Let \(\varphi: \PP^n_k \to \PP^m_k\) be a morphism. Then
either \(\varphi(\PP^n) = \pt\), or \(m \geq n\) and \(\dim \varphi(\PP^n) = n\);
in the second case, let \(\PP^r\subseteq\PP^m\) be the linear span of the image. Show that \(\varphi\) can be obtained as the composition of
- a \(d\)-uple embedding \(\PP^n \to \PP^N\) for a uniquely determined \(d \geq 1\),
- a linear projection \(\PP^N - L \to \PP^r\),
- the standard linear inclusion \(\PP^r\hookrightarrow\PP^m\), and
- an automorphism of \(\PP^m\) carrying that coordinate subspace to the image’s linear span.
When the image is nondegenerate, \(r=m\) and the inclusion is unnecessary. Also, \(\varphi\) has finite fibres.
(Projection onto the linear span)
The retained statement of Exercise II.7.3 projects directly onto \(\PP^m\), which requires the image to be nondegenerate when the \(d\)-uple embedding uses the complete degree-\(d\) system. For example, \([x:y]\mapsto[x:y:0]\) from \(\PP^1\) to \(\PP^2\) pulls \(\OO(1)\) back to \(\OO(1)\). It therefore has \(d=1\) and \(N=1\), and no surjective linear map \(k^2\to k^3\) exists to define the claimed projection. Even when \(m=N\), the map \([x:y]\mapsto[x^2:y^2:0]\) has \(d=2\) and image a line, whereas a linear projection \(\PP^2\to\PP^2\) is an automorphism and cannot carry the nondegenerate Veronese conic into a line. Projection onto \(\PP^r\), followed by a linear inclusion, covers these cases without changing the morphism or its uniquely determined \(d\).
Solution
All morphisms are over \(k\). For \(n=0\) the source is \(\Spec k\), so the image is one point and only the first alternative of (a) applies. Assume \(n\ge1\), and put \(S=k[x_0,\ldots,x_n]\) with its usual grading.
1. The morphism is defined by forms \(f_0,\ldots,f_m\in S_d\) with no common zero, for a unique integer \(d\ge0\).When \(d=0\), it is constant.
The Picard group of \(\PP_k^n\) is generated freely by \(\OO(1)\) [@Har10a, Proposition II.6.4 and Corollary II.6.16]. Consequently \(\varphi^*\OO_{\PP^m}(1)\cong\OO_{\PP^n}(d)\) for a unique integer \(d\). Pulling back the coordinate sections gives \(m+1\) sections generating this invertible sheaf [@Har10a, Theorem II.7.1]. The global sections of \(\OO(d)\) are \(S_d\) for \(d\ge0\) and zero for \(d<0\) [@Har10a, Proposition II.5.13]. Generation excludes \(d<0\) and gives the stated forms and their empty common zero locus. The morphism has coordinates \([f_0:\cdots:f_m]\). For \(d=0\), these are constants in \(k\), not all zero, so the morphism factors through their \(k\)-rational point.
2. For \(d>0\), the Veronese ring \(R=\bigoplus_{q\ge0}S_{qd}\) is a finite module over \(A=k[f_0,\ldots,f_m]\), where both rings are graded with \(f_i\) of degree one.
Let \(I=(f_0,\ldots,f_m)\subseteq S\). The empty projective common zero locus implies that each \(x_j\) has a power in \(I\). Indeed, on \(D_+(x_j)\) the functions \(f_i/x_j^d\) generate the unit ideal; clearing denominators in a unit-ideal expression gives \(x_j^{b_j}\in I\) for some \(b_j\ge1\). Every monomial of sufficiently large total degree is divisible by one of these powers. There is therefore \(B\ge d\) with \(S_q\subseteq I\) for every \(q\ge B\).
If \(h\in S_{qd}\) and \(qd\ge B\), homogeneity gives \[ h=\sum_{i=0}^m f_i h_i,\qquad h_i\in S_{(q-1)d}. \] Induction on \(q\) expresses \(h\) as an \(A\)-linear combination of monomials of degrees divisible by \(d\) and smaller than \(B\). There are finitely many such monomials, proving that \(R\) is a finite graded \(A\)-module.
3. If \(d>0\), the morphism is finite onto its image, and that image has dimension \(n\).
The surjection \(k[y_0,\ldots,y_m]\to A\), \(y_i\mapsto f_i\), identifies \(Y=\operatorname{Proj}A\) with a closed integral subscheme of \(\PP^m\). Also \(\operatorname{Proj}R\cong\PP^n\) by the Veronese construction [@Har10a, Exercise II.5.13]. The graded inclusion \(A\subseteq R\) induces the morphism \(\varphi\) with target restricted to \(Y\).
