Mahir Bilen Can - Problems and Solutions in Commutative Algebra [expository notes] (2015)

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PROBLEMS AND SOLUTIONS IN
COMMUTATIVE ALGEBRA
Mahir Bilen Can
mcan@tulane.edu
Disclaimer: This file contains some problems and solutions in commutative algebra as
well as in field theory. About first hundred problems are those that we encountered at some
point probably between years 2003 and 2005. We do not claim correctness of those solutions
(neither of the other solutions). Read them at your own risk. However, we do appreciate if
you send us corrections and suggest new problems and solutions.1
Notation: Unless otherwise stated all rings are assumed to be commutative with unity.
1. Find two ideals I and J in a ring R such that I · J 6= I ∩ J.
Solution.
Let I = J be the ideal generated by x in the polynomial ring R = k[x]. Then
I ∩ J = I = (x). Since the product of two ideals consists of finite sum of products of
elements of I and J , the ideal product I ·J is equal to (x2) which is different from (x).
2. Definition: ideals I and J from R are called co-prime, if their sum I+J is equal to R.
Show that if I and J are two co-prime ideals in a ring R, then I · J = I ∩ J.
Solution.
For all ideals I and J the inclusion I · J ⊆ I ∩ J is clear. To prove the other inclusion
we observe the simple fact that, for any ideal K of R, the following equality KR = K
is true. Hence, assuming I and J are co-prime, on the one hand we have I ∩ J =
(I ∩ J)(I + J) = (I ∩ J) · I + (I ∩ J) · J . On the other hand, (I ∩ J) · I ⊆ J · I and
(I ∩ J) · J ⊆ I · J . Equality is now obvious.
3. Definition: A multiplicative subset S of a ring R is a multiplicative submonoid of R.
Let S be a multiplicative subset in a ring R and I be an ideal.
1We thank Professor Lex Renner for his comments and critical eye on some of the problems with faulty
solutions. We thank Şafak Özden, also.
1
(a) Show that S−1I := {a/s : a ∈ I, s ∈ S} is an ideal in the localized ring
S−1R = {r/s : r ∈ R, s ∈ S}.
(b) Show that the localization commutes with quotients: S−1R/S−1I ∼=
S−1(R/I). Here we are abusing the notation on the right: of course
R/I is localized at the image of S in R/I.
Solution.
(a) Let a1/s1 and a2/s2 be two elements from S−1I. And let r/s ∈ S−1R . Then
r/s · a1/s1 + a2/s2 = (s2ra1 − s1sa2)/s1s2 ∈ S−1I. Therefore S−1I is an ideal.
(b) Elements of S−1(R/I) are of the form r̄/s̄ where bar denotes the images of elements
of S in R/I. If we start with an element r/s of S−1R, then r̄/s̄ makes sense. So we
can define the homomorphism φ : S−1R −→ S−1(R/I) by φ(r/s) = r̄/s̄. By its
construction φ is surjective. How about its kernel? Suppose r̄/s̄ = 0 in S−1(R/I).
Then there exists s̄′ in the image of S in R/I such that r̄ · s̄′ = 0 that is rs′ ∈ I. But
it is always the case that rs′/ss′ = r/s in S−1R. Therefore r/s ∈ S−1I. This shows
that the kernel of φ is S−1I, hence we get the desired isomorphism.
Remark 0.1. Let S be the complement of a prime ideal P in a ring R and I be an
ideal contained in P so that I ∩ S = ∅. By the above proven fact, RP/IP ∼= (R/I)P .
Some authors write IRP for what we are calling IP ; the ideal generated by the image
of I in RP .
4. Let φ : A −→ B be a ring homomorphism by which B has a finitely gen-
erated A-module structure. It is easy to verify that for any multiplicative
submonoid S ⊂ A the image φ(S) ⊂ B is a multiplicative submonoid also.
Show that the induced homomorphism φS : AS −→ Bφ(S) gives Bφ(S) a finitely
generated AS-module structure. (Here, without loss of generality we as-
sume that 0 /∈ φ(S). Otherwise, Bφ(S) = 0, which is a finitely generated
AS-module.)
Solution.
Let {b1, . . . , br} be a generating set for B as an A-module. Let a/b be an element of
the localized ring AS. For c/φ(d) ∈ Bφ(S) with c ∈ B, d ∈ S the action of AS on Bφ(S)
is defined by
a
b
· c
φ(d)
=
φ(a) · c
φ(bd)
.
2
Since B is generated by {b1, . . . , br} as an A-module, c is of the form a1 ·b1 + · · · ar ·br =
φ(a1)b1 + ·+ φ(ar)br for some a1, . . . , ar ∈ A. Therefore,
c
φ(d)
=
φ(a1)b1 + ·+ φ(ar)br
φ(d)
=
r∑
i=1
ai
d
· bi
1
proving that {b1/1, b2/1, . . . , br/1} is a generating set for Bφ(S) as an AS-module.
5. Let P1, . . . , Pm be a finite set points from Cn such that P1 6= Pj for all j ∈
{2, . . . ,m}. Find an explicit polynomial F (x1, . . . , xn) ∈ C[x1, . . . , xn] which
takes constant value 1 on P1 and 0 on Pj for all j ∈ {2, . . . ,m}.
Solution.
Let (ak1, . . . , akn) ∈ Cn denote the coordinates of Pk for k = 1, . . . , n. Since P1 6= Pj,
there exists smallest index rj ∈ {1, . . . , n} such that ajrj 6= a1rj . Define gj(x1, . . . , xn)
by
gj(x1, . . . , xn) :=
xrj − ajrj
a1rj − ajrj
It is clear that gj(Pj) = 0 and gj(P1) = 1. The product of gj’s for j = 2, . . . ,m is the
desired polynomial F .
6. Definitions: Given two ideals I and J , the ideal quotient I : J is defined to be the ideal
{h ∈ R : hJ ⊆ I}. The radical of an ideal I, denoted by rad(I), is the ideal consisting
of elements r ∈ R such that some power rn, n ∈ N of r lies in I. Logically, we call an
ideal I radical if rad(I) = I.
Notation: Given an ideal I of a polynomial ring k[x1, . . . , xn], V (I) ⊆ kn denotes the
set of points a ∈ kn such that all polynomials from I vanishes on a.
Let I and J be two radical ideals. Show that the ideal of the Zariski closure
V (I) \ V (J) coincides with the quotient ideal I : J.
Remark 0.2. (a) Hilbert’s Nullstellensatz in commutative algebra says that for an
algebraically closed field k, and for any finitely generated polynomial ideal J
the ideal of the vanishing locus of J is equal to radical of J . In other words,
I(V (J)) = rad(J).
(b) The complement V (I) \ V (J) of V (J) in V (I) need not to be an algebraic set (it
is an open subset of V (I)). It doesn’t make sense to talk about the ideal of an
open subset. We must take its closure so that we can talk about the ideal of the
closed set.
3
(c) The quotient I : J need not be a radical ideal in general. But if I and J are
radicals then so is I : J ; suppose fn ∈ I : J for some n. Then fnhn ∈ I for any
h ∈ J . But I being radical, fh ∈ I hence f ∈ I : J .
(d) f ∈ R is not a zero divisor in R/I if and only if I = I : f . In this case, the variety
of the ideal generated by f and I has dimension one less than V (I).
For obvious reasons we assume that V (I) is not equal to V (J) (otherwise there is
nothing to prove).
Let α ∈ V (I) \ V (J) be a point, hence there exists a polynomial f in J such that
f(α) 6= 0.
We claim that each element h of I : J vanishes on α, that is to say h(α) = 0. Indeed,
hf ∈ I. But f(α) 6= 0, so every element h of I : J vanishes on α ∈ V (I) \ V (J).
It follows that I : J lies in the ideal of the closure of the complement. Conversely,
if we take a polynomial f vanishing on the closure of V (I) \ V (J), then obviously it
vanishes on V (I) \ V (J). Then for any h ∈ J , fh vanishes on all of V (I); f vanishes
on the complement and h vanishes on V (J). Thus fh ∈ I since I is radical. But then
f belongs to the quotient ideal I : J .
7. Let M be a finitely generated R-module and a ⊂ R an ideal. Suppose
ψ : M → M is an R-module map such that ψ(M) ⊆ aM . Find a monic
polynomial p(t) ∈ R[t] with coefficients from a such that p(ψ) = 0. Here, aM
is the module consisting of all finite sums of elements of the form bm, where
b ∈ a and m ∈M .
Solution.
The solution technique is important here. Let {x1, . . . , xm} be a generating set for M
as an R-module. By hypothesis, for each i = 1, . . . ,m we have
ψ(xi) =
∑
ai,jxj, (1)
for some ai,j ∈ a. We define Ai,j to be the operator δi,jψ−ai,je, where e is the identity
endomorphism of M and δi,j is the Kronecker’s delta function.
It is clear from (1) that
∑
j=1Ai,jxj = 0 for all i = 1, . . . ,m. In other words, the matrix
of operators A := (Ai,j)mi,j=1 annihilates the column vector v = (xj)mj=1. Notice that
we can consider M as an R[ψ]-module, and that Ai,j ∈ R[ψ]. Thus A is a matrix over
R[ψ]. Therefore, its adjugate makes sense and multiplying Av = 0 on the left by the
adjugate gives us (detA)xj = 0 for all j = 1, . . . ,m. Consequently, detA annihilates
all ofM . Expandingthe determinant we obtain a monic polynomial p in ψ with entries
from a. Furthermore, p(ψ) = 0 on M .
4
8. If M is a finitely generated R-module such that aM = M for some ideal a,
then there exists x ∈ R such that 1− x ∈ a and xM = 0.
Solution.
By the previous problem we observe that the identity operator id on M satisfies a
monoic polynomial: p(id) = idr + a1idr−1 + · · ·+ arid = 0 for some aj ∈ a. Therefore,
if we define x = 1 + a1 + · · · + ar, then x − 1 ∈ a and furthermore xm = 0 for all
m ∈M .
9. If a ⊂ R is an ideal such that every element of 1 + a is invertible, M is a
finitely generated R-module, and aM = M , then M = {0}.
Solution.
Let x ∈ R be as in the previous problem; 1 − x ∈ a and xM = {0}. In particular
x− 1 ∈ a, hence x = 1 + x− 1 is invertible. It follows from xM = {0} that M = {0}.
10. Let Jac(R) denote the intersection of all maximal ideals in R. (Jac(R) is
called the Jacobson radical of R.) Show that x ∈ Jac(R) then 1 − xy is
invertible for any y ∈ R. Conversely, if 1− xy is invertible for all y ∈ R, then
x belongs to all maximal ideals.
Solution.
Suppose x is from Jac(R). If 1−xy is not invertible, then it is contained in a maximal
idealm of R. In particular, since x is fromm, we see that 1 ∈ m which is a contradiction.
Conversely, suppose that 1−xy is invertible for all y ∈ R. If x does not lie in a maximal
ideal m, then the ideal generated by x and m is equal to R. Hence, xy + m = 1 for
some y ∈ R and m ∈ m. In this case, m = 1−xy ∈ m, so it is not a unit, contradicting
with our initial assumption. Therefore, x ∈ m.
Definition: Have a taste of Zorn’s lemma: A Noetherian ring is a ring in which every
non-empty set of ideals has a maximal element. For other definitions and properties of
Noetherian rings see 0.60.
Fact 0.3. Artin-Rees Lemma: Let a be an ideal in a Noetherian ring R and
letM be a finitely generated R-module. If N ⊂M is a submodule, then there
exists a positive integer k such that for all n ≥ k, anM ∩N = an−k((akM)∩N).
11. Let R be a Noetherian ring and a be an ideal such that every element of
1 + a is invertible in R. Show that ∩n>0an = (0).
5
Solution.
