Download Solutions for the Suggested Problems 1. Suppose that R and S are

Solutions for the Suggested Problems
1. Suppose that R and S are rings with identity and that ϕ : R → S is a surjective ring
homomorphism. Let 1R and 1S denote the multiplicative identity elements of R and S,
respectively. Prove that ϕ(1R ) = 1S .
Solution. Let e = ϕ(1R ). Thus, e ∈ S. Furthermore, suppose that s ∈ S. Since ϕ is
surjective, there exists an element r ∈ R such that ϕ(r) = s. Note that
es = ϕ(1R )ϕ(r) = ϕ(1R r) = ϕ(r) = s and se = ϕ(r)ϕ(1R ) = ϕ(r1R ) = ϕ(r) = s .
Therefore, e ∈ S and es = s = se for all s ∈ S. Thus, e is a multiplicative identity element
for the ring S. However, so is 1S . We proved in class that the multiplicative identity element
of a ring (if it exists) is unique. Therefore, we have e = 1S . That is, we have ϕ(1R ) = 1S .
2. Suppose that R is a ring and that I is an ideal of R. Suppose also that R/I ∼
= Z, where
Z is the ring of integers. Suppose that J is an ideal of R, that I ⊆ J, and that I 6= J. Prove
that R/J is a finite ring.
Solution. As pointed out in class one day, there exists a surjective ring homomorphism
ϕ : R → R/I. By assumption, there exists a ring isomorphism ψ : R/I → Z. Let χ = ψ ◦ ϕ,
which is a surjective ring homomorphism from R to Z. Furthermore, Ker(χ) = Ker(ϕ)
because ψ is an isomorphism. Thus, Ker(χ) = I.
We will consider the surjective homomorphism χ : R → Z and use the correspondence
theorem. We are assuming that J is an ideal of R containing I and that J 6= I. The
corresponding ideal in Z is χ(J). Denote χ(J) by K. Since J 6= I, K = χ(J) is not the zero
ideal of the ring Z. Thus, J = mZ for some positive integer m.
Now we use proposition 4 on the handout about homomorphisms and ideals. With the
above notation, it follows that
R/J ∼
= Z/K .
Now K = mZ for some positive integer m and hence Z/K = Z/mZ. This ring has finitely
many elements. To be precise, it has m elements. The ring R/J is isomorphic to Z/K and
therefore also has a finite number of elements. Thus, R/J is indeed a finite ring.
3. Suppose that R and S are rings and that ϕ : R → S is a ring homomorphism. Suppose
that e is an idempotent of R. Prove that ϕ(e) is an idempotent of S. By making use of this
result, determine all possible ring homomorphisms from Z to Q. Determine all possible ring
homomorphisms from Z to Z/10Z.
Solution. Let s = ϕ(e), which is an element in S. Since e is an idempotent of R, we have
ee = e. Thus, we have
ss = ϕ(e)ϕ(e) = ϕ(ee) = ϕ(e) = s .
This proves that ss = s and hence that s is an idempotent in the ring S.
Now suppose that R = Z and that ϕ : Z → S is a ring homomorphism. Note that 1 is an
idempotent in Z. Hence ϕ(1) must be an idemptotent in S, using the result proved above.
Let s = ϕ(1). Thus, s ∈ S and ss = s.
Now ϕ is a group homomorphism from the underlying additive group of Z to the underlying additive group of S. The additive group of Z is a cyclic group and 1 is a generator of that
group. Thus a group homomorphism ϕ from Z to another group is completely determined
if one knows ϕ(1). Thus, the ring homomorphism ϕ is completely determined if one knows
s = ϕ(1).
To complete the description of the ring homomorphisms ϕ from Z to S, we use the
following lemma. It shows that for any idempotent s, there is indeed a ring homomorphism
ϕ : Z → S such that ϕ(1) = s.
Lemma: Suppose that s is any idempotent in S. Define a map ϕ : Z → S by
ϕ(n) = ns
for all n ∈ Z. The map ϕ is a ring homomorphism from Z to S. Also, we have ϕ(1) = s.
Note that ns does not refer to ring multiplication. The element ns of S is defined in
terms of the underlying additive group of S. (Recall that if G is a group, g ∈ G, and n ∈ Z,
then one can define g n . But if the group operation is written additively, one would write ng
instead of g n .)
We will give most of the proof of the above lemma. The fact that ϕ(1) = s is clear.
Suppose that n, m ∈ Z. Then
ϕ(n + m) = (n + m)s = = ns + ms = ϕ(n) + ϕ(m) .
We have used the ”law of exponents” from group theory. To prove that ϕ(nm) = ϕ(n)ϕ(m),
we must use the distributive law many times. If n and m are positive, then
ϕ(n)ϕ(m) = (ns)(ms) = (s + s + s... + s)(s + s + s...s) ,
where the first parenthesis contains n terms, the second contains m terms. Using the distributive law, we obtain ss + ss + ... + ss, with nm terms. Thus,
ϕ(n)ϕ(m) = (nm)(ss) = (nm)s = ϕ(nm)
as we wanted to show. Note that we are using the fact that s is an idempotent of S when
we replace ss by s.
