Table Of ContentPROBLEMS AND SOLUTIONS IN
COMMUTATIVE ALGEBRA
Mahir Bilen Can
[email protected]
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 (cid:54)= 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 a /s and a /s be two elements from S−1I. And let r/s ∈ S−1R. Then
1 1 2 2
r/s·a /s +a /s = (s ra −s sa )/s s ∈ S−1I. Therefore S−1I is an ideal.
1 1 2 2 2 1 1 2 1 2
(b) Elements of S−1(R/I) are of the form r¯/s¯where bar denotes the images of elements
of S in R/I. If we start with an element r/s of S−1R, then r¯/s¯ makes sense. So we
can define the homomorphism φ : S−1R −→ S−1(R/I) by φ(r/s) = r¯/s¯. By its
construction φ is surjective. How about its kernel? Suppose r¯/s¯ = 0 in S−1(R/I).
Then there exists s¯(cid:48) in the image of S in R/I such that r¯·s¯(cid:48) = 0 that is rs(cid:48) ∈ I. But
it is always the case that rs(cid:48)/ss(cid:48) = 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, R /I = (R/I) .
P P P
Some authors write IR for what we are calling I ; the ideal generated by the image
P P
of I in R .
P
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 φ : A −→ B gives B a finitely
S S φ(S) φ(S)
generated A -module structure. (Here, without loss of generality we as-
S
sume that 0 ∈/ φ(S). Otherwise, B = 0, which is a finitely generated
φ(S)
A -module.)
S
Solution.
Let {b ,...,b } be a generating set for B as an A-module. Let a/b be an element of
1 r
the localized ring A . For c/φ(d) ∈ B with c ∈ B, d ∈ S the action of A on B
S φ(S) S φ(S)
is defined by
a c φ(a)·c
· = .
b φ(d) φ(bd)
2
Since B is generated by {b ,...,b } as an A-module, c is of the form a ·b +···a ·b =
1 r 1 1 r r
φ(a )b +·+φ(a )b for some a ,...,a ∈ A. Therefore,
1 1 r r 1 r
r
c φ(a )b +·+φ(a )b (cid:88) a b
1 1 r r i i
= = ·
φ(d) φ(d) d 1
i=1
proving that {b /1,b /1,...,b /1} is a generating set for B as an A -module.
1 2 r φ(S) S
5. Let P ,...,P be a finite set points from Cn such that P (cid:54)= P for all j ∈
1 m 1 j
{2,...,m}. Find an explicit polynomial F(x ,...,x ) ∈ C[x ,...,x ] which
1 n 1 n
takes constant value 1 on P and 0 on P for all j ∈ {2,...,m}.
1 j
Solution.
Let (a ,...,a ) ∈ Cn denote the coordinates of P for k = 1,...,n. Since P (cid:54)= P ,
k1 kn k 1 j
there exists smallest index r ∈ {1,...,n} such that a (cid:54)= a . Define g (x ,...,x )
j jrj 1rj j 1 n
by
x −a
g (x ,...,x ) := rj jrj
j 1 n
a −a
1rj jrj
It is clear that g (P ) = 0 and g (P ) = 1. The product of g ’s for j = 2,...,m is the
j j j 1 j
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[x ,...,x ], V(I) ⊆ kn denotes the
1 n
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. If I is a radical ideal,
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) (cid:54)= 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) (cid:54)= 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 {x ,...,x } be a generating set for M
1 m
as an R-module. By hypothesis, for each i = 1,...,m we have
(cid:88)
ψ(x ) = a x , (1)
i i,j j
for some a ∈ a. We define A to be the operator δ ψ−a e, where e is the identity
i,j i,j i,j i,j
endomorphism of M and δ is the Kronecker’s delta function.