Let M denote ∩n>0an. Obviously, M is a R-submodule of a. By Artin-Rees lemma,
there exists n such that an+i ∩M = ai(M ∩ an) for all i ≥ 0. Since set theoretically
we have M ∩ an+i = M for any n+ i, we get M = aiM for any i ≥ 0. By Problem 9
above, we get M = 0.
12. If a is an ideal such that every element of 1 + a is invertible, M a fininitely
generated R-module and M ′ ⊆ M a submodule, then M ′ + aM = M implies
that M ′ = M .
Solution.
We consider the R-module M/M ′. Our assumption M ′ + aM = M implies that
aM/M ′ = M/M ′. Hence, by Problem 9 M/M ′ = {0}, or equivalently, M = M ′.
13. If a is an ideal such that every element of 1 + a is invertible, M a fininitely
generated R-module. Show that the elements u1, . . . , un generate M if and
only if the images u1, . . . , un generate M/aM as an R-module
Solution.
The implication (⇒) is obvious. We prove the converse. Suppose u1, . . . , un generate
M/aM as an R-module. Let {u1, . . . , un} denote a set of preimages of ui’s, and let
M ′ denote the submodule generated by ui’s. It is clear that M ′ + aM = M , hence by
Problem 12 it follows that M ′ = M .
Definition: A local ring is a ring with unique maximal ideal.
14. Let (R,m) be a Noetherian local ring and suppose that the images of the
elements a1, . . . , an ∈ m generate m/m2 as a vector space. Show that a1, . . . , an
generate m as an ideal.
Solution.
We denote by M the maximal ideal m viewed as an R-module, and denote by a the
maximal ideal m viewed as an ideal. The solution is now an application of Problem 13.
15. In the notation of the previous problem, a1, . . . , an generates m/m
2 as a vector
space, then a1, . . . , an generates m minimally, that is to say none of ai’s is
redundant.
6
Solution.
Towards a contradiction, without loss of generality, assume that a1 redundant; a1 =
r2a2 + · · ·+ rnan for some ri ∈ R. Then, modulo m2, a1, . . . , an are linearly dependent
which is a contradiction.
Definition. An ideal I 6= (1) is primary if fg ∈ I implies either f ∈ I, or gm ∈ I for
some m ∈ N.
16. Prove that if Q is primary, then rad(Q) is a prime ideal. Furthermore, in
this case, rad(Q) is the smallest prime ideal containing Q.
Notation: If P denotes the prime ideal rad(Q), then Q is called P -primary.
Solution.
Let fg ∈ rad(Q), hence (fg)n ∈ Q for some n ∈ N. Since Q is primary, either fn ∈ Q,
or gnm ∈ Q for some m ∈ N. In other words, either f ∈ rad(Q), or g ∈ rad(Q)
implying that rad(Q) is a prime ideal.
If M is a prime ideal such that Q ⊆ M , then rad(Q) ⊆ M because of the following
two things: First, for any two ideals I and J , I ⊆ J implies rad(I) ⊆ rad(J). Indeed,
f ∈ rad(I), then fn ∈ I for some n ∈ N, hence fn ∈ J . In particular, f ∈ rad(J).
Secondly, if an ideal J is prime, then J is equal to its own radical. To see this it is
enough to show that rad(J) ⊂ J whenever J is prime. Let f ∈ rad(J), hence fn ∈ J
for some n ∈ N. It follows primeness that f ∈ J .
We apply this observation to our original problem. If M is a prime ideal containing
Q, then rad(Q) ⊂ M . Therefore, we conclude that rad(Q) is the smallest prime ideal
containing Q whenever Q is primary.
17. LetM be a finitely generated R-module and S ⊂ R be a multiplicative subset
of R. Show that MS = 0 if and only if an element s of S annihilates M , that
is to say, sM = 0.
Solution.
If s annihilates M , then for any m/r ∈ MS, we have m/r = sm/sr = 0/sr = 0, thus
MS = 0. Conversely, assume that MS = 0, that is m/r = 0 for every m ∈ M and
r ∈ S. By definition this holds if there is an s ∈ R such that s(1 · m − 0 · r) = 0.
Thus s · m = 0. Now, this s is specific to m. Since M is finitely generated, there
exists a finite generating set {m1, . . . ,mn} of M , and there exists a corresponding set
{s1, . . . , sn} of annihilators. Since R is commutative, the product s1 · · · sn annihilates
all mi’s, hence it annihilates whole of M .
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Definition: Support of an R-module M is the set of all prime ideal p such that Mp
is non-trivial. Here, Mp is the localization of the module at the multiplicative subset
R− p.
18. Let M be a finitely generated R-module. Show that a prime ideal p is in the
support of M if and only if the annihilator ideal ann(M) of M is contained
in p.
Solution.
By Problem 17 we see that Mp = 0 if and only if there exists s ∈ R − p such that
sM = 0. But then ann(M) cannot be contained in p. This proves one implication.
For the converse, suppose that p is in the support ofM . Then by the previous problem
again, there cannot be any s ∈ R \ p such that sM = 0. Hence ann(M) must be
contained in p.
19. Show that a short exact sequence of R-modules:
0 −→ A1
α1−→ A2
α2−→ A3 −→ 0 (2)
gives rise to a left exact sequence
0 −→ HomR(A3, N)
α∗2−→ HomR(A2, N)
α∗1−→ HomR(A1, N) (3)
Solution.
Given a homomorphism f : A3 −→ N , we pull it back to a homomorphism from
A2 to N by fα2 : A2 −→ N . Hence we get a homomorphism HomR(A3, N)
α∗2−→
HomR(A2, N).
Next, let us see that α∗2 is injective: suppose α∗2f = α∗2g for some f and g from
HomR(A3, N). Then, α2f = α2g. But α2 is surjective by exactness (1). Thus, f and
g agree on every point of A3 showing that they are the same functions. Therefore α∗2
is injective.
Next, let us see that im(α∗2) ⊆ ker(α∗1), namely α∗1α∗2 = 0. Let f ∈ HomR(A3, N). Then
α∗1α
∗
2f = fα2α1 = f · 0 = 0 again by exactness of (1). Therefore im(α∗2) ⊆ ker(α∗1).
Finally, let us see that im(α∗2) ⊇ ker(α∗1): let f : A2 −→ N be in the kernel of α∗1,
namely fα1 = 0. Define g : A3 −→ N as follows, let g(a3) be the value f(a2) for
any a2 ∈ A2 such that a2 and g is well-defined. Notice that gα2 = f . Therefore
im(α∗2) = ker(α∗1) and (2) is left exact.
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20. Give an example of a module N and a short exact sequence such that
HomR(−, N) does not give a short exact sequence.
Solution.
Let p ∈ Z be a prime and consider the following exact sequence of Z-modules:
0 −→ Z p−→ Z −→ Z/p−→ 0
Apply HomZ(−,Z) and check that the result is not an exact sequence.
21. Let G be an abelian group and write G ' Zn⊕Gtorsion. Show that HomZ(G,Z) ∼=
Zn.
Solution.
Let f ∈ HomZ(G,Z), then f is determined by the images to the generators of the
copies of Z in G. The reason for not affected by the torsion part of G is the following:
if a ∈ Gtorsion, then n · a = 0 for some n ∈ Z. Then 0 = f(0) = f(n · a) = nf(a) since
f is a Z-module homomorphism. Therefore effect of the torsion part of G is 0 showing
that an f is determined by the images of the generators of the copies of Z in G. So,
the result follows.
22. In the category of R = k[x1, . . . , xn]-modules show that HomR(R(−a), R) '
R(a).
Solution.
Note that 1 in R(−a) is an element of degree a. Therefore, a homomorphism f ∈
HomR(R(−a), R) ' R(a) of graded R-modules maps 1 to a degree a element in R that
is f(1) is of degree a. Now, since an R-module homomorphism R is determined by its
value on 1; we have an isomorphism of R-modules f ∈ HomR(R(−a), R) −→ R via
f 7−→ f(1). Thus if we declare 1 in R to be of degree a, namely, if we regard the image
as the graded module R(−a), then we get an isomorphism of graded R-modules.
23. What is Z/2⊗Z Z/3?
Solution.
Let a⊗ b ∈ Z/2⊗Z Z/3. If b is 0 or 2, then a⊗ b = 0⊗ 0. If b = 1, then we can replace
it by 4 (since 4 ≡ 1 mod 3) and get a⊗ b = 0⊗ 0 again. Therefore Z/2⊗Z Z/3 = 0.
9
24. More generally, show that Z/a⊗Z Z/b ' Z/ gcd(a, b).
Solution.
First of all remember that gcd(a, b) of two integers is the largest (hence unique) integer
that divides both a and b. Furthermore, there are integers x and y such that gcd(a, b) =
ax+ by.
Now, that said, define φ : Z/a⊗Z Z/b −→ Z/ gcd(a, b) by r⊗ s 7−→ rs mod gcd(a, b).
This map is a well-defined homomorphism of Z-modules. It is injective because: if
rs = 0 mod gcd(a, b), then rs = n(ax + by) for some n ∈ Z. Thus r ⊗ s = 1 ⊗ rs =
1⊗ n(ax+ by) = 1⊗ nax = nax⊗ 1 = 0⊗ 1 = 0.
Surjectivity is clear by 1 ⊗ r 7−→ r mod gcd(a, b). Therefore φ is an isomorphism of
Z-modules.
25. Show that − ⊗R M is a right exact functor on the category of R-modules,
but it need not be an exact functor.
Solution.
Let
0 −→ A1
α1−→ A2
α2−→ A3 −→ 0
be a short exact sequence of R-modules and let M be some other R-module. We want
to show that the following is a right exact sequence:
A1 ⊗M −→ A2 ⊗M −→ A3 ⊗M −→ 0
Since the maps are defined by ai⊗m 7−→ αi(ai)⊗m, right exactness is straightforward
to check.
A counter example to exactness of tensoring is the following:
0 −→ Z p−→ Z −→ Z/p −→ 0
This is an exact sequence of Z-modules. Assume that − ⊗Z Z/p gives us an exact
sequence:
0 −→ Z⊗Z Z/p
p−→ Z⊗Z Z/p −→ Z/p⊗Z Z/p −→ 0
But Z ⊗Z Z/p ∼= Z/p
p−→ Z ⊗Z Z/p ∼= Z/p is the 0 map which, on the contrary, was
supposed to be an injection. Contradiction.
10
26. For R-modules M , N , and P , prove that
HomR(M ⊗R N,P ) ' HomR(M,HomR(N,P )).
Solution.
Consider the map f ∈ HomR(M ⊗R N,P )
φ−→ HomR(M,HomR(N,P )) 3 φ(f),
defined by φ(f)(m) = f(m⊗−) ∈ HomR(N,P ). Here, it should be clear that f(m⊗−)
is a homomorphism from N to P for every m ∈M .
Let us prove that φ is an isomorphism of R-modules: If φ(f) = 0, then f(m⊗ n) = 0
for every m⊗n ∈M ⊗RN , hence f is the zero homomorphism in HomR(M ⊗RN,P ).
So, φ is injective.
If g ∈ HomR(M,HomR(N,P )), then for every m ∈ M , g(m) is an R-homomorphism
from N to P . So, define f : M ⊗ N −→ P by f(m ⊗ n) = g(m)(n). It is clear that
we get a homomorphism. Also, by definition, φ(f) = g. So, φ is surjective, hence an
isomorphism of R-modules
27. It is easy to see that every direct sum of modules gives an exact sequence.
Prove that the converse is not true; there is an exact sequence
0 −→ A −→ B −→ C −→ 0
such that B � C
⊕
A
Solution.
Consider
0 −→ Z/2 −→ Z/4 −→ Z/2 −→ 0
with obvious maps. This is an exact sequence. However, Z/4 � Z/2
⊕
Z/2.
