Download Solutions Sheet 8

D-MATH
Prof. Richard Pink
Algebraic Geometry
FS 2017
Solutions Sheet 8
Projective Schemes, Fiber Products
Exercise 3 is partly from Algebraic Geometry by Hartshorne.
1. Consider an ungraded ring R and a Z>0 -graded R-algebra S. Prove:
(a) If S is finitely generated over R, then Proj S is quasi-projective over R.
(b) If S is finitely generated over R by homogeneous elements of degree > 0, then
Proj S is projective over R.
Solution: Suppose that S is generated by homogeneous elements f0 , . . . , fn of
degrees > 0 and homogeneous elements g1 , . . . , gm of degree 0, with m = 0 in the
case (b). Let d be the least common multiple of the L
degrees of f0 , . . . , fn . Then
Proj S does not change if we replace S by the subring n>0 Snd . If we also replace
the generators f0 , . . . , fn by all monomials in them of total degree d, we can assume
that all of f0 , . . . , fn have the same degree d. Then S is isomorphic to a factor ring
of the graded ring S 0 := R0 [X0 , . . . , Xn ] for the ungraded ring R0 := R[Y1 , . . . , Ym ].
Thus Proj S is isomorphic to a closed subscheme of Proj S 0 = PnR0 .
This proves the desired statement in the case (b), because then R0 = R. In the
case (a) observe that PnR0 is isomorphic to the fiber product of Spec R0 = Am
R
with Proj R[X0 , . . . , Xn ] = PnR over Spec R. It is therefore isomorphic to an open
n
subscheme of the fiber product of Pm
R and PR over Spec R. By the Segre embedding,
mn+m+n
the latter is isomorphic to a closed subscheme of PR
. Thus the composite
mn+m+n
morphism Proj S → PR
is a locally closed embedding, proving (a).
2. Consider a Z>0 -graded ring S.
(a) Prove that Proj S = ∅ if and only if S+ ⊂ rad(S).
(b) Show that for S noetherian this is equivalent to Sn = 0 for all n 0.
(c) Deduce that the projective scheme over a field K defined by a homogeneous
ideal a ⊂ K[X0 , . . . , Xn ] is empty if and only if there exists m > 0 such that
X0m , . . . , Xnm ∈ a.
(d) Does (b) hold without the noetherian hypothesis?
Solution: (a) If S+ ⊂ rad(S), then S+ is contained in every prime ideal of S;
hence Proj S = ∅. Conversely
Pn suppose that S+ contains an element f that is
not nilpotent. Expand f = i=1 fi into homogeneous parts fi of degree i. Then
not all the fi can be nilpotent, because the nilradical of S is an ideal. Thus S+
1
contains a homogeneous element g that is not nilpotent. Then the localization Sg
is non-zero; hence so is its subring Sg,0 , which therefore possesses a prime ideal,
which in turn corresponds to a point in the standard open subset Dg ∼
= Spec Sg,0
of Proj S; hence Proj S 6= ∅.
(b) Suppose there is an integer k such that Sn = 0 for all n > k. Then for any
f ∈ S+ , each homogeneous part of f k has degree at least k, thus f k = 0 and f is
nilpotent. Conversely, if S is noetherian, the ideal S+ is generated by finitely many
homogeneous elements, say by s1 , . . . , sn of degrees d1 , . . . , dn > 0. If S+ ⊂ rad(S),
we can choose k > 1 such that each ski = 0. With d := max{d1 , . . . , dn } it then
follows that every homogeneous element of degree at least k(d + 1) is zero.
(c) Set R := K[X0 , . . . , Xn ]. We have X0m , . . . , Xnm ∈ a for some m > 0 if and only
if Rd ⊂ a for all d 0; hence if and only if Rd /ad = (R/a)d = 0 for all d 0. By
(b) this is equivalent to Proj(R/a) = ∅.
(d) No, counterexample: Let S be the polynomial ring in infinitely many variables
K[X1 , X2 , . . .], graded by total degree, modulo the homogeneous ideal (X12 , X23 , . . .).
Then S+ = rad(S), but for each n the residue class of Xnn is non-zero in Sn .
3. (a) Let ϕ : S → T be a graded homomorphism of Z>0 -graded rings (preserving
degrees). Show that U := {p ∈ Proj T | p + ϕ(S+ )} is an open subset of
Proj T and that ϕ determines a natural morphism f : U → Proj S.
(b) Work out U and f for the embedding R[X0 , . . . , Xm ] ,→ R[X0 , . . . , Xn ], both
f
graded by total degree, for n > m. The resulting morphism PnR ⊃ U −→ Pm
R
is an example of a projection to a linear subspace.
