18.705 HOMEWORK 2 – SOLUTION
Due Friday, September 20 in class
Ex. 1. (commutatively modified [FC, 1.7]) Let B1 , . . . , Bn be ideals in a ring R.
Show that R = B1 ⊕ · · · ⊕ Bn iff there exist idempotents e1 , . . . , en with sum 1 such
that ei ej = 0 whenever i 6= j, and Bi = Rei for all i. If R = B1 ⊕ · · · ⊕ Bn , then
each Bi is a ring with identity ei , and we have an isomorphism between R and the
direct product of rings B1 × · · · × Bn .
Proof.
PnSuppose that R = B1 ⊕ · · · ⊕ Bn . Then there exist ei ∈ Bi such that
1 = i=1 ei . If r ∈ Bi , then rej = 0 for j 6= i since rej ∈ Bi ∩ Bj = 0. Thus in
particular ei ej = 0 for i 6= j. Also,
r =r·1=r·
n
X
ej = rei ,
(1)
j=1
since rej = 0 for j 6= i. So r ∈ Rei and thus Bi = Rei . Taking r = ei gives ei = e2i .
Thus the ei have the desired properties.
Conversely, suppose theP
ei exist with the properties
stated in the second P
sentence
P
of
the
exercise.
Since
1
=
e
,
we
have
r
=
re
for
any
r
∈
R,
so
R
=
Rei =
i
i
P
P
P
Bi . If r ∈ Bi ∩ ( j6=i Bj ), then r = ri ei = j6=i rj ej for some rk ∈ R. Then
P
rei = ri e2i = ri ei = r, but then r = rei = ( j6=i rj ej )ei = 0. So R = ⊕Bi .
Now for the last sentence of the exercise. Since Bi is an ideal, Bi is an abelian
group and closed under multiplication. All ring axioms, except for the existence of
a unit element, are inherited from R. Now if r ∈
QBi , then r = rei , as seen in (1), so
ei is a unit element for Bi . The map ϕ : R → Bi given by ϕ(s) = (se1 , . . . , sen )
is an abelian group isomorphism Q
since R = ⊕Bi as an R-module. Since ϕ(1) =
(e1 , . . . , en ) is the unit element of Bi and
ϕ(s)ϕ(t) = (se1 , . . . , sen )(te1 , . . . , ten ) = (ste21 , . . . , ste2n ) = (ste1 , . . . , sten ) = ϕ(st),
we see that ϕ is a ring isomorphism.
Ex. 2. (commutatively modified [FC, 1.8]) Let R = B1 ⊕ · · · ⊕ Bn , where the Bi ’s
are ideals of R. Show that any ideal I of R has the form I = I1 ⊕ · · · ⊕ In where,
for each i, Ii is an ideal of the ring Bi . Describe the prime ideals of R.
Proof.
ei as in the previous exercise. Let Ii = Iei . Then since
P Let Bi = Rei , P
1 = ei , we have I = Ii . Since each Ii ⊂ Bi , the sum is direct, that is, I = ⊕Ii .
Each Ii is an ideal in Bi since Ii is an abelian group and Bi Ii ⊂ RIi = RIei =
Iei = Ii .
I claim that a prime ideal of R is of the form
P = B1 ⊕ · · · ⊕ Bj−1 ⊕ Pj ⊕ Bj+1 ⊕ · · · ⊕ Bn ,
where Pj is a prime ideal of Bj . That is, each Pi = P ei = Bi except P
for exactly
one
P
which
is
prime
in
B
.