28. For a short exact sequence of R-modules
0 −→ A1
α1−−→ A2
α2−−→ A3 −→ 0
show that A2 ∼= A1 ⊕ A3 if and only if there is a homomorphism β2 (or a
homomorphism β1) with α2β2 = idA3 (with α1β1 = idA1).
Solution.
Let A2
φ−−→ A1⊕A3 be an isomorphism. We have the following commutative diagram:
11
0 A1 A2 A3 0
0 A1 A1 ⊕ A3 A3 0
id
α1
i
φ
α2
π
id
An element of A1 ⊕ A3 is of the form (a1, a3) for unique a1 ∈ A1 and a3 ∈ A3. Define
β2 : A3 −→ A2 as follows: given a3 ∈ A3 there exists unique a2 ∈ A2 such that φ(a2)
= (0, a3) since φ is an isomorphism. Set β2(a3) = a2. Then by the commutativity of
the diagram it is clear that α2β2(a3) = a3.
Conversely, suppose that we have a homomorphism β2 : A3 −→ A2 such that α2β2 =
idA3 . Then define ψ : A1 ⊕ A3 −→ A2 by ψ((a1, a3)) = α1(a1) + β2(a3). This is a well
defined R-module homomorphism. Let us see why it is an isomorphism.
Suppose ψ((a1, a3)) = α1(a1) + β2(a3) = 0. Then α1(−a1) = β2(a3). Compose this
equality with α2 to get 0 = a3. Then α1(−a1) = 0. But α1 is injective, so a1 = 0.
Therefore ψ is injective.
Now, let a2 ∈ A2. Then α2(a2 − β2α2(a2)) = α2(a2) − α2(a2) = 0, therefore a2 −
β2α2(a2) ∈ im(α1), say, α1(a1) = a2 − β2α2(a2). Then ψ((a1, α2(a2)) = α1(a1) +
β2(α2(a2)) = a2, therefore A1 ⊕ A3
ψ−→ A2 is surjective and hence an isomorphism.
The case of β1 is proven similarly.
Definition 0.4. We recall some very basic definitions from homological algebra. A
chain complex is a sequence of abelian groups with homomorphisms between them:
· · · ∂4−→ C3
∂3−→ C2
∂2−→ C1
∂1−→ C0
∂0−→ 0
satisfying
∂n+1∂n = 0 for all n ≥ 0.
Last condition ensures that the image of ∂n+1 lies in the kernel of ∂n, hence next
definition makes sense: ith Homology group of the chain complex (C, ∂) is defined by
Hi(C) = ker ∂i/im ∂i+1.
These groups measure how far the chain complex from being exact. Note that the
chain complex (C, ∂) does not need to be of infinite length, however, a finite length
chain complex can be extended by adding trivial groups and trivial connecting homo-
morphisms in between them.
12
29. Compute the homology of the complex 0 −→ V1
φ−→ V0 −→ 0 where V1 = V0 = k3
and φ is:  1 0 −1−1 1 0
0 −1 1

Solution.
Kernel of φ is:

ab
c
 ∈ V1 : a− c = 0−a+ b = 0
−b+ c = 0
 which is spanned by
11
1
. So, H1 is iso-
morphic to k. For H0, note that the image of φ is spanned by the image of the basis
10
0
 ,
01
0
 ,
00
1
 under φ. It is straightforward to see that the image of this basis
is nothing but the same basis. Thus, H0 is trivial.
30. Prove that for a complex V : · · · −→ Vn −→ . . . −→ V0 −→ 0 of finite dimen-
sional vector spaces, the following equality is always true:∑
i=0
(−1)i dimVi =
∑
i=0
(−1)i dimHi(V )
This alternating sum is called the Euler characteristic of the complex. It is
clear that the complex is exact then its Euler characteristic is 0.
Solution.
This follows from the basic linear algebra fact that if Vi+1 and Vi are two finite dimen-
sional vector spaces and φi+1 is a linear map between them, then
dim(kerφi+1) + dim(imφi+1) = dim(Vi+1).
Alternating sum of these equations (for i ≥ 0) gives us the desired equality.
31. Let k be a field. Suppose k[x0, x1, . . . , xn] is graded by degree. Show that ith
graded component has vector space dimension
(
n+i
i
)
.
Solution.
The vector space basis for the ith graded piece is {xp00 x
p1
1 . . . x
pn
n : t
∑
pj = i}. So, the
problem is equivalent to finding the number of ways of distributing i candies to n+ 1
children named x0, x1, . . . , xn. Of course, this is done in
(
n+1
i
)
different ways.
13
Definition 0.5. Let M = ⊕n∈ZMn be a Z-graded R-module. The numerical function
HM : Z −→ Z defined as
HM(n) = dimk(Mn)
is called the Hilbert function of M .
The power series
PM(z) =
∑
HM(t)z
t
is called the Hilbert series of M.
32. What is the Hilbert series of k[x1, . . . , xn]?
Solution.
Using Problem 31, we see that it is nothing but 1/(1− t)n.
33. Find the Hilbert series of k[x1, x22, . . . , xnn].
Solution.
The set of monomials contained in the ith graded piece is
{xi11 x2i22 . . . xninn : i1 + 2i2 + · · ·+ nin= i}
Therefore, the Hilbert series is given by
∏
r≥1
1
1−tr .
34. Let P be a function P : N −→ Z such that the associated difference function
∆P (i) := P (i) − P (i − 1) is a polynomial with rational coefficients (for suffi-
ciently large i). Show that P itself is a polynomial with rational coefficients
and has degree one greater than ∆P .
Hint: Use induction on the degree s of the difference polynomial. The base case is
trivial. If the leading coefficient of ∆P is as, then define h = ass!
(
i
s+1
)
, and compute
∆h. By construction, ∆P −∆h will have degree s− 1.
Solution.
As given in the hint, obviously, h is of degree s− 1, and ∆h(i) = ass!
(
i
s+1
)
− ass!
(
i−1
s+1
)
.
Computing this we find ∆h(i) = ass!
(
i−1
s
)
. Clearly ∆h is of degree s. At the same
time, the leading coefficient of ∆h is as. Thus, the degree of ∆h−∆P = ∆(P − h) is
s− 1. By the induction hypothesis, this says that P − h is a polynomial with rational
coefficients and of degree s. But h is a polynomial with rational coefficients of degree
s+ 1, hence P is has the same type. This finishes the induction.
14
35. Let M be a finitely generated, Z-graded module. Show that there exists a
polynomial f(x) ∈ Q[x] such that for i� 0, HM(i) = f(i).
Definition 0.6. The polynomial f(i) is called the Hilbert polynomial of M , written
HP (M, i).
Solution.
Applying induction on the number of variables in the ring over whichM is defined, the
base case is trivial. So, suppose it is true for n− 1 variables {x1, . . . , xn−1}. Using the
homomorphism, multiplication by xn, M(−1)
xn−→M , we can build an exact sequence:
0 −→ ker(xn) −→M(−1)
xn−→M −→ coker(xn) −→ 0
Since multiplication by xn kills both ker and coker, they can be regarded as modules
over the polynomial ring of n − 1 variables. (Of course, they are finitely generated).
By exactness, HM(i)−HM(i− 1) ∈ Q[i]. We are done by the previous exercise.
Definition 0.7. Let I be a homogenous ideal in R = k[x0, . . . , xn], hence it corresponds
to a projective variety Y = V (I) in Pn with ring of global sections R/I. Let us write
the Hilbert polynomial of Y (or that of I) as the Hilbert polynomial of the coordinate
ring R/I:
HP (R/I, i) =
am
m!
im +
am−1
(m− 1)!
im−1 + . . .
Then we define the dimension of the projective variety Y ⊂ Pn as m, and its degree as
am.
36. Compute the Hilbert polynomial of Pn and find its degree.
Solution.
The coordinate ring of Pn is k[x0, . . . , xn]. Thus, for large i’s we have HP (Pn, i) =
HR(i) = dimk(k[x0, . . . , xn]i) =
(
n+i
i
)
=
(
n+i
n
)
. Obviously this is a polynomial in i with
leading term 1
n!
in. Recall that for a Hilbert Polynomial as
s!
is + . . . the degree is the
dimension of the variety and as is the degree of the variety. Thus, the degree of Pn is
1.
Definition 0.8. Given n, d > 0, let M0, . . . ,MN be all the monomials of degree d in
the n+1 variables x0, . . . , xn where N =
(
n+d
n
)
−1. We define a mapping ρd : Pn −→ PN
by sending the point P = (a0, . . . , ad) to the point ρd(P ) = (M0(a), . . . ,MN(a)). This
is called d-uple embedding of Pn in PN .
15
37. Let θ : k[y0, . . . , yN ] −→ k[x0, . . . , xn] be the homomorphism defined by sending
yi to Mi (in the notation of the above definition). Let a be the kernel of
θ. Show that a is a homogenous prime ideal, and V (a) is an irreducible
projective variety in PN .
Solution.
Recall that an ideal is called homogenous if it is generated by homogenous polynomials.
A relation among Mi’s is zero if and only if that relation is a homogenous polynomial.
For example, ifM0 = x20,M1 = x0x1,M2 = x22, thenM0M2−M21 = 0 which is of course
y0y2 − y21. This is because each Mi has the fixed degree d. Therefore, the kernel of
θ is generated by homogenous elements. The image of θ is a subring of k[x0, . . . , xn]
generated by Mi’s. Obviously being an integral domain is preserved for the subrings
of integral domains. Therefore the quotient k[y0, . . . , yN ]/ ker θ is an integral domain
showing that ker θ is a prime ideal.
38. Show that the variety, ρd(Pn) defined in Problem 37 has the ring of global
sections equal to k[y1, . . . , yN ]/ ker θ, namely it is given as the locus of ker θ.
Solution.
The kernel of θ is generated by the polynomials that are zero when evaluated in the
monomials M0, . . . ,MN in place of y0, . . . , yN . Therefore these are exactly the genera-
tors of the vanishing ideal on the set of points (M0(a); . . . ;MN(a)) in PN . Therefore
the variety ρd(Pn) is cut out by the kernel of θ.
39. Show that ρd is a homeomorphism of Pn onto V (ker θ).
Solution
First, let us see that ρd is an injection. Suppose (M0(a); . . . ;MN(a)) = (M0(b); . . . ;MN(b)),
namely ρd(a) = ρd(b) for some a, b ∈ Pn. Observe that ai’s cannot be all zero (oth-
erwise Mi(a) = 0 for all i). Say ai 6= 0. Among monomials of degree d we have
adi and a
d−1
i aj for j = 0, . . . , n. Since (M0(a); . . . ;MN(a)) = (M0(b); . . . ;MN(b)) in
PN , we have (M0(a)/adi , . . . ;MN(a)/adi ) = (M0(b)/bdi ; . . . ;MN(b)/bdi ) in AN . There-
fore we have ad−1i aj/adi = b
d−1
i bj/b
d
i for j = 0, . . . , n. But then aj/ai = bj/bi hence
(a0; . . . ; an) = (b0; . . . ; bn) in Pn. Thus ρd is injective. Since a map is surjective onto
its image. Therefore ρd is a bijection. It is clear that it is continuous (on each open
piece yi 6= 0 in PN it is given by continuous maps). It is enough to show that it has a
continuous inverse. The inverse map ρ−1d maps (M0(a); . . . ;MN(a)) to a. Since at each
affine open xi 6= 0, this reduces to ρ−1d ((M0(a)/adi , . . . ,MN(a)/adi )) = (a0/ai, . . . , an/ai)
and amongMj(a)/adi we have aj/ai for j = 0, . . . , n, we see that the inverse is basically
a projection. Therefore, it is continuous hence ρd is a homeomorphism.