(c) The morphism f can be an isomorphism even when ϕ is not. For example,
suppose that ϕd : Sd → Td is an isomorphism for all d > d0 , where d0 > 0 is
an integer. Show that U = Proj T and the morphism f : Proj T → Proj S is
an isomorphism.
Solution: (a) Let p ∈ U . By definition of U , there exists an s ∈ S+ such that
ϕ(s) ∈
/ p. Choosing s homogeneous, we obtain an open neighborhood Dϕ(s) of p
contained in U ; hence U is open. Then ϕ yields a ring homomorphism Ss,0 → Tϕ(s),0
and hence a morphism Dϕ(s) ∼
= Spec Tϕ(s),0 → Spec Ss,0 ∼
= Ds ,→ Proj S. Check
that this behaves right on changing s; then by gluing we obtain a natural morphism
U → Proj S.
(b) Note that S+ is the ideal (X0 , . . . , Xm )S ⊂ S. Thus ϕ∗ (X0 , . . S
. , Xm )S is just the
ideal (X0 , . . . , Xm )T ⊂ T , and so U = PnR r V (X0 , . . . , Xm )T = m
i=0 D(Xi )T . For
any 0 6 i 6 m the morphism f |D(Xi )T corresponds to the ring homomorphism
n
0
0
, . . . , XXmi ] ,→ R[ X
,..., X
] = TXi ,0 . In projective coordinates that is
SXi ,0 = R[ X
Xi
Xi
Xi
described by the formula (x0 : . . . : xn ) 7→ (x0 : . . . : xm ), which is independent
of i.
(c) Clearly U ⊂ Proj T . Conversely, let p ∈ Proj T and suppose p ⊃ ϕ(S+ ). Let
t ∈ Tk for some k > 0. By assumption ϕd is an isomorphism for d > d0 , thus
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there exists s ∈ Skd0 such that td0 = ϕ(s) ∈ ϕ(S+ ) ⊂ p. Since p is prime, this
implies that t ∈ p, hence Tk ⊂ p. Varying k > 0, we conclude that T+ ⊂ p, which
is impossible since p ∈ Proj T . Thus p + ϕ(S+ ) and so p ∈ U . To show that
f is an isomorphism, let {ti }i be a set of generators of T+ , so that {Dti }i is an
open cover of Proj T . Then {Dtd0 }i is also a cover of Proj T . Set si := ϕ−1 (tdi 0 ),
i
which exists because ϕd is an isomorphism for d > d0 . Then the Dsi cover Proj S
and for each i the map fi = f |D d0 : Dtd0 → Dsi is a morphism of affine schemes
t
i
i
because each D d0 ∼
= Spec T d0 and Ds ∼
= Spec S(s ) and fi corresponds to the ring
ti
(ti )
i
i
homomorphisms ϕi : S(si ) → T(td0 ) induced by ϕ. Since each si has degree at least
i
d0 and ϕd is an isomorphism for all d > d0 , each ϕi is also an isomorphism, hence
so is each fi . Since the Dsi cover Proj S, this implies that f [ is an isomorphism,
and that f is surjective. For injectivity, let q, q0 ∈ Proj T with f (p) = f (q). Then
ϕ−1 (p) = ϕ−1 (q) ∈ Proj S and for each d > d0 we have pd = qd . Pick x ∈ p and
let s ∈ S+ such that s ∈
/ q, which exists because S+ * q. Then sd0 x ∈ p is of
degree at least d0 and thus lies in q. Since q is prime, this implies that x ∈ q.
Interchanging the roles of p and q shows the reverse inclusion, hence p = q and
thus f is injective.
4. (a) Show that
Pn (Q) ∼
= Pn (Z) ∼
= (x0 , . . . , xn ) ∈ Zn+1 gcd(x0 , . . . , xn ) = 1 /{±1}.
(b) Translate Fermat’s Last Theorem into a statement about the set of rational
points C(Q) of the projective curve C := V (X n + Y n − Z n ) ⊂ P2Q .
*5. Show that any projective scheme X over a field K is isomorphic to a closed subscheme of PnK for some n 0 whose ideal is generated by homogeneous polynomials of degree 2.
Solution (sketch): By assumption X ∼
= Proj S/a for some homogeneous ideal
a ⊂ S := K[X0 , . . . , Xn ]. Pick generators f1 , . . . , fm of a ∩ S+ and let d be the
least common multiple of their degrees. Then for any d0 > d we have ad0 = Sd0 −d ad .
