaces.
Solution
Put \(V=k^{n+1}\) and \(S=\operatorname{Sym}(V^*)=k[x_0,\ldots,x_n]\). For a vector subspace \(W\subseteq V\), let \(\PP(W)\) be its set of one-dimensional subspaces, with \(\PP(0)=\varnothing\), and let \[ W^\perp=\{\ell\in V^*: \ell|_W=0\} \] be its annihilator. The intersection of an empty family of hyperplanes is understood to be \(\PP^n\).
1. For a nonzero subspace \(W\subseteq V\) of dimension \(r+1\), the set \(\PP(W)\) is a linear projective variety of dimension \(r\), and\[ I(\PP(W))=(W^\perp). \]
Choose a basis of \(W\) and extend it to one of \(V\). In the dual coordinates \(y_0,\ldots,y_n\), the subspace \(W\) is given by \(y_{r+1}=\cdots=y_n=0\). The same linear coordinate change defines a projective automorphism; it and its inverse are regular on the standard affine charts by their coordinate-ratio formulas. Consequently \(\PP(W)\) is identified with the coordinate copy of \(\PP^r\) and has dimension \(r\) [@Har10a, Chapter I, §2].
The ideal \(J=(y_{r+1},\ldots,y_n)\) is prime, since its quotient is the polynomial ring \(k[y_0,\ldots,y_r]\). Its projective zero set is nonempty and equals \(\PP(W)\). By P-AGH24CORRESPONDENCE, \(I(\PP(W))=J\). Its degree-one part is precisely \(W^\perp\). Primality also proves that \(\PP(W)\) is irreducible.
2. Conditions (i) and (ii) in (a) are equivalent.
If \(I(Y)\) is generated by linear forms \(\ell_i\), then \[ Y=Z(I(Y))=\bigcap_i Z(\ell_i). \] Zero generators can be omitted, and every remaining \(Z(\ell_i)\) is a hyperplane. Since the space \(S_1=V^*\) is finite-dimensional, the span of the \(\ell_i\) has a finite basis, so the same intersection is defined by finitely many hyperplanes.
Conversely, suppose \(Y\) is an intersection of hyperplanes with equations \(\ell_i=0\). Set \(W=\bigcap_i\ker\ell_i\) in \(V\). The projective intersection is exactly \(\PP(W)\). Since a variety is nonempty, \(W\ne0\). Step 1 gives \(I(Y)=(W^\perp)\), which is generated by linear forms. This proves (a).
3. For the variety in (b), the minimum number of ideal generators is \(\boxed{n-r}\), and a basis of \(W^\perp\) is a minimal generating set.
By steps <1>1–<1>2, write \(Y=\PP(W)\) with \(\dim W=r+1\). Linear algebra gives \(\dim W^\perp=n-r\), so a basis of this annihilator gives \(n-r\) generators of \(I(Y)\).
For the lower bound, put \(\mathfrak m=S_+=(x_0,\ldots,x_n)\). The ideal \(I(Y)\) is generated in degree one, and \(\mathfrak m I(Y)\) consists of its components in degrees at least two. Hence the natural map gives a vector-space isomorphism \[ I(Y)/\mathfrak m I(Y)\cong I(Y)_1=W^\perp. \] Any ideal generating set maps to a spanning set of this quotient over \(S/\mathfrak m=k\). It must therefore have at least \(n-r\) elements, even if nonhomogeneous generators are allowed. When \(r=n\), the ideal is zero and the empty generating set gives the same conclusion.
4. The nonemptiness and dimension assertions in (c) hold.
Write \(Y=\PP(W)\) and \(Z=\PP(T)\), where \(\dim W=r+1\) and \(\dim T=s+1\). A projective point belongs to both precisely when its one-dimensional subspace lies in \(W\cap T\). Thus \[ Y\cap Z=\PP(W\cap T). \] The vector-space dimension identity gives \[ \dim(W\cap T)=\dim W+\dim T-\dim(W+T) \ge r+s+2-(n+1)=r+s-n+1. \] If \(r+s-n\ge0\), this dimension is at least one, so the projective intersection is nonempty. Whenever \(W\cap T\ne0\), step 1 makes its projectivization a linear variety and gives \[ \dim(Y\cap Z)=\dim(W\cap T)-1\ge r+s-n. \] This proves both assertions of (c), including intersections with the whole projective space.
5. Q.E.D.
Step 2 proves (a), step 3 proves the minimal-generation assertion in (b), and step 4 proves (c).