16
40. Find the degree of d-uple embedding of Pn in PN .
Solution
By definition, we want to compute the Hilbert polynomial of the k-vector space
k[y0, . . . , yN ]/ ker θ,
where θ(yi) = Mi(x). This ring is isomorphic to the subring k[M0, . . . ,MN ] of k[x0, . . . , xn]
generated by all the monomials of degree d. Therefore it suffices to compute the Hilbert
polynomial of the graded (by degree) vector space k[M0, . . . ,MN ] =
⊕
Si. Obviously
Si = 0 if i is not a multiple of d. Let i = rd for some nonzero r, then Si is equal to the
ith piece of k[x0, . . . , xn]. Thus the Hilbert polynomial is equal to
(
n+rd
rd
)
=
(
n+rd
n
)
as a
function of r = 0, 1, . . . . This is a polynomial in r;(
n+ rd
n
)
=
(n+ rd)(n+ rd− 1) · · · (rd+ 1)
n!
.
The leading coefficient of this polynomial is dn/n!. Therefore, the degree (leading term
times n!) of ρd(Pn) is dn.
Fact 0.9. If partially ordered set S has the property that every chain has an upper
bound in S, then the set S contains at least one maximal element.
Definition 0.10. The nilradical of a ring is the set of all nilpotent elements of the
ring.
41. Show that a nilradical is an ideal. Furthermore, show that the nilradical is
equal to the intersection of all prime ideals of the ring.
Solution.
Let n denote the nilradical of the ring R. Let x and y be two elements from n,
and suppose n,m ∈ N are such that xn = ym = 0. Binomial theorem implies that
(x+ y)n+m = 0, hence n is closed under addition. Let a ∈ R. Since R is commutative,
(ax)n = 0, hence ax ∈ n. Therefore, n is an ideal. To prove that n is the intersection
of all prime ideals, we first make the following simple observation: If P is prime ideal
P ⊂ R, then the quotient ring R/P (which is an integral domain) does have any
nilpotent elements. Therefore, n ⊆ P . This proves that n ⊆ ∩P : primeP .
For the opposite inclusion, it suffices to show that for any non-nilpotent element x,
there is some prime ideal that does not contain x. Towards this contradiction we
assume that ∩P : primeP − n is non-empty. Let S be the set of all ideals that does not
contain any power of x. Since n ∈ S, we know that S is non-empty. Furthermore,
if I1 ⊂ I2 ⊂ · · · is a nested sequence of elements from S, then ∪Ii ∈ S. Hence, by
Zorn’s lemma (Fact0.9) S has a maximal element M ∈ S. Notice that if we prove M
17
is a prime ideal, then we succeed in our goal that there is a prime ideal that does not
contain x.
Assuming M is not prime, we take two elements a, b ∈ R − M such that ab ∈ M .
The ideals Ma generated by a and M , and Mb, generated by b and M both properly
contain M , hence both of them contains a power of x: ay+m1 = xp and bz+m2 = xq.
Since (ay + m1)(bz + m2) = abyz + aym2 + bzm1 + m1m2 lies in M we conclude that
xpxq = xp+q is contained in M also. This contradiction shows that M is a prime ideal,
hence our proof is complete.
Fact 0.11. Recall that the Jacobson radical Jac(R) of a ring R is the intersection of
all maximal ideals of R. Given an ideal I ⊂ Jac(R) and a finitely generated R-module
M , Nakayama’s Lemma says that
IM = M implies M = 0.
An R-module P is projective if there exists a module K such that P
⊕
K ' F for
some free module F . Equivalently, every short exact sequence
0 −→ N −→M f−→ P −→ 0
splits; there exists P h−→M such that f ◦ h = idP .
Remark 0.12. Being projective is transitive in the following sense: ifM is a projective
B-module (henceM is a direct summon of a free B-module F ) and if B is an A-algebra
with free A-module structure, then M is a projective A-module as well. The reason
for this transitivity is that F is a free A-module, hence M is a direct summand of a
free A-module.
42. Show that the following are equivalent for a finitely generated Z-module M :
(a) M is projective;
(b) M is torsion free;
(c) M is free.
Solution. (c) =⇒ (b) and (c) =⇒ (b) are obvious. Suppose thatM is torsion free and
let {m1, . . . ,mr} be a generating set for M . Define ψ : M −→ Zr by ψ(mi) = ei and
extend by linearity, where ei is the ith standard vector in Zr. Since none of the mi’s are
torsion, this is a well defined map of Z-modules. Furthermore, it is an isomorphism.
So (b) =⇒ (c). Finally, since M is a direct summand of a free module, it is torsion
free; (a) =⇒ (b). So, we are done.
18
Definition 0.13. For an integral domain R, a fractional ideal is an R-submodule A
of the fraction field K of R such that for some nonzero element d ∈ R the following
inclusion holds:
d · A =
{
da
b
:
a
b
∈ A
}
⊂ R.
Equivalently, A = d−1I for some ideal I ⊂ R and a nonzero d ∈ R. In particular,
any ideal of R is a fractional ideal. The product of two fractional ideals A = d−1I,
B = r−1J is another fractional ideal AB = (dr)−1IJ . This operation defines a monoid
structure on the set of all fractional ideals of R. Indeed, the identity fractional ideal is
the ring R itself.
The unit group of this monoid contains all principal ideals. Obviously, the set of all
fractional principal ideals forms a subgroup of the unit group of the monoid. Invertible
fractional ideals modulo its subgroup of principal fractional ideals is called the class
group of R. Thus the class group of an integral domain R measures how far is R from
being a PID.
43. Show that in an integral domain R with fraction field K, if a fractional ideal
A is invertible, then it is a projective R-module.
Solution.
Assume that A is an invertible fractional ideal. Let A−1 be its inverse. Then a1a′1 +
· · · + ana′n = 1 for some ai ∈ A and a′i ∈ A−1 since AA−1 = R. Let S be a free R
module of rank n say generated by y1, . . . , yn. Define φ : S −→ A by φ(yi) = ai and
extend it by linearity. Define also ψ : A −→ S by ψ(c) = c(a′1y1 + · · · + a′nyn). This
makes sense because ca′i ∈ R. Obviously, φψ = idA, so A is a direct summand of F .
In other words, A is a projective module.
Remark 0.14. The converse of this problem is true also: if a fractional ideal A is
projective, then it is invertible
44. Show that the ideal (2, 1+
√
−5) of Z[
√
−5] is a projective Z[
√
−5]-module but
not free.
Solution.
Note that Z[
√
−5] = 1 · Z⊕
√
−5 · Z, therefore Z[
√
−5] is a free Z-module of rank 2.
If I := (2, 1 +
√
−5) were free Z[
√
−5]-submodule of Z[
√
−5], then it would be of rank
one. Thus there would be a single generator; I ' αZ[
√
−5] for some α ∈ I. But it
is easy to see that 2 and 1 +
√
−5 are Z[
√
−5]-linearly independent; αZ[
√
−5] cannot
generate I. Therefore I cannot be a free Z[
√
−5]-module. However, I2 is the ideal (2).
19
Since (2) is an invertible ideal with inverse (1/2)Z[
√
−5], I is an invertible ideal with
inverse I · (1/2)Z[
√
−5], hence I is a projective module.
45. Use Nakayama’s Lemma to prove that a finitely generated projective mod-
ule over a local ring is free.
Solution.
Let {w1, . . . , wn} be a minimal set of generators for a finite projective module M and
let F be a free module of rank n. Consider the following surjection φ : F →M :
φ(
∑n
i=1
aiei) =
∑n
i=1
aiwi,
where ei’s are the generators for F . Since we have declared wi’s to be the minimal
generating set for M , if
∑n
i=1 aiwi = 0, then we should have ai’s in the maximal ideal
m of the local ring. Otherwise, they would be units and that would contradict the
minimality of wi’s. Thus kerφ ⊂ mF . Now, for M being a projective module and
having a surjection φ on it, we get a splitting of F via an injection ϕ : M −→ F
such that φϕ = idM . Therefore we can write F = ϕ(M) ⊕ kerφ. Since we have
the containments mϕ(M) ⊂ ϕ(M) and kerφ ⊂ mF and since kerφ is disjoint from
ϕ(M), we see that kerφ ⊂ m kerφ. So, kerφ = m kerφ. Now, by Nakayama’s Lemma,
kerφ = 0, hence φ is an isomorphism.
Definition 0.15. Left Derived Functors: Let C denote the category of R-modules
and suppose F is a right exact, covariant, additive (preserving addition of homomor-
phisms) functor from C to C. The left derived functors LiF (N) of an R module N are
defined as follows: take a projective resolution of N (for instance, a free resolution)
and apply the functor F to the exact sequence omitting N . The new sequence is a
complex and LiF (N) is defined as the ith homology of this complex.
46. Let C denote the category of graded C[x]-modules and let F stand for the
functor − ⊗C[x] (C/x) on C. Compute the left derived functors LiF of N =
C[x]/x2 for all i.
Solution.
Here is a projective resolution for N :
0→ (x2) ·x
2
−→ C[x]→ N → 0
Applying F to it, we get the following complex, say P :
0⊗C[x] C[x]/x
0⊗1−−→ (x2)⊗C[x] C[x]/x
·x2⊗1−−−→ C[x]⊗C[x] C[x]/x→ 0
20
Then
L0F (N) = H0(P ) = (C[x]⊗C[x] C[x]/x)/x2 · C[x]⊗ C[x]/x
= (C[x]/x)/(x2 · C[x]/x)
∼= (C[x]/x)
because (x2 · C[x]/x) = 0. Therefore L0F (N) = (C[x]/x).
Next,
L1F (N) = H1(P ) = ker(·x2 ⊗ 1)/im(0 · 1)
= ((x2)⊗C[x] C[x]/x)/0
= (x2C[x]/x)/0 = 0.
Of course, LiF (N) = Hi(P ) = 0 for all i > 1.
Definition 0.16. Tor. Left derived functors of the tensor products are called Tor.
What we have computed in the previous problems are ToriC[x](N,C[x]/x).
47. Let M be an R-Module and let r ∈ R. Compute the R-modules ToriR(R/r,M).
Solution.
Consider the free resolution
0→ R ·r−→ R→ R/r → 0
Apply −⊗RM to this by omitting R/r to get
0→ R⊗RM
·⊗1−−→ R⊗RM → 0
This is equal to 0 → M ·r−→ M → 0. Therefore ToriR(R/r,M) = M/r · M and
Tor1R(R/r,M) = {m ∈M : r ·m = 0}.
48. Let M be an R-module and let r ∈ R. Compute the R-modules ToriR(R/r,M).
Solution.
Consider the free resolution
0→ R ·r−→ R→ R/r → 0
Apply −⊗RM to this by omitting R/r to get
0→ R⊗RM
·r×1−−→ R⊗RM → 0
21
This is equal to 0 → M ·r−→→ M → 0. Therefore Tor0R(R/r,M) = M/r · M and
Tor1R(R/r,M) = {M ∈M : r ·m = 0}.
49. Prove that for a homogenous polynomial f of degree d and a homogenous
ideal I ⊂ R, there is a graded exact sequence:
0→ R(−d)/I : f → R/I → R/〈I, f〉 → 0.
Hint: Clearly, 0 → 〈I, f〉/I → R/I → R/〈I, f〉 is exact. How can you get a graded
map from R to 〈I, f〉/I?
Solution.
Obviously the exact sequence given in the hint is graded. Consider the graded map
R(−d) ·f−→ 〈I, f〉/I. The kernel is {g ∈ R : fg ∈ I} : f after a glance at the definition
of I : J. Thus, R(−d)/I : f ∼= 〈I, f〉/I. Thus, replacing 〈I, f〉 by R(−d)/I : f in the
exact sequence given in the hint, we get the graded exact sequence:
0→ R(−d)/I : f → R/I → R/〈I, f〉 → 0
50. Show that a given f ∈ R is a nonzerodivisor on R/I if and only if I : f = I.
Solution.
