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Algebraic Geometry
By: Robin Hartshorne
Solutions
Solutions by Joe Cutrone and Nick Marshburn
1
Foreword:
This is our attempt to put a collection of partially completed solutions scattered
on the web all in one place. This started as our personal collection of solutions
while reading Hartshorne. We were stuck (and are still) on several problems,
which led to our web search where we found some extremely clever solutions by
[SAM] and [BLOG] among others. Some solutions in this .pdf are all theirs and
just repeated here for convenience. In other places the authors made corrections
or clarifications. Due credit has tried to be properly given in each case. If you
look on their websites (listed in the references) and compare solutions, it should
be obvious when we used their ideas if not explicitly stated.
While most solutions are done, they are not typed at this time. I am trying
to be on pace with one solution a day (...which rarely happens), so I will update
this frequently. Check back from time to time for updates. As I am using this
really as a learning tool for myself, please respond with comments or corrections.
As with any math posted anywhere, read at your own risk!
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1 Chapter 1: Varieties
1.1 Affine Varieties
1. (a) Let Y be the plane curve defined by y = x2. Its coordinate ring A(Y )
2 ∼ 2 ∼
is then k[x,y]/(y − x ) = k[x,x ] = k[x].
(b) A(Z) = k[x,y]/(xy−1) ∼ k[x, 1], which is the localization of k[x] at
= x
x. Any homomorphism of k-algebras ϕ : k[x, 1] → k[x] must map x
x
into k, since x is invertible. Then ϕ is clearly not surjective, so in
particular, not an isomorphism.
(c) Let f(x,y) ∈ k[x,y] be an irreducible quadratic. The projective
closure is defined by z2f(x, y) := F(x,y,z). Intersecting this variety
z z
withthehyperplaneatinfinityz = 0givesahomogeneouspolynomial
F(x,y,0) in two variables which splits into two linear factors. If
F has a double root, the variety intersects the hyperplane at only
one point. Since any nonsingular curve in P2 is isomorphic to P1,
Z(F)\∞=P1\∞∼A1. So Z(f)∼A1. If F has two distinct roots,
= =
say p,q, then the original curve is P1 minus 2 points, which is the
same as A1 minus one point, call it p. Change coordinates to set
p = 0 so that the coordinate ring is k[x, 1].
x
1 2 3
2. Y is isomorphic to A via the map t 7→ (t,t ,t ), with inverse map being
the first projection. So Y is an affine variety of dimension 1. This also
shows that A(Y) is isomorphic to a polynomial ring in one variable over
2 3
k. I claim that the ideal of Y,I(Y) is (y − x ,z − x ). First note that
2 2 3
for any f ∈ f[x,y,z], I can write f = h (y − x ) + h (z − x ) + r(x), for
1
r(x) ∈ k[x]. To show this it is enough to show it for an arbitrary monomial
α β γ α 2 2 β 3 3 γ α 2β
x y z = x (x + (y − x )) (x + (z − x )) = x (x + terms with
2 3γ 3 2 3 α+2β+3γ
y−x )(x +termswithz−x )=h1(y−x )+h2(z−x )+x , for
2 3
h ,h ∈k[y,y,z]. Now, clearly (y−x ,z−x ) ⊆ I(Y). Soshowthereverse
1 2
inclusion, let f ∈ I(Y ) and write f = h (y−x2)+h2(z−x3)+r(x). Using
1
2 3 2 3
the parametrization (t,t ,t ),0 = f(t,t ,t ) = 0 + 0 + r(t), so r(t) = 0.
3. Let Y ⊆ A2 be defined by x2 − yz = 0 and xz − x = 0. If x = 0,
then y = 0 and z is free, so we get a copy of the z-axis. If z = 0,
then y is free, so we get the y-axis. If x 6= 0,z = 1,y = x2. So Y =
2
Z(x −y,z −1)∪Z(x,y)∪Z(x,z). Since each piece is isomorphic to
A1, (see ex 1), the affine coordinate ring of each piece is isomorphic to a
polynomial ring in one variable.