First,
suppose
P
has
such
a
form
and
a
=
ai ei , b =
j
j
P
bi ei ∈ R with ab ∈ P . Then ai bi e2i ∈ Pi for all i. Since Pj is prime, without
loss of generality, aj ej ∈ Pj . Since Pi = Bi for i 6= j, we have ai ei ∈ Pi for all i,
1
2
18.705 HOMEWORK 2 – SOLUTION
hence a ∈ P . Now suppose P is a (proper) ideal not of that form. There are two
non-exclusive cases. Case 1: some Pj is not a prime ideal of Bj and Pj 6= Bj . Then
there exists aj , bj ∈ Bj \ Pj such that aj bj ∈ Pj ⊂ P . But then aj , bj ∈ R \ P , so
P is not prime. Case 2: there exist Pi , Pj , i 6= j such that Pi ( Bi and Pj ( Bj .
Then ei ej = 0 ∈ P , but ei , ej 6∈ P , so P is not prime.
Ex. 3. [R, 1.4] Two ideals I and J are strongly coprime if I + J = A. Check that
this is the usual notion of coprime for A = Z or k[x] (k a field). (Comment: Recall
that two elements in a UFD are said to be “strongly coprime” or “relatively prime”
if they share no common factors.) Prove that if I and J are strongly coprime then
IJ = I ∩ J and A/IJ ∼
= (A/I) × (A/J).
Prove also that if I and J are strongly coprime then so are I n and J n for n ≥ 1.
Proof. Let A = Z or A = k[x]. Then A is a PID, so if I, J are ideals, then
I = (f ), J = (g), some f, g ∈ A. Then I + J = A if and only if 1 = af + bg for some
a, b ∈ A. But 1 = af + bg if and only if the gcd(f, g) = 1 since gcd(f, g) divides
af + bg.
Now suppose I + J = A, then
IJ ⊂ I ∩ J = (I ∩ J)(I + J) = (I ∩ J)I + (I ∩ J)J ⊂ IJ
so IJ = I ∩ J.
Since IJ ⊂ I, J, there is a well-defined ring homomorphism f : A/IJ → A/I ×
A/J given by a + IJ 7→ (a + I, a + J). If (a + I, a + J) = 0, then a ∈ I ∩ J = IJ,
so f is injective. If (a + I, b + J) ∈ A/I × A/J, then since I + J = A, we have
a = aI + aJ with aI ∈ I, aJ ∈ J and similarly for b = bI + bJ . Then f (aJ + bI ) =
(aJ + bI + I, aJ + bI + J) = (a + I, b + J). So f is an isomorphism.
Finally, let n be given and suppose I n + J n ( A. Then there exists a maximal
ideal M such that I n + J n ⊂ M . But then I n , J n ⊂ M , so I, J ⊂ M since M is
prime. But then I + J = A ⊂ M , which is absurd.
Ex. 4. (∗) [R, 1.13] If A is a reduced ring and has finitely many minimal prime
ideals Pi , then A ,→ ⊕ni=1 A/Pi ; moreover, the image has nonzero intersection with
each summand.
Proof. Let f be the map given above. If f (a) = 0, then a ∈ Pi for all i, so
a ∈ ∩Pi = 0. So f is injective. Since all the Pi are minimal, for each i we
have Pi 6⊃ ∩j6=i Pj (otherwise Pi ⊃ Pj , some j by [AM, Prop. 1.11]). So choose
ai ∈ ∩j6=i Pj \ Pi . Then f (ai ) is non-zero in the A/Pi component and zero in all the
other components. So f (A) ∩ A/Pi 6= 0, as desired.
Ex. 5. DSK4. (Comment: Let f : X → Y be a function between two topological
spaces. One of the equivalent conditions for f to be continuous is “if V is any
closed subset of Y , then f −1 (V ) is a closed subset of X.”)
Let ϕ : A → B be a ring homomorphism. By HW 1, Exer. 2, we can define a
map ϕ∗ : Spec B → Spec A given by ϕ∗ (P ) = ϕ−1 (P ) for P ∈ Spec B. Show that
ϕ∗ is a continuous function. (Comment: so the Zariski topology is defined in such
a way that “algebraically defined” maps are continuous.)