Recall that I : J = {h ∈ R : hJ ⊆ I}. Suppose now that I : f = I, therefore if for
some h ∈ R it happens that h〈f〉 ⊆ I, then h ∈ I. So, f · (h mod I) = fh mod I ≡ 0
mod I implies that h ∈ I. Thus, f is a nonzero divisor. Conversely, if f is a nonzero
divisor on R/I then for any h ∈ R with the property that h〈f〉 ⊆ I we must have
h ∈ I.
Remark 0.17. We defined the Hilbert polynomial of a ring in Problem 35. Suppose
the initial term of the Hilbert polynomial of a quotient ring R/I is given by
HP (R/I, i) =
am
m!
im + · · ·
If f is a homogenous linear form which is a nonzero divisor on R/I, then we have
HP (R/〈I, f〉, i) = HP (R/I, i)−HP (R/I, i− 1)
=
am
(m− 1)!
im−1 + . . .
We deduce that, by slicing with the hyperplane defined by f the dimension drops by
one while preserving the degree.
22
Definition 0.18. Let M be an R module. An element x of R is called M -regular if it
is not a zero divisor on M . An ordered sequence of elements x = x1, . . . , xn from R is
called M -regular if the following two conditions hold:
(a) xi is M/(xi, . . . , xi−1)-regular for i = 1, · · · , n.
(b) M/xM 6= 0
A weak M -regular sequence is defined by requiring only the first condition.
51. Show that given an R-module M over a local ring (R,m), a weak M-regular
sequence x ⊂ m is always M-regular.
Solution.
Let x ⊂ m be a weak M -regular sequence and assume M/xM = 0 that is M = xM .
But then by Nakayama’s Lemma M must be trivial.
52. Show that x, y(1− x), z(1− x) is an R-sequence, but y(1− x), z(1− x), x is not
where R = k[x, y, z] a polynomial ring. Here, by an R-sequence we mean an
R-regular sequence for R regarded as a module over itself.
Solution.
Multiplication by y(1 − x) on R/xR is the same as multiplication by y on k[y, z](∼=
R/xR). So y(1 − x) is R/xR regular. And multiplication by z(1 − x) on R/(x, y(1 −
x))R is the same as multiplication by z on k[z](∼= R/(x, y(1 − x))R. So, z(1 − x) is
R/(x, y(1− x))R-regular. Furthermore, the ideal I = (x, y(1− x), z(1− x)) is equal to
(x, y, z). Therefore R/IR ∼= k 6= 0. Thus x, y(1− x), z(1− x) is an R-sequence.
Since (y(1 − x), z(1 − x), x) is the same ideal with I above, we must show that this
sequence fails to be a weak R sequence. Multiplication by z(1−x) on R/(y(1−x))R is
not regular: z(1− x) · y = 0 in R/(y(1− x))R. Therefore (y(1− x), z(1− x), x) cannot
be an R-sequence.
53. Let R be a Noetherian local ring, M a finite R-module, and let x be an
M-sequence. Show that every permutation of x is an M-sequence.
Solution.
Since every permutation is the product of adjacent transpositions, it suffices to show
that x1, . . . , xi−1, xi, · · · , xn isM -regular. Note that for the moduleM/(x1, · · · , xi−1)M ,
(xi, . . . xn) is a regular sequence. Therefore, by induction, it will suffice to handle the
case of n = 2. We want to show that x2, x1 is M -regular assuming X1, x2 is M -regular.
Of course, we may assume that xi is not a unit, otherwise x1M = M , hence we would
23
be done. Now, look at the kernel of multiplication by x2 on M . Since x2 must be
regular on M/xiM , ker(·x2) ⊂ xiM . If z ∈ ker(·x2), then z − x1m for some m ∈ M.
Since xi(x2m) = x2(x1m) = 0, and x1 is regular on M, (x2m) must be 0. But then
m ∈ ker(·x2). Therefore ker(·x2) = xiker(·x2). Since x1 is a non unit, it is in the
maxima ideal, hence any element 1 + (x1) is regular on M. Finally we have to show
that x1 is regular on M/x2M . Assume otherwise; there exists m ∈ M/x2M such that
xim ∈ x2M hence xim = x2m′ for some m′ ∈M . Here m′ cannot be in xiM otherwise
the equation would imply m ∈ x2M. So, x2 is not regular on M/x1M, contradiction.
Thus, x2, x1 is a regular sequence too.
Definition 0.19. Let A be a subring of a ring B and b ∈ B. Then b is call integral
over A if b is a root of a monic polynomial with coefficients from A, that is if there is
a relation of the form bn + a1bn − 1 + · · ·+ an = 0 with ai ∈ A.
54. Let A ⊆ B be rings. Show that an element b ∈ B is integral over A if and
only if there exists a ring C between A and B such that b ∈ C and C is
finitely generated as an A-module
Solution.
(⇒) Let b ∈ B be integral over A. The let C be the ring generated by A and b, that is
C := A[b]. Obviously A ⊂ CsubsetB. Let us see that C is indeed a finitely generated
module over A. Obviously any element of C is a polynomial in b with coefficients of
A. Since bn = −(a1bn − 1 + · · ·+ an), any element of C can be written as an A-linear
combination of 1, b, b2, . . . , bn−1 hence C is finitely generated A-module.
(⇐) Suppose there exists an intermediate ring C which is a finitely generated Amodule.
Then C = Awi + . . . Awn for some wi ∈ C. Let b ∈ C. Then wib =
∑
j aijwj for
i = 1, . . . , n. But then we get a relation bn + a1bn−1 + · · · + an = 0 (by the Cayley’s
theorem; expanding det(bδij − aij)).
55. Let B be an integral domain and A ⊂ B a subring such that B is integral
over A. Then A is a field if and only if B is a field.
Solution.
(⇒) Suppose B is an integral domain which is integral over a field A. Then any element
b in B satisfies a monic polynomial over A i.e, bn + a1bn−1 + · · · + an = 0 for some
ai ∈ A. Since an is invertible, from this equation we see that b is invertible. Therefore,
every element of B is invertible, hence it is a field.
(⇐) Suppose B is a field and A ⊂ B a subring such that B is integral over A. It is
enough to show that for a ∈ A, the inverse a−1 of a also lies in A. Assume otherwise;
b := a−1 6∈ A. But then it satisfies a monic polynomial over A, that is bn + a1bn−1 +
24
· · · + an = 0 for some ai ∈ A. Multiplying this relation by an−1, we see that b ∈ A
which is a contradiction. Therefore A is a field, also.
56. Let B be an extension ring of A which is integral over A. Let P ⊂ B be a
prime ideal. Then P is maximal if and only if P ∩ A is maximal in A.
Solution.
P ∩A is obviously a prime ideal. The composition A ↪→ B → B/P gives the injection
A/A∩ P → B/P. Note that B/P is integral over A/A∩ P. Therefore, by the previous
problem that A/A∩B is a field if and only if B/P is a field. Hence A∩ P is maximal
in A if and only if P is maximal in B.
57. Let A ⊂ B be rings and suppose B is integral over A. Let m be a maximal
ideal in A. Show that there exists a prime P in B lying over m, that is
P ∩ A = m. Furthermore, any such P is maximal ideal of B.
Solution.
Let us first see that mB 6= B. Assume contrary that mB 6= B, then there exists bi ∈ B
and πi ∈ m such that
∑n
i=1 biπi = 1. Set C := A[b1, ...bn], then C is finite over A and
mC = C. Let C = Au1 +· · ·+Aur for some uj ∈ C. Then, we get ui =
∑
πijuj for some
πij ∈ m. Therefore ∆ := det(δij − πij) satisfies ∆uj = 0 for every j. Hence ∆C = 0.
But since 1 ∈ C,∆ = 0. On the other hand ∆ ≡ mod m simply by its expansion.
Hence 1 ∈ m, a contradiction. Now, since mB 6= B, it is contained in a prime ideal,
say P. By the previous problem we know that P must be a maximal ideal.
58. Let A ⊂ B be rings ans suppose B is integral over A. Let p be a prime ideal
of A. Show that there exists a prime p ⊂ B such that P ∩A = p. Furthermore,
there is not inclusion between primes in B that lie of p.
Localizing the exact sequence 0→ A→ B at p, we get an exact sequence
0→ Ap → Bp = B ⊗A Ap
in which Bp is an extension ring of Ap and integral over Ap. Using the following com-
mutative diagram:
Ap Bp
A B
25
we see that the prime ideals of B lying over p corresponds bijectively with the maximal
ideals of Bp lying over the maximal ideal pAp ⊂ Ap. Hence by the previous problem
we are done.
59. Let A ⊂ B be rings and suppose B is a finitely generated A-module( hence
integral over A.) Show that for a prime ideal of A, there are only finite
number of prime ideals in B that lie over p.
Solution.
First we assume A to be a local ring with maximal ideal m. Since B is finitely generated
module over A,B = Awi + · · · + Awr for some wi ∈ B. Then, we the quotient ring
B/mB becomes a vector space over A/m with a generating set {w̄i}. Every prime
P ⊂ B containing mBgives a vector space P/mB ↪→ B/mB. Also note that by the
previous problems we know that if any prime lying over a maximal must be maximal
too. Therefore these primes are coprime to each others. But then by the Chines
Remainder Theorem we see that B/mB ∼=
∏
i Pi/mB. Since each Pi/mB contributes
to the dimension of the vector space, there must be finitely many Pi′s. So, in the local
case we are done. For the general case we make use of the diagram:
Ap Bp
A B
Since primes containing p ⊂ A correspond to the maximal ideals of Bp over pAp in the
local ring Ap, we reduce to the local case. Hence we are done.
26
Note: Next two problems have been added to the file on January 2015.
Definition 0.20. Let p be a prime number and consider the ring of p-adic integers
Zp, which is constructed as the inverse limit lim←−Z/p
iZ. Let Ai denote Z/pi+1Z for
simplicity. By definition, this is the ring of sequences (ai)i≥0 ∈
∏∞
i=0Ai which satisfy
the compatibility condition with respect projections µji : Aj → Ai defined by
µji(aj mod p
j+1) = ai mod p
i+1 for all j > i.
60. Prove that there exists a bijection between Zp and the set of all formal
power series of the form
α0 + α1p+ α2p
2 + · · · , where αi ∈ {0, 1, . . . , p− 1}.
Solution.
Let (ai)∞i=0 be an element from Zp. As ai ∈ Ai = Z/pi+1Z, i = 0, 1, . . . , we assume that
ai is a positive integer less than pi+1. Set α0 = a0 ∈ {0, . . . , p− 1}.
Since a1 = a0 mod p, a1 − a0 is divisible by p. Since a1 < p2 and a0 < p, there exists
unique α1 ∈ {0, . . . , p− 1} such that a1 = a0 + pα1.
Similarly, by using projections µji, we have that a2 = a1 mod p2, or equivalently
that a2 − a1 = α2p2 for a unique integer α2. Since 0 ≤ a2 < p3, we see that α2
has to be less than p. Therefore, we see that a2 = a1 + pα2 = α0 + pα1 + p2α2.
Continuing in this manner, we conclude that the nth term of the sequence (ai)∞i=0 is
equal to
∑n
i=0 αip
i for some non-negative integers 0 ≤ αj < p, uniquely determined
by a0, a1, . . . , an−1. Therefore, the data of the element (ai) ∈ Zp is represented by the
infinite series α0 + α1p+ α2p2 + · · · in a unique way.
Let us revert this process to compute αi’s in terms of ai’s. We already know that
a0 = α0. α1 = (a1 − a0)/p. More generally
αn =
an − (α0 + · · ·αn−1pn−1)
pn
.