i,j
(cid:80)
It is clear from (1) that A x = 0 for all i = 1,...,m. In other words, the matrix
j=1 i,j j
of operators A := (A )m annihilates the column vector v = (x )m . Notice that
i,j i,j=1 j j=1
we can consider M as an R[ψ]-module, and that A ∈ R[ψ]. Thus A is a matrix over
i,j
R[ψ]. Therefore, its adjugate makes sense and multiplying Av = 0 on the left by the
adjugate gives us (detA)x = 0 for all j = 1,...,m. Consequently, detA annihilates
j
all of M. Expanding the 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 +a idr−1 +···+a id = 0 for some a ∈ a. Therefore,
1 r j
if we define x = 1 + a + ··· + a , then x − 1 ∈ a and furthermore xm = 0 for all
1 r
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
idealmofR. Inparticular,sincexisfromm,weseethat1 ∈ mwhichisacontradiction.
Conversely, supposethat1−xy isinvertiblefor ally ∈ R. Ifxdoes notliein amaximal
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
let M 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 ∩ an = (0).
n>0
5
Solution.
Let M denote ∩ an. Obviously, M is a R-submodule of a. By Artin-Rees lemma,
n>0
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(cid:48) ⊆ M a submodule, then M(cid:48) +aM = M implies
that M(cid:48) = M.
Solution.
We consider the R-module M/M(cid:48). Our assumption M(cid:48) + aM = M implies that
aM/M(cid:48) = M/M(cid:48). Hence, by Problem 9 M/M(cid:48) = {0}, or equivalently, M = M(cid:48).
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 u ,...,u generate M if and
1 n
only if the images u ,...,u generate M/aM as an R-module
1 n
Solution.
The implication (⇒) is obvious. We prove the converse. Suppose u ,...,u generate
1 n
M/aM as an R-module. Let {u ,...,u } denote a set of preimages of u ’s, and let
1 n i
M(cid:48) denote the submodule generated by u ’s. It is clear that M(cid:48) +aM = M, hence by
i
Problem 12 it follows that M(cid:48) = 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 a ,...,a ∈ m generate m/m2 as a vector space. Show that a ,...,a
1 n 1 n
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, a ,...,a generates m/m2 as a vector
1 n
space, then a ,...,a generates m minimally, that is to say none of a ’s is
1 n i
redundant.
6
Solution.
Towards a contradiction, without loss of generality, assume that a redundant; a =
1 1
r a +···+r a for some r ∈ R. Then, modulo m2, a ,...,a are linearly dependent
2 2 n n i 1 n
which is a contradiction.
Definition. An ideal I (cid:54)= (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. Let M be a finitely generated R-module and S ⊂ R be a multiplicative subset
of R. Show that M = 0 if and only if an element s of S annihilates M, that
S
is to say, sM = 0.
Solution.
If s annihilates M, then for any m/r ∈ M , we have m/r = sm/sr = 0/sr = 0, thus
S
M = 0. Conversely, assume that M = 0, that is m/r = 0 for every m ∈ M and
S S
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 {m ,...,m } of M, and there exists a corresponding set
1 n
{s ,...,s } of annihilators. Since R is commutative, the product s ···s annihilates
1 n 1 n
all m ’s, hence it annihilates whole of M.
i
7
Definition: Support of an R-module M is the set of all prime ideal p such that M
p
is non-trivial. Here, M is the localization of the module at the multiplicative subset
p
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 M = 0 if and only if there exists s ∈ R − p such that
p
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 of M. 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 −→ A −α→1 A −α→2 A −→ 0 (2)
1 2 3
gives rise to a left exact sequence
α∗ α∗
0 −→ Hom (A ,N) −→2 Hom (A ,N) −→1 Hom (A ,N) (3)
R 3 R 2 R 1
Solution.
Given a homomorphism f : A −→ N, we pull it back to a homomorphism from
3
α∗
A to N by fα : A −→ N. Hence we get a homomorphism Hom (A ,N) −→2
2 2 2 R 3
Hom (A ,N).
