### SOLUTIONS TO THE FIRST PROBLEM SHEET FOR

```SOLUTIONS TO THE FIRST PROBLEM SHEET FOR
COMMUTATIVE ALGEBRA M4P55
´
AMBRUS PAL
1. The answer to the first question is yes. There is a commutative diagram:
R1


y
R1
I∩R1
−−−−→ R2


y
−−−−→
R2
I
where the top horizontal map is the inclusion and the vertical maps are the quotient
R1
maps. Moreover the bottom horizontal map is injective. In particular I∩R
is
1
R1
isomorphic to a subring of R2 /I. Because R2 /I is a domain so is I∩R1 . So I ∩ R1
is a prime ideal of R1 .
The answer to the second question is no. For example (x) is maximal in C[x]
but Z ∩ (x) = {0} is not a maximal ideal in Z ≤ C[x].
2. (a) Since IJ ⊆ I, we get I ⊆ (I : J).
(b) For every x ∈ (I : J) we have xJ ⊆ I, so (I : J)J ⊆ I.
(c) Now ((I : J) : K)JK = (((I : J) : K)K)J ⊆ (I : J)J ⊆ I, so ((I : J) : K) ⊆
(I : JK). As ((I : JK)K)J = (I : JK)JK ⊆ I, we get ((I : JK)K) ⊆ (I : J). But
the latter implies that (I : JK) ⊆ ((I : J) : K), and hence ((I : J) : K) = (I : JK).
A similar computation
shows
that ((I : K) : J) = (I : JK). T
T
T
T
(d) T
Since ( i Ii : J)J ⊆ i Ii ⊆ Ii for every i, we get that
⊆ i (Ii : J).
T ( i Ii : J)T
As i (Ii : J)J
P ⊆ (Ii : J)J ⊆ Ii for every i, we getPthat i (Ii : J) ⊆ ( i Ii : J).
(e) As Ji ⊆ P
i Ji for each
T index i, we have (I : i Ji ) ⊆ (I : Ji ) of every such i,
and hence (I : i Ji ) ⊆ i (I : Ji ). On the other hand
X
X
X\
X
(I :
Ji )(
Ji ) ⊆
( (I : Ji )Ji ) ⊆
I ⊆ I,
i
T
i
i
i
P
i
so i (I : Ji ) ⊆ (I : i Ji ) holds, too.
3. (a) If x ∈ r(r(I)) then xn ∈ r(I) for some n ∈ N. Hence xnm = (xn )m ∈ I for
some m ∈ N.
(b) Since IJ ⊂ I ∩ J we have r(IJ) ⊆ r(I ∩ J) ⊆ r(I) ∩ r(J). If x ∈ r(I) ∩ r(J)
then xn ∈ I and xm ∈ J for some n, m ∈ N. Then xn+m ∈ IJ so x ∈ r(IJ), and
hence R(I) ∩ r(J) ⊆ r(IJ).
(c) We have the following equivalences: r(I) = R ⇔ 1 ∈ r(I) ⇔ 1 = 1n ∈ I for
some n ∈ N ⇔ 1 ∈ I ⇔ I = R.
(d) By definition I ⊆ r(I n ). Since r(I n ) ⊆ r(I) we only need to show that r(I) ⊆ I.
If x ∈ r(I) then xn ∈ I for some n ∈ N. Hence x ∈ I since I is prime.
Date: January 31, 2014.
1
´
AMBRUS PAL
2
(e) Let a ∈ r(I) and b ∈ r(J) be such that a + b = 1. Then an ∈ I and bm ∈ J for
some n, m ∈ N. Since in every term in the sum on the right:
n+m
X n + m
1 = (a + b)n+m =
ai bn+m−i
i
i=0
either a has exponent at least n, or b has exponent at least m, the expression is in
I + J. Hence I + J = R.
(f ) By Krull’s theorem applied to the ring R/I we have
\
p.
r(I) =
p/p R
I⊆p
If I = r(I) then
I=
\
p
p/p R
I⊆p
and hence I is the intersection of prime ideals. On the other hand if I is the
intersection of prime ideals then
\
I=
p
p/p R
I⊆p
since the right hand side always contains I. Therefore I = r(I) in this case.
4. Assume that the claim is not true and let x ∈ A[x] be an element of the Jacobson
radical which is not nilpotent. By Krull’s theorem there is a prime ideal p / A[x]
which does not contain x. By the first problem q = p ∩ A is a prime ideal of A.
Let q0 be the ideal generated by q in A[x]. It is contained by p, so it is does not
contain x. The quotient of A[x] by q0 is A/q[x], which is an integral domain, as
A/q is. The image of x under the quotient map A[x] → Aq[x] is non-zero, and it is
still an element of the Jacobson radical. Therefore we reduced to the case when A
is an integral domain.
Note that A[x]∗ = A∗ in this case. Indeed the leading term of the product of a
non-constant polynomial multiplied with any non-zero polynomial is the product of
the leading terms, as A has no zero divisors, and so it will has positive degree. This
implies that 1 − xf is not invertible for any non-zero f ∈ A[x]. So the Jacobson
radical of A[x] is zero according to the last problem, which is a contradiction.
5. Let φα : R → C be the homomorphism given by the rule p(x1 , . . . , xn ) 7→
p(α1 , . . . , αn ). This is clearly surjective, and hence its kernel is a maximal ideal. It
will be enough to show that mα = Ker(φα ). Since xi − αi ∈ Ker(φα ) we get that
mα ⊆ Ker(φα ). On the other hand let p(x1 , . . . , xn ) ∈ Ker(φα ). Write p(x1 , . . . , xn )
as a polynomial in x1 − α1 , . . . , xn − αn :
X
p(x1 , . . . , xn ) =
ce1 e2 ···en (x1 − a1 )e1 · · · (xn − αn )en (ce1 e2 ···en ∈ C).
e1 ,e2 ,...,en ∈N
The constant term of this expansion vanishes by assumption. Therefore every term
is an element of mα , and hence mα ⊆ Ker(φα ).
In order to show that J(R) = 0 it will be enough to note that it follows immediately from the previous problem, as R is an integral domain.
6. Let a ∈ J(R) and let x ∈ R be arbitrary. Assume that 1 − ax is not a unit.
Then the ideal (1 − ax) / R is proper, so there is a maximal ideal m / R which
SOLUTIONS TO THE FIRST PROBLEM SHEET FOR COMMUTATIVE ALGEBRA M4P55 3
contains it. By the definition of J(R) we have a ∈ m. Therefore ax ∈ m and hence
1 = (1 − ax) + ax ∈ m, too. But this is a contradiction.
On the other hand let a ∈ R be such that 1 − ax is a unit for every x ∈ R.
Assume that a 6∈ J(R). Then there is a maximal ideal m / R such that a 6∈ m by the
definition of the Jacobson radical. In this case the ideal (a) + m is strictly larger
than m, and hence it is R, since m is maximal. Therefore 1 = ax + m for some
x ∈ R and m ∈ m. So 1 − ax = m ∈ m, which means that 1 − ax is not invertible,