Proof. Let V be closed in Spec A. So V = V(I) for some ideal I ⊂ A. Let f = ϕ∗ .
I claim that f −1 (V ) = V(ϕ(I)B). This is because P ∈ f −1 (V ) if and only if
there exists Q ∈ V such that f (P ) = ϕ−1 (P ) = Q. This happens if and only if
18.705 HOMEWORK 2 – SOLUTION
3
ϕ−1 (P ) ⊃ I if and only if P ⊃ ϕ(I) (since P ⊃ ϕ(ϕ−1 (P )) and ϕ−1 (ϕ(I)) ⊃ I).
And since P is an ideal, P ⊃ ϕ(I) if and only if P ⊃ ϕ(I)B.
Thus if V is closed, then f −1 (V ) is closed. So f is continuous.
Ex. 6. [AM, 1.18(iv)] (Comment: If X is a topological space, then U ⊂ X is open
if and only if X \ U is closed. A neighborhood of x ∈ X is an open U such that
x ∈ U ⊂ X.)
DSK5. Show that Spec A is T1 if and only if every prime ideal of A is maximal.
(Comment: T1 is defined in [M], but I couldn’t find T0 .)
Proof. For [AM, 1.18(iv)], note that a space X is T0 if and only if for all x, y ∈ X,
x 6= y, there exists closed V such that (x ∈ V and y 6∈ V ) or (y ∈ V and x 6∈ V ).
This is because U is open if and only if V = X \ U is closed.
So choose x, y ∈ Spec A, x 6= y. Without loss of generality, x 6⊂ y. Thus x ∈ V (x)
but y 6∈ V (x). Thus Spec A is T0 .
For DSK5, note that a space X is T1 if and only if each point x is a closed point.
To see this, if X is T1 and x ∈ X, then for each y 6= x, there exists a neighborhood
Uy 3 y with x 6∈ Uy . So {x} = X \ ∪y6=x Uy is closed. Conversely, if each point
x ∈ X is closed, and x 6= y, then X \ {y} is a neighborhood of x which does not
contain y. In the case X = Spec A, In Class Proposition 1.3.9 says that the closure
{x} = V(x) for any x ∈ X. Since every prime ideal is contained in a maximal ideal,
V(x) = {x} if and only if x is maximal. Thus X is T1 if and only if every prime
ideal of A is maximal.
Ex. 7. [AM, 1.7]
Proof. Let P be a prime ideal of A and let x 6= 0 ∈ A/P . Since x is the image
of some x ∈ A, we have for some n, x(xn−1 − 1) = 0. Since A/P is a domain,
xn−1 = 1, so x is invertible. Hence A/P is a field and P is maximal.
Ex. 8. [AM, 2.9]
f
g
Proof. Let 0 → M 0 → M → M 00 → 0 be the given exact sequence, let x1 , . . . , xn
generate M 0 , and let y1 , . . . , yk generate M 00 . Choose zi ∈ g −1 (yi ) for each i.
We claim that {f (x1 ), . . . , f (xn ), z1 , . . . , zk } generates M . To see this, let m ∈
Pk
Pk
M . Then g(m) =
ai yi for some ai ∈ A. So g(m − i=1 ai zi ) = 0, so
i=1
Pk
Pk
Pn
m − i=1 ai zi ∈ Ker g = Im f , and hence m = i=1 ai zi + i=1 bi f (xi ) for some
bi ∈ A. So we have proved our claim.
References
AM. M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison-Wesley
Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969. MR 39 #4129
FC. T. Y. Lam, A first course in noncommutative rings, Springer-Verlag, New York, 2001. MR
2002c:16001
M. James R. Munkres, Topology: a first course, 2nd edition. MR 57 #4063
R.
Miles Reid, Undergraduate commutative algebra, Cambridge University Press, Cambridge,
1995. MR 98c:13001
E-mail address: [email protected]
URL: http://www.mit.edu/~dskeeler/705/