0
∼
In
d . Setting S :=
L particular, by a theorem of the lecture we have X = Proj S/Sa
0
0
∼
k>0 Skd , another theorem of the lecture then implies that X = Proj S /S ad . But
by the d-uple embedding we have S 0 ∼
= S̃/b where S 0 is a polynomial ring in n+d
d
variables over K and the ideal b is generated by certain homogenous elements of
degree 2. Moreover, the elements of ad correspond to linear polynomials in the
variables of S̃. Thus the gradedring S 0 /S 0 ad is isomorphic to the factor ring of
a polynomial ring in n0 := n+d
− dimK (ad ) variables modulo an ideal that is
d
generated by homogeneous elements of degree 2. Therefore X is isomorphic to the
0
closed subscheme of PnK given by these polynomials.
*6. Show that arbitrary finite limits exist in the category of schemes.
Solution: See for instance http://stacks.math.columbia.edu/tag/04AS
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7. Consider homogeneous ideals a and b of a Z>0 -graded ring S, giving rise to closed
subschemes X and Y of P := Proj S. Prove that the scheme-theoretic intersection
X ∩ Y := X ×P Y is the closed subscheme corresponding to the homogeneous
ideal a + b.
Solution: For any homogeneous f ∈ S+ , the intersections of X and Y with the
standard open affine Df ∼
= Spec Sf,0 of P are the closed subschemes Spec Sf,0 /(Sf a)0
and Spec Sf,0 /(Sf b)0 . Their fiber product over Df is the spectrum of
Sf,0 /(Sf a)0 ⊗Sf,0 Sf,0 /(Sf b)0 ∼
= Sf,0 /((Sf a)0 + (Sf b)0 )
∼
= Sf,0 /(Sf a + Sf b)0
∼
= Sf,0 /(Sf (a + b))0 ,
that is, the intersection fo Df with the closed subscheme of P associated to the
ideal a + b.
8. For any morphism of schemes f : X → Y and any open or closed or locally closed
subscheme Y 0 ⊂ Y we call f −1 (Y 0 ) := X ×Y Y 0 the scheme-theoretic inverse image
of Y 0 under f , which is an open, resp. closed, resp. locally closed subscheme of X.
Show that f factors through a morphism X → Y 0 if and only if f −1 (Y 0 ) = X.
Solution (sketch): Generally, any locally closed embedding of schemes is a monomorphism. The definition of the fiber product as a limit implies that for any monomorphism S 0 → S and any schemes X and Y over S 0 the natural morphism X ×S 0 Y →
X ×S Y is an isomorphism. In our situation, if f factors through Y 0 , it follows
that f −1 (Y 0 ) = X ×Y Y 0 ∼
= X ×Y 0 Y 0 ∼
= X. Conversely, if f −1 (Y 0 ) = X, then f
proj
factors through the composite morphism X = f −1 (Y 0 ) = X ×Y Y 0 −→ Y 0 . Done.
9. Consider the quadric surface X := V (X 2 +Y 2 −Z 2 −W 2 ) ⊂ P3R , whose intersection
with the standard affine chart P3R r V (W ) ∼
= A3R is the one-sheeted hyperboloid Y
2
2
2
with the equation X + Y = Z + 1.
(a) Show that a change of coordinates induces an isomorphism X ∼
= V (XY −ZW )
1
1
∼
= PR × R PR .
(b) Show that the subschemes corresponding to {x} × P1R and P1R × {x} for all
x ∈ P1 (R) correspond to lines in X.
(c) Find a nice visualization of these lines on Y in the literature or the web.
*(d) Show that all lines contained in X are of this form.
Solution: (a) The change of coordinates (X, Y, Z, W ) 7→ (X + Z, Y + W, X − Z, W − Y )
does the trick. The second isomorphism is from the lecture.
(b) For a point (a0 : a1 ) ∈ P1R , the image of {(a0 : a1 )} × P1R under the Segre embedding P1R ×R P1R → V (XY − ZW ) is given by the ideal (a1 X − a0 Z, a1 Y − a0 W ).
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Since the ideal is defined by two independent linear equations, the associated subscheme is linear and of dimension 1, i.e., a line. The argument for the subscheme
corresponding to P1R × {(a0 : a1 )} is analogous.
(c) For instance https://en.wikipedia.org/wiki/Hyperboloid
*(d) Idea: Reduce to the statement that a plane V̄ (aX + bY + cZ + dW ) ⊂ P3R
intersects V (XY − ZW ) in a pair of lines if and only if ad = bc.
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