61. What are the ring operations on the series representation of p-adic integers?
Solution.
Suppose (ai), (bi) are two p-adic integers represented by the series
∑
αip
i and
∑
βip
i,
respectively. We first analyze what happens to addition in the series notation. Let
(ci) = (ai) + (bi) = (ai + bi) be represented by the summation
∑
γip
i.
27
Since α0 = a0, β0 = b0, we see that the ‘constant term’ γ0 of the series representation
of (ci) has to be equal to α0 + β0 mod p. Note that α0 + β0 = γ0 + pδ0 for a unique
δ0 ∈ {0, 1}.
Next we determine γ1. Since c1 = a1 + b1 = (α0 + β0) + p(α1 + β1) ∈ Z/p2Z is equal
to γ0 + γ1p, by the uniqueness of the power series coefficients we see that γ1 has to be
equal to δ0 +α1 +β1 mod p. Since δ0 +α1 +β1 < 2p, we see that δ0 +α1 +β1 = γ1 +pδ1
for a unique δ1 ∈ {0, 1}. Thus we write α1 + β1 = γ1 − δ0 + pδ1.
Similarly,
a2 + b2 = α0 + β0 + (α1 + β1)p+ (α2 + β2)p
2 mod p3
= γ0 + (δ0 + α1 + β1)p+ (α2 + β2)p
2 mod p3
= γ0 + (δ0 + γ1 − δ0 + δ1p)p+ (α2 + β2)p2 mod p3
= γ0 + γ1p+ (δ1 + α2 + β2)p
2 mod p3
Similar to the previous case, we write γ2 for δ1 + α2 + β2 mod p, hence the equality
γ2 + δ2p = δ1 + α2 + β2 implies that δ2 ∈ {0, 1}. Thus
a2 + b2 = γ0 + γ1p+ γ2p
2 mod p3,
where γ2 is equal to δ1 +α2 + β2 mod p, and δ1 is found from δ0 +α1 + β1 = γ1 + pδ1.
More generally, if an =
∑n
i=0 αip
i and bn =
∑n
i=0 βip
i, then an + bn =
∑n
i=0(αi + βi)p
i,
and γn is equal to δn−1 + αn−1 + βn−1 mod p, where δn−1 is found (inductively) from
δn−2 + αn−2 + βn−2 = γn−1 + pδn−1.
Next we look at what happens to the nth term of the product (ai)(bi) = (aibi). Suppose∑
γip
i corresponds to this product. Clearly γ0 = a0b0 mod p = α0β0 mod p. For γ1
we look at a1b1 modulo pZ, which has to be equal to a0b0 modulo pZ. Since a1b1 =
α0β0 + (α1β0 +α0β1)p in Z/p2, the equality a1b1 = a0b0 mod p is straightforward. On
the other hand, to compute γ1 we need to look at the carry over from α0β0. Indeed,
writing α0β0 as γ0 +pu1, we see that γ1 must be α0β1 +α1β0 +u1 mod p. Similarly, γ2
must be u2 +
∑2
i=0 αiβ2−i mod p, where u2 is the carry-over from the previous parts.
Indeed,
a2b2 = α0β0 + (α0β1 + α1β0)p+ (α0β2 + α1β1 + α2β0)p
2 mod p3
= γ0 + (α0β1 + α1β0 + u1)p+ (α0β2 + α1β1 + α2β0)p
2 mod p3
= γ0 + γ1p+ (u2 + α0β2 + α1β1 + α2β0)p
2 mod p3,
where u2 is found from the equation α0β1 + α1β0 + u1 = u2p + γ1. Let γ2 denote
u2 +α0β2 +α1β1 +α2β0 mod p
2, hence we write γ2 = u2 +α0β2 +α1β1 +α2β0 + u3p2.
Iterating this process we see how to multiply power series representations of p-adic
integers by carry-overs.
28
Gröbner Bases:
The monomials xa: = xa11 · · ·xann in a polynomial rings k[x1, . . . , xn] over any field k
can be identified with the lattice points (a1, . . . an) ∈ Nn. A total order ≺ on Nn is
a term order if the zero vector 0 is the unique minimal element, and a ≺ b implies
a+c ≺ b+c for all a, b, and c in Nn. Given a term order ≺, every nonzero polynomial
f ∈ k[x] has a unique initial monomial, denoted in≺(f). If I is an ideal in k[x], then its
initial ideal is the monomial ideal generated by the initial monomials of the elements
of I. The monomials which do not lie in in≺(I) is a Grobner Basis for I with respect
to ≺, if in≺(I) is generated by {in≺(g) : g ∈ G}. Given any polynomial f ∈ k[x], by
division algorithm, we can write f = h + r uniquely for h ∈ I and for an r ∈ k[x]
with no term of r is divisble by any in≺(g) for g ∈ G. Therefore, any polynomial f
can be uniquely written as a linear combination of the standard monomials. It is clear
that the set of standard monomials makes a k-vector space basis for the quotient ring
k[x]/in≺(I). Two important term orderings are
• Pure Lexicographic Order: α > β if the leftmost nozero entry of α−β is positive.
• Graded Lexicographic Order: α > β if
∑
α >
∑
β or
∑
α =
∑
β and the
leftmost nonzero entry of α− β is positive.
62. Fix a term order ≺ . Show that a finite subset G = {g1, . . . gn} ⊂ I is a Gröbner
basis for I if and only if S(gi, gj) reduces 0 mod G where
S(f, g) =
l.c.m(in≺(f), in≺(h))
in≺(f)
· f − l.c.m(in≺(f), in≺(h))
in≺(h)
· h
Solution. Type it later..
63. Fix a term ordering ≺ on R := k[x1 . . . xn]. Let I be an ideal in R. Show that
there exists a basis Bµ of the vector space R/I consisting of the images of
the standard monomials,the monomials that do not lie in in≺(I).
Solution.
Let G be a Gröbner for I, so, by definition in≺(I) is generated by the monomials n≺(I)
for g ∈ G. We know by the division algorithm that any polynomial f is congruent
modulo I to a polynomial r which does not posses any term divisible by in≺(I) for
g ∈ G (hence with none of in≺(I) for f ∈ (I)). Note that this implies that r is in the
span of the standard monomials (monomials that do not lie in in≺(I)). therefore any
polynomial f ∈ R, the image R/I is a linear combination of the images of the standard
monomials. Assume for a second that a linear combination of standard monomials lie
in I; s :=
∑
α aαx
α ∈ I. Then in≺(I) would be a standard monomial inside in≺(I)
which is absurd. Therefore we must have aα = 0 for all α ∈ Nn. This shows that
29
the standard monomials are linearly independent mod I. So we have shown that the
images of the standard monomials is a basis for the vector space R/I.
64. Show that a Gröbner Basis G of an ideal I generates the ideal.
Solution.
Recall that a finite subset G ∈ I is called a Gröbner basis for I if the initial ideal of I
is generated by the initial terms of G. Denote by the ideal generated by G. We want
to show that G = I. Observe that (in≺(g) : g ∈ G) = in≺(G) = in≺(I). Now, by the
previous problem we know that R/I is a vector space spanned by the monomials that
do not lie in in≺(I) = in≺(G). Therefore R/I = R/G. Since G ⊂I, this shows that
I = G.
65. Let I = 〈x2 + y, x+ xy〉. Find a Gröbner basis for the ideal I.
Solution.
This is an application of Buchberger’s Algorithm. I will use the Pure Lex with x > y.
Let fi = x2 +y and f2 = xy+y. And let G0 = {f1, f2}. Using pure lex, in(f1) = x2 and
in(f2) = xy. Then S(f1, f2) = y · f1 − x · f2 = y2 − x2. Reduce y2 − x2 mod G0 : since
y2−x2 + f1 = y2 + y and in(y2 + y) = y2 is not divisible by the initial monomials of f1
and f2, the result is f3 = y2 − y. Now, let G1 = {f1, f2, f3}. Then S(f1, f3) = y3 − yx2
and modulo G1 this is 0. and S(f2, f3) = 0. Therefore algorithm stops here and we get
a Gröbner basis {f1, f2, f3}.
30
USEFUL GEOMETRIC INTERPRETATIONS
We repeat; all rings are assumed to be commutative unless otherwise stated.
Given a ring R we denote by Spec(R) the set of all prime ideals of R. Note that we
exclude the whole ring itself from this set, but if R is an integral domain, the zero-ideal
is assumed to be an element of Spec(R). There is a useful topology on Spec(R), called
the Zariski topology defined as follows:
For each ideal I ⊂ R, let V (I) ⊂ Spec(R) denote the set of all prime ideals containing
I. A closed set in Spec(R), by definition, is either
• intersection of arbitrarily many sets of the form V (I), or
• union of finitely many sets of the form V (I).
66. Show that (a) V (
∑
Ii) = ∩V (Ii); (b) V (I ∩ J) = V (I) ∪ V (J); (c) V (IJ) =
V (I) ∪ V (J) .
Solution.
We start with the first claim. If P contains the sum of the ideals Ii, then it contains each
ideal Ii as well. Therefore, P ∈ V (Ii) for all i, hence V (
∑
Ii) ⊆ ∩V (Ii). Conversely,
if a prime ideal contains all Ii’s, then it contains their sum as well, hence the equality
V (
∑
Ii) = ∩V (Ii) follows.
For (b), let P ∈ V (I ∩ J) be a prime ideal containing I ∩ J . Assume that P does not
contain neither I nor J . Let a ∈ I and b ∈ J be two elements such that a ∈ I −P and
b ∈ J − P . Since ab belongs to both of I and J , it lies in I ∩ J , hence it lies in P . P
is a prime ideal, therefore, either a ∈ P , or b ∈ b, both of which gives a contradiction.
Therefore, either P ∈ V (I), or P ∈ V (J). Conversely, if P lies in V (I), then it of
course lies V (I ∩ J). This finishes the proof.
The proof of (c) is similar to that of (b).
Observe that a prime ideal is pulled back to a prime ideal by ring homomorphisms:
Let φ : A→ B be a ring homomorphism, P ⊆ B be a prime ideal, and let a, b be two
elements from A. If ab ∈ ψ−1(P ), then ψ(ab) ∈ P , hence either ψ(a) ∈ P , or ψ(b) ∈ P .
It follows that either a ∈ ψ−1(P ) or b ∈ ψ−1(P ). Therefore, there exists an induced
map in the opposite direction: ψ∗ : Spec(B)→ Spec(A), defined by ψ∗(P ) = ψ−1(P ).
Definition 0.21. A basic open set in a spectrum Spec(R) is, by definition, the com-
plement D(f) := Spec(R) − V ((f)), where (f) is the principal ideal generated by an
31
element f ∈ R. Thus D(f) is the set of all primes ideals which do not contain f . The
collection {D(f)}f∈R forms a basis for the Zariski topology.
67. Show that the induced map ψ∗ : Spec(B) → Spec(A) of the ring homomor-
phism ψ : A → B pullbacks distinguished open sets to distinguished open
sets.
Solution.
Let D(a) ⊂ Spec(A) be the distinguished open set associated with an element a of
A. The pre-image ψ∗−1(D(a)) consists of prime ideals Q in Spec(B) that are mapped
into D(a). If Q ∈ ψ∗−1(D(a)), then ψ∗(Q) = ψ−1(Q) is an element of D(a). Hence,
a /∈ φ−1(Q), or φ(a) /∈ Q. In other words, ψ∗−1(D(a)) = D(φ(a)) and this is what we
wanted to prove to begin with.
Remark 0.22. Our conclusion from Problem 67 is that the induced map ψ∗ : Spec(B)→
Spec(A) is continuous with respect to Zariski topology.