For every nonzero \(f_i\), its inverse image of \(D_+(f_i)\subseteq Y\) is the affine open \(D_+(f_i)\subseteq\operatorname{Proj}R\). The corresponding ring inclusion is \[ (A[f_i^{-1}])_0\longrightarrow(R[f_i^{-1}])_0. \] If homogeneous elements \(u_j\) of degrees \(e_j\) generate \(R\) as an \(A\)-module, then \(u_j/f_i^{e_j}\) generate the right-hand ring as a module over the left-hand ring. This follows by taking degree-zero parts of expressions in the localized graded module. Thus these ring maps are finite and injective, so their maps on spectra are surjective by lying over. The opens cover \(Y\); hence \(\PP^n\to Y\) is finite and surjective. Composing with \(Y\hookrightarrow\PP^m\) shows that \(\varphi\) is itself finite.
Integral extensions preserve Krull dimension: going up lifts chains and incomparability prevents a strict chain from collapsing under contraction [@AM18, Chapter 5]. Step 2 makes \(R\) integral over \(A\). The ring \(S\) is integral over \(R\), since each \(x_j\) satisfies the monic equation \(T^d-x_j^d=0\) with coefficient \(x_j^d\in R\). Therefore \(\dim A=\dim R=\dim S=n+1\). For a standard graded finite-type domain over \(k\), its Proj has dimension one less than the ring [@Har10a, Chapter I, §2]. It follows that \(\dim Y=n\), so \(m\ge n\). Every fibre of the finite morphism is the spectrum of a finite-dimensional algebra over its residue field, and thus has finitely many points. Together with step 1, this proves (a) and the fibre assertion of (b).
4. For \(d>0\), the morphism factors through the \(d\)-uple embedding, projection onto its image’s linear span, and the stated linear inclusion and automorphism.
Let \(V\) be the span of \(f_0,\ldots,f_m\) in \(S_d\), and write \(r+1=\dim_kV\). Choose a basis \(g_0,\ldots,g_r\) of \(V\) and extend it to a basis \(g_0,\ldots,g_N\) of \(S_d\), where \[ N=\binom{n+d}{d}-1. \] The complete set of degree-\(d\) monomials gives the \(d\)-uple closed embedding; replacing that monomial basis by the \(g_i\) changes it by an ambient projective linear automorphism [@Har10a, Exercises I.2.12 and I.3.4]. Denote the resulting embedding by \(\nu_d\). The projection \[ p:\PP^N\setminus L\longrightarrow\PP^r,\qquad [z_0:\cdots:z_N]\longmapsto[z_0:\cdots:z_r], \quad L=V_+(z_0,\ldots,z_r), \] is defined on \(\nu_d(\PP^n)\) because the \(g_0,\ldots,g_r\) span the same globally generating space as the \(f_i\).
Write \(f_i=\sum_{j=0}^r c_{ij}g_j\). The matrix \(C=(c_{ij})\) has rank \(r+1\), since the \(f_i\) span \(V\). Extend its columns to a basis of \(k^{m+1}\), producing an invertible matrix and hence an automorphism \(\alpha\) of \(\PP^m\). For the coordinate inclusion \(\iota:\PP^r\hookrightarrow\PP^m\), the resulting factorization is \[ \boxed{\varphi=\alpha\circ\iota\circ p\circ\nu_d.} \] Both sides have the same generating sections \(f_i\), so this is equality of morphisms by Theorem II.7.1, not only equality on points.
No nonzero linear form vanishes on \(p\nu_d(\PP^n)\): its pullback would be a zero linear combination of the basis elements \(g_0,\ldots,g_r\). Consequently \(\alpha\iota(\PP^r)\) is exactly the linear span of the image. When \(r=m\), \(\iota\) is the identity and the factorization has the three maps in the nondegenerate statement.
5. The integer in the factorization is uniquely determined by \(\varphi^*\OO(1)\cong\OO(d)\).
The pullback of \(\OO(1)\) under a linear projection is \(\OO(1)\) on its domain, because its defining linear forms generate that restricted line bundle. Linear inclusions and projective linear automorphisms also preserve \(\OO(1)\), and the \(e\)-uple embedding pulls it back to \(\OO(e)\). Thus any factorization of the stated kind gives \(\varphi^*\OO(1)\cong\OO(e)\). The uniqueness in step 1 forces \(e=d\).
6. Q.E.D.
Steps <1>1–<1>3 prove the alternatives and finite fibres. Steps <1>4–<1>5 give the general factorization and the uniqueness of its degree.