R 2
Next, let us see that α∗ is injective: suppose α∗f = α∗g for some f and g from
2 2 2
Hom (A ,N). Then, α f = α g. But α is surjective by exactness (1). Thus, f and
R 3 2 2 2
g agree on every point of A showing that they are the same functions. Therefore α∗
3 2
is injective.
Next, letusseethatim(α∗) ⊆ ker(α∗), namelyα∗α∗ = 0. Letf ∈ Hom (A ,N). Then
2 1 1 2 R 3
α∗α∗f = fα α = f ·0 = 0 again by exactness of (1). Therefore im(α∗) ⊆ ker(α∗).
1 2 2 1 2 1
Finally, let us see that im(α∗) ⊇ ker(α∗): let f : A −→ N be in the kernel of α∗,
2 1 2 1
namely fα = 0. Define g : A −→ N as follows, let g(a ) be the value f(a ) for
1 3 3 2
any a ∈ A such that a and g is well-defined. Notice that gα = f. Therefore
2 2 2 2
im(α∗) = ker(α∗) and (2) is left exact.
2 1
8
20. Give an example of a module N and a short exact sequence such that
Hom (−,N) does not give a short exact sequence.
R
Solution.
Let p ∈ Z be a prime and consider the following exact sequence of Z-modules:
0 −→ Z −→p Z −→ Z/p −→ 0
Apply Hom (−,Z) and check that the result is not an exact sequence.
Z
21. LetGbeanabeliangroupandwriteG (cid:39) Zn⊕Gtorsion. ShowthatHomZ(G,Z) ∼=
Zn.
Solution.
Let f ∈ Hom (G,Z), then f is determined by the images to the generators of the
Z
copies of Z in G. The reason for not affected by the torsion part of G is the following:
if a ∈ G , then n·a = 0 for some n ∈ Z. Then 0 = f(0) = f(n·a) = nf(a) since
torsion
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[x ,...,x ]-modules show that Hom (R(−a),R) (cid:39)
1 n R
R(a).
Solution.
Note that 1 in R(−a) is an element of degree a. Therefore, a homomorphism f ∈
Hom (R(−a),R) (cid:39) R(a) of graded R-modules maps 1 to a degree a element in R that
R
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 ∈ Hom (R(−a),R) −→ R via
R
f (cid:55)−→ 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/3?
Z
Solution.
Let a⊗b ∈ Z/2⊗ Z/3. If b is 0 or 2, then a⊗b = 0⊗0. If b = 1, then we can replace
Z
it by 4 (since 4 ≡ 1 mod 3) and get a⊗b = 0⊗0 again. Therefore Z/2⊗ Z/3 = 0.
Z
9
24. More generally, show that Z/a⊗ Z/b (cid:39) Z/gcd(a,b).
Z
Solution.
First of all remember that gcd(a,b) of two integers is the largest (hence unique) integer
thatdividesbothaandb. Furthermore, thereareintegersxandy suchthatgcd(a,b) =
ax+by.
Now, that said, define φ : Z/a⊗ Z/b −→ Z/gcd(a,b) by r⊗s (cid:55)−→ rs mod gcd(a,b).
Z
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 (cid:55)−→ r mod gcd(a,b). Therefore φ is an isomorphism of
Z-modules.
25. Show that − ⊗ M is a right exact functor on the category of R-modules,
R
but it need not be an exact functor.
Solution.
Let
0 −→ A −α→1 A −α→2 A −→ 0
1 2 3
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:
A ⊗M −→ A ⊗M −→ A ⊗M −→ 0
1 2 3
Since the maps are defined by a ⊗m (cid:55)−→ α (a )⊗m, right exactness is straightforward
i i i
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/p gives us an exact
Z
sequence:
0 −→ Z⊗ Z/p −→p Z⊗ Z/p −→ Z/p⊗ Z/p −→ 0
Z Z Z
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