Definition 0.23. Let X be a topological space. A presheaf F (of rings, groups,
modules, or anything) on X is an assignment (of rings, groups, modules, or anything)
to each open set and satisfying certain compatibility conditions with respect to restric-
tions: For each pair of open sets U1, U2 with U2 ⊂ U1, there exists a ‘restriction’ map
r1,2 = rU1,U2 : F (U1)→ F (U2) that satisfies
(a) If U1 = U2, then r1,2 is the identity map;
(b) If U3 ⊂ U2 ⊂ U1, then the following diagram of restriction maps commute:
F (U1) F (U2)
F (U3)
r1,2
r1,3 r2,3
A sheaf is a presheaf F such that for every collection of open sets {Ui} with U =
⋃
Ui,
the following two conditions are satisfied:
(a) If x1, x2 are two elements from F (U) and rU,Ui(x1) = rU,Ui(x2) for all Ui’s, then
x1 = x2,
(b) If a collection elements xi ∈ F (Ui) satisfies rUi,Ui∩Uj(xi) = rUj ,Ui∩Uj(xj) for all
i, j, then there exists x ∈ F (U) such that xi = rU,Ui(x) for all i.
32
As an example, we define a sheaf of rings OR on Spec(R). For each prime ideal P , let
RP denote the localization of R at the multiplicative set (submonoid) R−P . For each
open set U of Spec(R), we define OR(U) to be the set of functions
s : U →
⊔
P∈U
RP (4)
such that
• s(P ) ∈ RP for all P ∈ U ;
• for each P ∈ U there exists an open neighborhood P ∈ U ′ ⊂ U and two elements
a, b in R satisfying 1) b is not contained in any prime Q ∈ U ′, hence a/b lives in
RQ, 2) a/b from 1) is equal to s(Q) for all Q ∈ U ′.
It is straightforward to verify that OR is a sheaf on Spec(R).
Finally, let us define the notion of the ‘stalk at a point.’ Let F be a presheaf (of
groups, rings, ..) on a topological space X and let x ∈ X be a point. The stalk of F
at x, denoted by Fx is the set of germs of sections of F at x. In other words, each
element of Fx is an equivalence class pairs (U, s), where U is an open set containing
x and s ∈ F (U) a section; the pair (V, t) is equivalent to (U, s) if there exists an open
set W ⊆ U ∩ V such that s|W = t|W .
A locally ringed space is a ringed space for which stalks of the sheaf are local rings.
68. Show that (Spec(R),OR) is a locally ringed space and furthermore the stalk
of OR at P is nothing but the local ring RP .
Solution.
Let (OR)P denote the stalk ofOR at P and define ψ : (OR)P → RP by ψ((U, s)) = s(P ).
We claim that ψ is an isomorphism. To prove the surjectivity, let a/b be an element
of the local ring RP , hence, a, b ∈ R and b /∈ P . Then the distinguished open set
U = D(b) contains P . We define a section s ∈ OR(U) by s(Q) = a/b for all Q ∈ D(f).
In particular s(P ) = a/b. Therefore, ψ is surjective.
Next we prove that ψ is injective. Let (U, s) and (V, t) be two germs with the same
image a/b at P . Then we need to show that (U, s) and (V, t) represent the same
equivalent class. By definition given in (4) we know that for both section s and t,
there are open sets U ′ and V ′ around P and elements a1, b1 and a2, b2 from R such
that for all Q1 ∈ U ′ and Q2 ∈ V ′ we have b1 /∈ Q1, b2 /∈ Q2 with s(Q1) = a1/b1 and
t(Q2) = a2/b2. In particular, a1/b1 = s(P ) = t(P ) = a2/b2 in RP which is true if and
only if c(a1b2 − a2b1) = 0 for some unit c from R − P . This equality is true for all
prime ideals Q such that b2, b1, c /∈ Q, or equivalently, for all Q ∈ D(b1)∩D(b2)∩D(c)
which is an open set. Therefore (U, s) and (V, t) represents the same germ, hence ψ is
injective.
33
69. Given (Spec(R),OR) let OD(f) denote the restriction of the sheaf OR to the
open set D(f), f ∈ R. Show that the locally ringed space (D(f),OD(f)) is
isomorphic to (SpecRf ,ORf ).
Solution.
Let φ : R → Rf denote the homomorphism defined by φ(a) = a/1. A prime ideal of
Rf is of the form PRf where P is a prime ideal of R such that f /∈ P . It is clear
that φ−1(PRf ) = P . The image of the associated map φ∗ : SpecRf → Spec(R). The
image of φ∗ is the set of all primes in R that does not contain f . In other words,
φ∗(SpecRf ) = D(f) in Spec(R). The induced map on the stalks is given by the
localized homomorphism
φP : RP → (Rf )PRf .
Since P does not contain f , the localization (Rf )PRf is isomorphic to RP , hence the
canonical map φP is an isomorphism. Since we have isomorphisms between their stalks,
the isomorphism SpecRf → D(f) is an isomorphism between locally ringed spaces
(SpecRf ,ORf ) and (D(f),OD(f)).
Definition 0.24. An affine schemeis a locally ringed space (X,F ) which is isomorphic
to a pair (Spec(R),OR) for some ring R. The isomorphism here is a local isomorphism.
A scheme is a locally rings space such that around every point x ∈ X there exists a
neighborhood x ∈ U for which the pair (U,F |U) is an affine scheme. Here, F |U denotes
the restriction of the sheaf F to U .
Definition 0.25. A scheme (X,OX) is reduced if for every open U ⊆ X the ring
OX(U) has no nilpotent elements.
70. Show that reducedness is a local property: (X,OX) is reduced if and only if
for every p ∈ X the local ring (OX)p has no nilpotent element.
Solution.
Let (U, s) and (V, t) be two germs from the stalk at a point P ∈ X. Multiplication
and/or addition of these germs are done in in O(W ), where W ⊆ U ∩ V is a neighbor-
hood of P . Therefore, a germ (U, s) is nilpotent if and only if sn = 0 in a sufficiently
small neighborhood W of P . It follows that OX(U) has nilpotent elements if and only
if the stalk (OX)P has nilpotent elements.
34
71. Let (X,O) denote an affine scheme and Ored denote the sheaf associated
with the presheaf U 7→ (O(U))red. Here, the subscript red means we are
taking the quotient of the ring O(U) by the ideal of nilpotent elements.
Show that (X,Ored) is a scheme. Let us denote it by Xred. Show that there
is a morphism π : Xred → X which a homeomorphism of the underlying
topological spaces.
Solution.
It is straightforward to verify that U 7→ (O(U))red is a sheaf. The second claim
that Xred is homeomorphic to X boils down to the fact that the nilradical (ideal
of nilpotent elements) of a ring is the intersection of all prime ideals in the ring.
See Problem 41. Indeed, let n denote the nilradical of R. The canonical quotient
homomorphism p : R −→ R/n gives a homeomorphism between topological spaces
π : Spec(R/n) ∼−→ Spec(R). (5)
72. Let R be a reduced ring and suppose we have a homomorphism f : R′ → R
from another ring R′ into R. Show that there exists a unique ring homomor-
phism g : R′red → R such that f = g ◦ p, where p : R′ → R′/n is the canonical
projection into quotient of R′ by its nilradical.
Solution.
We claim that the map g : R′/n → R defined by g(p(a)) = f(a) is a well-defined
homomorphism. if p(a) = p(b), then a − b is a nilpotent element in R′. Since R has
no nilpotent elements, a− b is mapped to 0 ∈ R via f . Therefore, g(p(a))− g(p(b)) =
f(a) − f(b) = f(a − b) = 0. In other words, g is well defined. g is a homomorphism
because
• g(p(a) + p(b)) = g(p(a + b)) = f(a + b) = f(a) + f(b) = g(p(a)) + g(p(b)) for all
a, b ∈ R′;
• g(p(a))g(p(b)) = f(a)f(b) = f(ab) = g(p(ab)) = g(p(a)p(b)) for all a, b ∈ R.
Uniqueness is clear from the definition.
73. Let f : Xred → Y be a map from a reduced scheme into another scheme. Show
that there exists unique a scheme map g : Xred → Yred such that f = πg.
Solution.
Reducedness is a local property. The solution follows from Problem 72.
35
74. Let x be an element of a ring R. Show that the distinguished basic open
set D(x) ⊂ Spec(R) is empty if and only if x is nilpotent.
Solution.
Recall that D(x) is the set of all prime ideals that do not contain x. Therefore,
D(x) = ∅, then x lies in all prime ideals of R, hence it belongs to the nilradical.
Conversely, if x is nilpotent, it lies in all of the prime ideals, hence D(x) = ∅.
Definition 0.26. A topological space X is called irreducible if it is not a union of two
proper closed subsets.
75. Show that a scheme is irreducible if and only if every open subset is dense.
Solution.
Assume that X is irreducible and let U ⊆ X be an open subset. Since U∪(X−U) = X
and since X − U 6= X, we must have U = X. In other words, U is dense.
Conversely, suppose that every open subset in X is dense. If X = A∪B for two proper
closed subsets, then X = (X − A) ∪ (X − B). In this case (X − A) ∩ (X − B) 6= ∅
since open sets are dense. But taking complement once again we see that A ∪ B 6= X
which is absurd.
76. An affine scheme X = Spec(R) is reduced and irreducible if and only if R is
an integral domain.
Solution.
(⇒) First, assume that X is irreducible. Let a, b ∈ R be two non-zero elements such
that ab = 0. Since any prime ideal contains 0, we have D(a)∩D(b) = D(ab) = D(0) =
∅. Since D(a) and D(b) are open sets, we have a contradiction. Therefore, R is an
integral domain.
(⇐) If R is an integral domain it cannot have any nilpotent elements. Hence, Spec(R)
is a reduced affine scheme.
77. Spec(R) is irreducible if and only if R has a unique minimal prime ideal.
Solution.
Suppose X = Spec(R) is irreducible. If S is the set of all minimal prime ideals of
R, then ∪P∈SV (P ) = Spec(R). This contradicts with the irreducibility of X unless
S is a singleton. Conversely, if there exists unique minimal prime ideal P , then it is
36
necessarily the nilradical of R. In this case R/n is an integral domain. Since Spec(R/n)
is homeomorphic to Spec(R), the irreducibility of the latter scheme follows from Prob-
lem 76.
Definition 0.27. Let φ : F → G be a map (morphism) of sheaves. The presheaf
defined by U 7→ kerφ|U is a sheaf and denoted by kerφ. The morphism φ is called
injective if the associated kernel sheaf kerφ is the constant sheaf 0. Equivalently, φ is
injective if and only if for all open sets U the homomorphisms φ(U) : F (U) → G (U)
are injective.
78. Let ψ : A → B be a ring homomorphism and let f : Spec(B) → Spec(A)
denote the associated morphism between affine schemes. Show that if B
is an integral domain, then ψ is injective if and only if the corresponding
sheaf map f ] : OA → f∗OB is injective.
Solution.
(⇐) Suppose f ] : OA → f∗OB is injective. By definition, f ](U) is a ring map (in-
duced by φ) from OA(U) to OB(f−1(U)). If U = Spec(A), then OA(U) = A and
OB(f−1(U)) = OB(Spec(B)) = B, and the corresponding ring map ψ is injective.
(⇒) Suppose that ψ : A → B is an injective ring homomorphism. We are going to
show that any homomorphism f ](U) : OA(U) → OB(f−1(U)), where U ⊆ Spec(A)
is injective. Recall that the elements of the ring OA(U) are the s : U →
⊔
P∈U AP
satisfying for each P ∈ U , there exists a neighborhood D(g) and f ∈ A such that
s(P ) = f/g ∈ AP . After this refreshing, we observe that the value of f ](U) on a
section s ∈ OA(U) is defined by the compositions:
f−1(U)
f−→ U s−→
⊔
P∈U
AP
ψU−→
⊔
Q∈f−1(U)
BQ.