4. Let A2 = A1×A1. Consider the diagonal subvariety X = {(x,x)|x ∈ A1}.
This is not a finite union of horizontal and vertical lines and points, so it
is not closed in the product topology of A2 = A1 ×A1.
5. These conditions are all obviously necessary. If B is a finitely-generated
k-algebra, generated by t ,...,t , then B ∼ k[x ,...,x ]/I, where I is
1 n = 1 n
an ideal of the polynomial ring defined by some f1,...,fn. Let X ⊆ An
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be defined by f = ... = f = 0. We prove that I =I from which it
1 n X
will follow that k[X] ∼ k[x ,...,x ]/I ∼ B. If F ∈ I , then Fr ∈ I for
= 1 n = X
some r > 0 by the Nullstellensatz. Since B has no nilpotents, also F ∈ I,
thus IX ⊂ I, and since obviously I ⊂ IX, equality follows.
6. Let U ⊆ X be a nonempty open subset with X irreducible. Assume U
is not dense. Then there exists a nonempty open set V ⊆ X such that
V∩U =∅,namelyX\U. ThenX =Uc∪Vc,contradictingthefactthatX
is irreducible. So U is dense. If U were not irreducible, write U = Y ∪Y
1 2
where each Yi is closed inside of U and proper. Then for two closed
subsets X ,X ⊆ X, such that Y = U ∩X ,(X ∪X )∪Uc = X, so X is
1 2 i i 1 2
reducible. Contradiction, so U is irreducible. Suppose Y is an irreducible
subset of X and suppose Y = Y ∪ Y . Then Y = (Y ∩ Y) ∪ (Y ∩ Y),
1 2 1 2
so by irreducibility of Y, we have WOLOG Y = (Y1 ∩Y). Since Y is the
smallest closed subset of X containing Y, it follows that Y = Y , so Y is
1
irreducible.
7. (a) (i → ii) If X is a noetherian topological space, then X satisfies the
D.C.C for closed sets. Let Σ be any nonempty family of closed sub-
sets. Choose any X ∈ Σ. If X is a minimal element, then (ii)
1 1
holds. If not, then there is some X ∈ Σ such that X ⊂ X . If X
2 2 1 2
is minimal, (ii) holds. If not, chose a minimal X . Proceeding in this
3
way one sees that if (ii) fails we can produce by the Axiom of Choice
an infinite strictly decreasing chain of elements of Σ, contrary to (i).
(ii → i) Let every nonempty family of closed subsets contain a mini-
mal element. Then X satisfies the D.C.C. for closed subsets, so X is
noetherian.
(iii → iv) and (iv →iii) Same argument as above.
(i ↔ iii) Let C ⊂ C ⊂ ... be an ascending chain of open sets. Then
1 2
taking complements we get Cc ⊃ Cc ⊃ ..., which is a descending
1 2
chain of closed sets. So X is noetherian iff the closed chain stabilizes
iff the open chain stabilizes.
(b) Let X = SU be an open cover. Pick U and U such that U ⊂
α 1 2 1
(U ∪U ) (strict inclusion). Pick U such that U ⊂ (U ∪ U ∪
1 2 3 2 1 2
U ). Continue in this fashion to produce an ascending chain of open
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subsets. By part a), since X is noetherian, this chain must stabilize
and we get a finite cover of X.
(c) Let Y ⊆ X beasubsetofanoetheriantopologicalspace. Consideran
open chain of subsets V ⊆ V ⊆ ... in Y. By the induced topology,
0 1
there exists open U ⊆ X such that U ∩ Y = V . Form the open
S i S i S i
sets W = k U. So W ∩Y = k = k V =V . The chain
i i=1 i k i=1 i=1 i k
W ⊆W ⊆...inX stabilizes since X is Noetherian. So the chain
0 1
V0 ⊆ V1 ⊆ ... in Y , stabilizes, so by part a), Y is noetherian.
which stabilizes since X is noetherian. Thus the original chain in Y
stabilizes, so by part (a), Y is Noetherian.
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