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SOLUTIONS TO THE SECOND TEST FOR COMMUTATIVE
ALGEBRA M4P55
AMBRUS PÁL
In all you could have collected 75 points. In order to get the maximal grade, it
was enough to get 20 points, so there was a fair bit of redundancy.
1. (a) An R-module is Artinian if it satisfies the descending chain condition, that
is, if any chain of R-submodules:
I0 ⊇ I1 ⊇ . . . ⊇ In ⊇ . . .
is stationary. A ring R is Artinian if it is Artinian as an R-module. (2 points)
Every finite ring has only finitely many ideals, so it must be Artinian. (1 point)
(b) For every natural number n the ring Z/nZ has cardinality n, so any n is
possible. (2 points)
Let pn be a prime power. Then Z/pn Z has cardinality pn and it has a unique
prime ideal generated by p, so any prime power is the cardinality of a finite ring
with a unique maximal ideal. Let R be a finite ring with a unique maximal ideal
m. There is a positive integer n such that mn = 0. Now each successive quotient
in the descending chain
R ⊇ m ⊇ . . . ⊇ m ⊇ . . . ⊇ mn = 0
is a finite dimensional vector space over the finite field R/m. The cardinality of
the latter is the power of a prime p, as the cardinalities of finite fields are prime
powers. So the cardinality of all successive quotients above is a power of p, so the
order of R is also a prime power. (8 points)
(c) A ring is local if it has a unique maximal ideal. (1 point)
An R-module is Noetherian if it satisfies the ascending chain condition, that is,
if any chain of R-submodules:
I0 ⊆ I1 ⊆ . . . ⊆ In ⊆ . . .
is stationary. (2 points)
(d) Let M be a Z-module is both Artinian and Noetherian. Then it is finitely
generated, so M = hm1 , . . . , mn i. We will prove the claim by induction on n. If
n = 1, then M ∼
= Z/(a) for some a ∈ Z. If a = 0 then M ∼
= Z which is not
Artinian. Otherwise a 6= 0 when M is finite. As quotients and submodules of
Artinian modules are Artinian, in the general case N = hm1 , . . . , mn−1 i is finite by
induction, and M/N is generated by one element, so it is also finite. So M is finite,
too. The other implication holds trivially. (6 points)
(e) It is true that a finitely generated R-module is Artinian, as every finitely
generated R-module is a quotient of R⊕n for some natural number n, which is the
Date: November 29, 2016.
1
2
AMBRUS PÁL
direct product of Artinian R-modules, and we saw in the lectures that Artinian Rmodules are closed under forming quotients and finite direct products. (4 points)
2. (a) Set S = R − p. Then S is a multiplicatively closed set, i.e. we have 0 6∈
S, 1 ∈ S, and if x, y ∈ S then xy ∈ S. (2 points)
Define a relation ∼ on the set R × S by (a, s) ∼ (b, t) if and only if there is a
u ∈ S such that u(at − bs) = 0. It is easy to see that ∼ is an equivalence relation.
Denote the equivalence class of (a, s) under ∼ by a/s = as . Let us denote the set
of all equivalence classes under ∼ by Rp . Define binary operations of addition and
multiplication on Rp as follows:
at + bs
a b
+ =
,
s
t
st
a b
ab
· = .
s t
st
It is easy to verify that these operations are well-defined and make Rp into a ring
with zero 0/1 and unity 1/1. The module Mp is defined similarly. (7 points)
(b) We have
a
Z(p) = { | a, b ∈ Z, p 6 |b}.
b
Note that every element of Q/Z(p) is torsion as b · ab ∈ Z(p) for every ab ∈ Q. As
Q/Z(p) is a Z(p) -module every element is actually p-power torsion. Consequently
the pn -torsion of Q/Z(p) is isomorphic to Z/pn Z. So Q/Z(p) is Artinian, but not
Noetherian. (5 points)
(c) First we prove (i) ⇒ (ii). Let p /p R be proper, and let as ∈ Rp and m
r ∈ Mp
=
0
in
M
.
Then
there
is
a
t
∈
R
−
p
such
that
tam
=
0. By
be such that as · m
p
r
assumption either ta = 0, which implies that as = ta
=
0,
or
m
=
0,
which
implies
ts
m
r = 0. (3 points)
The implication (ii) ⇒ (iii) is trivial, since every maximal ideal is prime. (1
point)
Finally we prove (iii) ⇒ (i). Let a ∈ R and m ∈ M be such that am = 0 in
M . Then for every m / R we have a1 · m
1 = 0 in Mm . If for any such m we have
a
=
0
in
R
then
there
is
a
t
∈
R
−
m
such
that ta = 0. Since t 6= 0 and R is an
p
1
integral domain we get that a = 0. Hence we may assume that for every m / R we
have m
1 = 0 in Mm . Therefore for every such m there is a tm ∈ R − m such that
tm m = 0, or in other words tm ∈ AnnR (m). We get that the ideal AnnR (m) is not
contained in any proper ideal, and hence AnnR (m) = R. Therefore 1 ∈ AnnR (m),
so m = 0. (7 points)
3. (a) The nilradical N (R) of R is the set of all nilpontent elements of R. The
Jacobson radical J(R) of R is the intersection of all proper maximal ideals of R.
(2 points)
(b) We are going to construct a sequence b0 , b1 , . . . , bm , . . . ∈ R by induction on
m such that
m
m
X
X
an tn ·
bn tn ∈ 1 + htm+1 i.
n=0
P∞
n=0
P∞
Then n=0 bn t will be the inverse of n=0 an tn . If m = 0 then choose b0 = a−1
0 .
If b0 , b1 , . . . , bm−1 has been already constructed, then set
n
bm = −a−1
0 ·
m
X
i=1
an bm−n . (5 points)
SOLUTIONS TO THE SECOND TEST FOR COMMUTATIVE ALGEBRA M4P55
3
(c) Note that
J (R[[t]])
= {a =
= {a =
= {a =
∞
X
n=0
∞
X
n=0
∞
X
an tn ∈ R[[t]] | 1 − ax ∈ R[[t]]∗ (∀x ∈ R[[t]])}
an tn ∈ R[[t]] | 1 − a0 x0 ∈ R∗ (∀x0 ∈ R)}
an tn ∈ R[[t]] | a0 ∈ J (R)},
n=0
using the characterisation of the elements of the Jacobson radical from the lectures
and the previous part. (5 points)
(d) By passing to R/N (R) we may assume that N (R)
= 0. In this case we need
P∞
to show that N (R[[t]]) = 0, too. Assume now that ( n=0 an tn )m = 0. We will
show that an = 0 by induction on n. At the n-the step:
∞
∞
X
X
k m
mn
(
ak t ) = (
ak tk )m = am
+ higher order terms,
nt
k=0
k=n
so an = 0. The converse is not true. If
R = Z[x1 , x2 , . . . , xn , . . .]/hxnn | n ∈ Ni,
then
∞
X
xn t(2n)!
n=1
is a counterexample. (7 points)
(e) The only maximal ideal of R[[t]] is
∞
X
m = hti = {
an tn ∈ R[[t]] | a0 = 0}.
n=0
Indeed the quotient ring R[[t]]/m is R, the isomorphism is given by the map:
∞
X
an tn → a0 .
n=0
Therefore m is a maximal ideal. On the other hand every element of R − m is
invertible by part (b), and hence m is the only maximal ideal. (5 points)