Show that an algebraic set \(Y\) of pure dimension \(r\), meaning that every irreducible component of \(Y\) has dimension \(r\), has degree \(1\) if and only if \(Y\) is a linear variety (Ex. 2.11).
Hint: First, use (7.7) and treat the case \(\dim Y = 1\). Then do the general case by cutting with a hyperplane and using induction.
Solution
1. A pure-dimensional algebraic set of degree one is irreducible.
Write the irreducible decomposition \[ Y=Y_1\cup\cdots\cup Y_m. \] All components have the same dimension \(r\) by purity, and distinct components meet in dimension strictly less than \(r\). Repeated application of Proposition I.7.6(b) gives \[ \deg Y=\sum_{i=1}^m\deg Y_i. \] Each \(\deg Y_i\) is a positive integer by Proposition I.7.6(a). If \(\deg Y=1\), this forces \(m=1\) and \(\deg Y_1=1\). Thus \(Y\) is a variety.
2. A zero-dimensional algebraic set of degree one is a point, hence a linear variety.
By step 1 it is irreducible. An irreducible zero-dimensional projective variety over the algebraically closed field \(k\) is a single closed point. A point is a projective linear subspace of dimension zero.
3. An irreducible projective curve of degree one is a line.
Let \(Y\subseteq\PP^n\) be an irreducible curve with \(\deg Y=1\). If \(n=1\), then the only one-dimensional closed irreducible subset is \(\PP^1\) itself, so the result is immediate.
Assume \(n\ge2\) and choose two distinct points \(P,Q\in Y\). Suppose some hyperplane \(H\) contains \(P\) and \(Q\) but does not contain \(Y\). Theorem I.7.7, applied to the degree-one hypersurface \(H\), gives \[ \sum_{R\in Y\cap H} i(Y,H;R)=\deg Y\cdot\deg H=1, \] because every zero-dimensional component is a point of degree one. But both \(P\) and \(Q\) occur in the sum with positive intersection multiplicity, a contradiction.
Hence every hyperplane containing \(P\) and \(Q\) contains all of \(Y\). The intersection of all hyperplanes containing \(P\) and \(Q\) is their projective span, the line \[ L=\overline{PQ}. \] Thus \(Y\subseteq L\). Since both are irreducible closed subsets of dimension one, \(Y=L\).
4. Assume inductively that every degree-one variety of dimension \(r-1\) is linear. Then every degree-one variety \(Y\) of dimension \(r\ge2\) is linear.
If \(Y=\PP^n\), then it is already linear, so assume \(r<n\). Choose a hyperplane \(H\) not containing \(Y\). The irreducible components \(Z_j\) of \(Y\cap H\) all have dimension \(r-1\) by the projective dimension theorem. Theorem I.7.7 gives \[ \sum_j i(Y,H;Z_j)\deg Z_j =\deg Y\cdot\deg H=1. \] Every intersection multiplicity and every degree in this sum is a positive integer. Consequently there is exactly one component \(Z\), with \[ i(Y,H;Z)=1, \qquad \deg Z=1. \] By the induction hypothesis, the reduced intersection is an \((r-1)\)-dimensional linear subspace; write it as \(L\).
Choose \(P\in Y\setminus H\) and let \[ M=\langle L,P\rangle. \] This is an \(r\)-dimensional linear subspace. We claim \(Y\subseteq M\). Suppose instead that \(Q\in Y\setminus M\). Then \(P\notin\langle L,Q\rangle\); otherwise \(Q\) would lie in \(\langle L,P\rangle=M\). Choose a hyperplane \(H'\) containing the \(r\)-plane \(\langle L,Q\rangle\) but not \(P\). Then \(H'\) does not contain \(Y\), while \[ L\cup\{Q\}\subseteq Y\cap H'. \]
Applying Theorem I.7.7 to \(H'\) again shows that \(Y\cap H'\) has exactly one irreducible component of dimension \(r-1\) and degree one. Since it contains the \((r-1)\)-plane \(L\), that unique component must be \(L\) itself. But then \(Q\in L\), contradicting \(Q\notin M\). Therefore \(Y\subseteq M\). Both \(Y\) and \(M\) are irreducible closed subsets of dimension \(r\), so \[ Y=M. \] Thus \(Y\) is linear.
5. Every linear variety has degree one.
A projective linear variety of dimension \(r\) is projectively equivalent to the standard coordinate subspace \(\PP^r\subseteq\PP^n\). Its homogeneous coordinate ring is a polynomial ring in \(r+1\) variables, so its Hilbert polynomial is \[ \binom{t+r}{r} =\frac{1}{r!}t^r+\text{lower terms}. \] Hence its degree is \[ r!\cdot\frac1{r!}=1. \]
6. Q.E.D.
Steps <1>1–<1>4 prove that every pure-dimensional degree-one algebraic set is a linear variety, and step 5 proves the converse.