The last map is defined as follows: For each Q ∈ f−1(U) ⊂ Spec(B) there exists
unique prime ideal P ∈ U ⊂ Spec(A) such that P = f(Q) = ψ−1(Q), or equivalently
ψ(P ) = Q. Since g /∈ P , and ψ is injective, we see that ψ(g) /∈ ψ(P ). Therefore
ψU(f/g) = ψ(f)/ψ(g) ∈ BQ. Now we are ready to check that f ](U)(s) = f ](U)(t)
implies that s = t. Indeed, if ψ(f)/ψ(g) = ψ(f ′)/ψ(g′), then cψ(fg′ − f ′g) = 0 for
some c ∈ B − Q. Since B is an integral domain, and since ψ is injective f/g = f ′/g′,
hence f ](U) is injective.
Definition 0.28. A morphism of schemes is called dominant if its image is dense.
37
79. Let φ : A → B be an injective ring homomorphism. Show that the induced
map f : Spec(B)→ Spec(A) is dominant.
Solution.
We are going to show that every non-empty distinguished open set D(x) in Spec(A)
intersects f(Spec(B)). Assume contrary that there exists x ∈ A such that D(x) ∩
f(Spec(B)) = ∅. Let Q ∈ Spec(B) be a prime ideal in B. Then f(Q) is the prime
ideal P = φ−1(Q). Notice that x must be contained in P , otherwise, P ∈ D(x). It
follows that φ(x) is contained in every prime ideal in B, therefore, φ(x) is nilpotent.
Since φ is injective we x must be nilpotent. Then D(x) = ∅ a contradiction to our
initial assumption.
80. Let φ : A→ B be a surjective ring homomorphism. Show that f : Spec(B)→
Spec(A) is a homeomorphism onto a closed subset of Spec(A) and further-
more f ] : OSpec(A) → f∗OSpec(B) is surjective.
Solution.
The image of a prime ideal of B under f is a prime ideal in A that contains the kernel
of φ. Therefore, the image of φ is nothing but the closed set V (kerφ) ⊆ Spec(A). It is
clear that f is homeomorphic onto its image.
To prove the surjectivity of the sheaf map we look at the correspondingclaim on stalks.
For P ∈ Spec(A) let Q ∈ Spec(B) be the prime ideal such that φ−1(Q) = P . Since the
stalk of OA at P is AP and the stalk of f∗OSpec(B) at Q is BQ, for each a/b ∈ BQ with
b /∈ Q, it is enough to find c/d ∈ AP such that φ̃(c/d) = φ(c)/φ(d) = a/b. To this end,
let d be an element from φ−1(b). It is clear that d /∈ P . Similarly, let c ∈ φ−1(a). Then
c/d lies in AP , and furthermore, φ̃(c/d) = a/b.
81. Show that the converse of the above problem is true: if f : Spec(B) →
Spec(A) is a homeomorphism onto a closed subset of Spec(A) and if f ] :
OSpec(A) → f∗OSpec(B) is surjective, then φ : A→ B is surjective.
Solution.
Type later.
82. Show that for any OX-module F , HomOX (E ,F) ∼= Ě ⊗OX F .
Solution.
38
Consider it locally; on the left hand side we have HomOXx (Ex,Fx) and on the right
HomOXx (Ex,OXx)⊗Fx. Thus
HomOXx (Ex,OXx)⊗Fx ∼= HomOXx (Ex,OXx ⊗Fx)
∼= HomOXx (Ex,Fx),
so we are done.
83. Show that for any OX-modules F , G, and E,
HomOX (E ⊗ F ,G) ∼= HomOX (F , HomOX (E ,G))
Solution. Think local once again. It from Problem 26.
84. If f : (X,OX) −→ (Y,OY ) is a morphism of ringed spaces, and if F is an
OX-module and E is a locally free OY -module of finite rank, then there is a
natural isomorphism
f∗(F ⊗OX f ∗E) ∼= f∗(F)⊗OY E .
Solution.
Type later.
Definition 0.29. A morphism f : X −→ Y is called "locally of finite type" if there
exists a covering of Y by open affine subsets Vi = Spec(Bi), such that for each i,
f−1(Vi) can be covered by open affine subsets Uij = Spec(Aij), where each Aij is a
finitely generated Bi-algebra. The morphism f is of finite type if in addition each
f−1(Vi) can be covered by a finite number of the Uij.
85. Show that a morphism f : X −→ Y is locally of finite type if and only if for
every open affine subset V = Spec(B) of Y , f−1(V ) can be covered by open
affine subsets Uj = Spec(Aj), where each Aj is a finitely generated B-algebra.
Solution.
Suppose that f is locally of finite type and V = Spec(B) is an affine open subset
of Y . Then it is clear from definition that there exists an open covering of the form
Vj = Spec(Aj) for which Aj is a finitely generated B-algebra.
39
For the converse first recall the definition of a scheme: a locally ringed space (X,OX)
in which every point has an open neighborhood U such that the topological space
U , together with the restricted sheaf OX |U is an affine scheme, that is isomorphic to
(Spec(B),OSpec(B)) for some ring B.
Thus by definition of a scheme we get an open covering consisting of affine open sub-
schemes Vj = Spec(Bj). By the hypothesis we have a covering of f−1(Vj) with affine
open subsets Uij = Spec(Aij), where Aij is a finitely generated Bj-algebra. But this is
exactly the definition of being locally of finite type. We are done.
86. Let f and g be two elements from a ring A. Show that the basic open
subsets D(fg) = D(f) ∩D(g).
Solution.
By definition D(fg) is the set of all primes in A that do not contain fg. But then
none of those primes can contain neither f nor g. Therefore D(fg) ⊆ D(f) ∩D(g).
Conversely, a prime that does not contain both f and g cannot contain fg. Therefore
D(f) ∩D(g) ⊆ D(fg) hence we get the equality.
87. Show that {D(fα)}, a family of distinguished open subsets of Spec(R), is a
covering if and only if 1 is in the ideal generated by fα’s.
Solution.
Suppose Spec(R) =
⋃
αD(fα). Then
Spec(R) =
⋃
α
D(fα)
=
⋃
α
(V (fα))
c
=
(⋂
α
(V (fα))
)c
= (V (〈fα〉))c
Therefore ∅ = V (〈fα〉). But then 〈fα〉 = R.
88. Show that Spec(R) is quasi compact, that is any cover has a finite subcover.
Solution.
40
Suppose {Uα} is an open cover for Spec(R). Each Uα is covered by basic opens Uα =⋃
βD(fαβ). Therefore {D(fαβ)} is an open covering for Spec(R). By the previous
problem, 1 is in the ideal geberated by fαβ that is 1 = a1fα1β1 + . . .+ arfαrβr for some
ai ∈ R. Then {D(fαiβi)}i=1...r hence {Uαi}i=1...r is an open covering by the previous
problem. Therefore Spec(R) is quasi compact.
89. Let A be a ring. Show that the following conditions are equivalent:
(a) Spec(A) is disconnected.
(b) There exists nonzero elements e1 and e2 in A such that e1e2 = 0, e21 = e1,
e22 = e2, and e1 + e2 = 1 (these elements are called orthogonal idempo-
tents).
(c) A is isomorphic to a direct product A1 × A2 of two nonzero rings.
Solution.
(a) ⇒ (b) Suppose Spec(A) is disconnected that there are two disjoint open (hence
closed) proper subsets U1 and U2 with U1 ∪ U2 = Spec(A). Therefore there are dis-
tinguished opens D(fα) that cover U1 and distinguished opens D(gα) that cover U2.
Since Spec(A) is quasi-compact, and totality of those basic opens is a covering for A,
we can choose a finite subcover: D(f1), . . . , D(fr), D(g1), . . . , D(gs). We can safely
assume that none of the fi’s or gj’s are nilpotent or unit. Otherwise corresponding
basic open would be the whole space or empty. One remark here is that for any ring
and for any element a of it, D(an) = D(a). This is because a prime containing some
power of a will also contain it. Thus we may replace each basic open D(fi) by D(fni )
if needed. Now since these basic opens cover Spec(A), 1 is in the ideal generated
by f1, . . . , fr, g1, . . . , gs. Say 1 = a1f1 + . . . + arfr + b1g1 + . . . + bsgs. Before go-
ing further, we observe that D(fi) and D(gj) are disjoint for any i and j. Therefore
D(fi) ∪D(gj) = D(figj) = ∅. Thus figj is nilpotent. Thus (figj)mij = 0 for some mij.
Let m be the maximum of those natural numbers for i = 1, . . . , r and j = 1, . . . , s.
By the previous remarks we can replace each fi and gj by fmi and gmj and still get a
covering of Spec(A). Now let e1 = a1f1 + . . .+ arfr and e2 = b1g1 + . . .+ bsgs so that
e1 + e2 = 1 and e1 · e2 = 0. Furthermore, e1e2 = e1(1 − e1) implies that e1 = e21 and
e2 = e
2
2.
(b) ⇒ (c) Since we can write ae1 + ae2 = a, we have the following homomorphism
of rings A φ−→ A/(e1) × A/(e2) via a 7−→ (ae2, ae1). This homomorphism is injective
because (ae2, ae1) = (be2, be1) implies that (a − b)e2 + (a − b)e1 = 0 thus a = b. This
homomorphism is surjective since for any (x, y) ∈ A/(e1)×A/(e2), we have the equality
(x, y) = (e2x, e1y) from e1 + e2 = 1 and hence φ(e2x + e1y) = (e2(e2x + e1y), e1(e2x +
e1y)) = (e2x, e1y). Therefore A is the direct sum of two nonzero rings A1 = A/(e1)
and A2 = A/(e2).
41
(c) =⇒ (a) Suppose A = A1 × A2, and let e1 = (1, 0) and e2 = (0, 1). Therefore
e1e2 = 0, e1 + e2 = 1, e21 = e1 and e22 = e2. Now let V1 = V (e1) and V2 = V (e2) that is
the closed subsets of Spec(A) defined by e1 and e2 respectively. If a prime p contains
e1 then it cannot contains e2 otherwise it would contain e1 + e2 = 1. Therefore V1 and
V2 are disjoint proper subsets of Spec(A). Thus Spec(A) is disconnected.
90. A morphism of schemes f : X −→ Y is quasi compact if there is a cover of Y
by open affines Vi such that f−1(Vi) is quasi compact for each i. Show that
f is quasi compact if and only if for every open affine subset V ⊂ Y , f−1(V )
is quasi compact.
Solution.
⇐ Suppose that for every affine open subset V ⊂ Y , the preimage f−1(V ) is quasi
compact. Since Y is a scheme, every point has an affine open neighborhood which is
isomorphic to a spectrum of a ring. These affine neighborhoods give us a covering that
is required.
⇒ Suppose that f is quasi compact so that we have an open affine cover {Vi =
Spec(Ri)} of Y such that f−1(Vi) is quasi compact. Let V ⊂ Y be an arbitrary
affine open subset. Since affine schemes are quasi compact, we can choose finitely
many {V1, . . . , Vn} affine opens such that
⋃n
i=1 Vi = V . Now if we can show that
each open f−1(Vi ∩ V ) is quasi compact, then we are done because a finite union of
quasi compact sets is quasi compact. Let Z := f−1(Spec(Ri)) be the preimage of
Vi in X and let f |f−1(Z) = g. Now we have a morphism g : Z −→ Spec(Ri) with
g−1(Spec(Ri)) = f−1(Spec(Ri)) = Z is quasi compact, and we would like to show
that any open subset V ′ ⊂ Spec(Ri) has quasi compact preimage. Since Z is quasi
compact we can cover it by finitely many open