SOLUTIONS TO THE FIFTH PROBLEM SHEET FOR
COMMUTATIVE ALGEBRA M4P55
AMBRUS PÁL
1. Note that
(tn ) = {
X
ak tk ∈ A[[t]]},
k≥n
n
so the map (t ) → A given by
X
ak tk → an
k≥n
furnishes the required isomorphism. We get that A is a quotient of A[[t]], i.e.
A[[t]]/(t) ∼
= A. As the quotients of Noetherian rings are Noetherian, we get that A
is Noetherian if A[[t]] is Noetherian.
2. Since I is an ideal t · I ⊆ I, so t · In ⊆ In+1 . Hence I(n) ⊆ I(n + 1). The last
claim now follows from the definition of Noetherian rings.
3. Write Sk = {xk1 , xk2 , . . . , xkrk }. If
X
am tm ∈ Ik
m≥k
then ak ∈ I(k), so there are ck1 , ck2 , . . . , ckrk ∈ A such that
ak ≡ ck1 xk1 + ck2 xk2 + · · · + ckrk xkrk
mod Ik+1
by the defining property of Sk , so
X
am tm − (ck1 xk1 + ck2 xk2 + · · · + ckrk xkrk ) ∈ Ik+1 .
m≥k
4. Note that the image of tm Sn ⊆ In+m under the isomorphism (tn+m )/(tn+m+1 ) ∼
=
A generates I(n + m) for every natural number m. So we may repeat the argument
to get that In+m = (tm Sn ) + In+m+1 for each such m. Now let a ∈ In be arbitrary.
Via recursion on m we can construct by the above elements
cji ∈ A,
(i = 0, . . . , m, j = 1, . . . , rn )
such that
a−(
m
X
ti (c1i xn1 + c2i xn2 + · · · + crn i xnrn )) ∈ In+m+1 .
i=0
Clearly
∞
∞
∞
X
X
X
i
i
a=(
t c1i )xn1 + (
t c2i )xn2 + · · · + (
ti crn i )xnrn .
i=0
i=0
i=0
Date: January 12, 2017.
1
2
AMBRUS PÁL
Now we can prove via induction on k (for k ≤ n) that
In−k = (
k
[
Sn−j ).
j=0
The case k = n is the last claim.
5. We will use the notation in the proof of Theorem 14.3 in the notes. From that
proof it is clear that l(Mn ) is the coefficient of tn in f (t) · (1 − t)−d for some
f (t) =
N
X
ak tk ∈ Z[t]
k=0
and d ∈ N. Since
−d
(1 − t)
∞ X
d+k−1 k
=
t ,
d−1
k=0
we get
l(Mn ) =
N
X
ak
k=0
d+n+k−1
d−1
for all n ≥ N . The sum on the right hand side is a polynomial in n.
6. The map A → S −1 A given by a 7→ a1 maps mn to S −1 mn = (S −1 m)n . So we
get a map of A-modules A/mn → S −1 A/(S −1 m)n . Therefore it will be sufficient to
prove that every a ∈ S −1 A can be written as a sum 1b + dc with a ∈ A, b ∈ mn and
d ∈ S. We are going to prove this by induction on n. If a = xy with x ∈ A, y ∈ S.
Since m is maximal there is an r ∈ S such that t = 1 − ry ∈ m. Then
x(1 − ry)
xt
x rx
−
=
=
∈ S −1 m.
y
1
y
y
In the induction step write a as
b
c
+
1 d
with b ∈ A, c ∈ mn and d ∈ S. Then there is an r ∈ S such that t = 1 − rd ∈ m.
Then
rc b + rc c(1 − rd)
b + rc ct b + rc
b + rc c
+
−
=
+
=
+
∈
+ S −1 mn+1 .
a=
1
d
1
1
d
1
d
1
7. Consider the chain:
a=
0 ( (x1 ) ( (x1 , x2 ) ( · · · ( (x1 , x2 , . . . , xn ).
It is a chain of prime ideals of length n, so dim(C[x1 , x2 , . . . , xn ]) ≥ n. In order to
show the reverse inequality, it will be enough to show that
d(C[x1 , x2 , . . . , xn ]m ) ≤ n
for every maximal ideal m / C[x1 , x2 , . . . , xn ]. By Nullstellensatz we know that
m = (x1 − c1 , x2 − c2 , . . . , xn − cn ) for some ci ∈ C. By taking a linear change of
b denote the unique maximal ideal
coordinates we may assume that all ci = 0. Let m
of C[x1 , x2 , . . . , xn ]m . By the previous problem
n+m
m+1
m+1
b
l(C[x1 , x2 , . . . , xn ]m /m
) ≤ l(C[x1 , x2 , . . . , xn ]/m
)=
,
m
SOLUTIONS TO THE FIFTH PROBLEM SHEET FOR COMMUTATIVE ALGEBRA M4P55 3
the latter being the dimension of polynomials in n variables of total degree at most
m, so the claim follows.