One must also consider the cases where n and m are not necessarily positive. We won’t
do every case. Suppose that n is positive and m is negative. Then, by the definitions in
group theory, ns is as before and ms = |m|(−s) = −|m|s. We will use the above calculation,
but applied to the positive integers n and |m|. Thus,
ϕ(n)ϕ(m) = (ns)(ms) = ns − |m|s = − (ns)(|m|s)
= − (n|m|)(ss) = (−n|m|)(ss) = (nm)s = ϕ(nm) .
We have used the elementary fact that a(−b) = −(ab) for a, b ∈ S in the third equality. We
have used the law of exponents from group theory for the fifth equality.
The above lemma is helpful in this problem. Consider first S = Q. Now Q is an integral
domain and hence the only idempotents in Q are 0 and 1. This was proved in the solution
to problem F from problem set 2. Thus either ϕ(1) = 0 or ϕ(1) = 1. The above discussion
and the lemma show that there are exactly two ring homomorphisms from Z to Q. However,
we can exhibit them explicitly and so we don’t have to use the lemma. One homorphism is
simply the inclusion map ϕ : Z → Q defined by ϕ(n) = n for all n ∈ Z. This map is clearly
a ring homomorphism. It has the property that ϕ(1) = 1.
Another ring homomorphism from Z to Q is defined by ϕ(n) = 0 for all n ∈ Z. This map
ϕ is clearly a ring homomorphism. We have ϕ(1) = 0.
Now we come to a description of all ring homomorphisms ϕ from Z to Z/10Z. As
above, such a ring homomorphism is determined if one knows ϕ(1). Also, ϕ(1) must be an
idempotent in the ring Z/10Z. Furthermore, by the lemma, each idempotent in Z/10Z will
correspond to a homomorphism.
From problem F in assignment 2, the ring Z/10Z has exactly four idempotents, namely
[0]10 , [1]10 , [5]10 , and [6]10 . By the lemma, we get the following four ring homomorphisms
from Z to Z/10Z (which we denoted by ϕi for 1 ≤ i ≤ 4):
ϕ1 (n) = [0]10 ,
for all n ∈ Z.
ϕ2 (n) = [n]10 ,
ϕ3 (n) = [5n]10 ,
ϕ4 (n) = [6n]10
,
4. Let R = M2 (R). Consider the following subset of R:
a b a, b ∈ R
S =
b a Show that S is a subring of R. Furthermore, show that S is a commutative ring with
identity, but not an integral domain. In addition, find all the idempotents and all the
nilpotent elements in S.
0 1
1 0
. We can describe the set S as follows:
and E =
Solution. Let I =
1 0
0 1
.
S =
aI + bE a, b ∈ R
We will use standard properties of matrix algebra. Notice that I is the identity matrix and
so II = I and IE = EI = E. Notice also that EE = I.
First of all, S is a subgroup of R under the operation of addition. For if a, b, c, and d are
in R, then
(aI + bE) + (cI + dE) = (a + c)I + (b + d)E ,
which is in the set S. Also, the zero matrix is in S and, for all a, b ∈ R, the additive inverse
of aI + bE is (−a)I + (−b)E, which is in S.
To show that S is closed under the multiplication operation of R, suppose again that
a, b, c, d ∈ R. We have
(aI+bE)(cI+dE) = acI+bcEI+adIE+bdEE = acI+bcE+adE+bdI = (ac+bd)I+(ad+bc)E ,
which is indeed in the set S. Therefore, we have proved that S is a subring of R.
The above multiplication rule shows that
(cI +dE)(aI +bE) = (ca+db)I +(da+cb)E = (ac+bd)I +(ad+bc)E = (aI +bE)(cI +dE) ,
and so S is a commutative ring. It has an identity element, namely I = 1I + 0E.
The idempotent elements of S. Suppose s ∈ S. Then s = aI + bE, where a, b ∈ R. We
have
ss = (aI + bE)(aI + bE) = (a2 + b2 )I + (2ab)E ,
which is equal to s if and only if a2 + b2 = a and 2ab = b. If b = 0, then the second equation
is satisfied and the first equation becomes a2 = a, which means that a = 0 or a = 1. On the
other hand, if b 6= 0, then the above equations mean that a = 1/2 and (1/2)2 + b2 = 1/2.
This gives two possibilities for b, namely b = 1/2 or b = −1/2. Altogether, we find four
idempotents in the ring S, corresponding to (a, b) = (0, 0), (1, 0), (1/2, 1/2), and (1/2, −1/2).
Thus, the four idempotents in S are
1/2 −1/2
1/2 1/2
1 0
0 0
.
,
,
,
−1/2 1/2
1/2 1/2
0 1
0 0
The nilpotent elements in S. Suppose that s = aI + bE, where a, b ∈ R.As above, we
have ss = (a2 + b2 )I + (2ab)E. This is the zero matrix if and only if a2 + b2 = 0 and 2ab = 0.
The first equation implies that a = b = 0. The second equation is then satisfied too. Thus,
we find exactly one nilpotent element in S, namely
0 0
.
0 0
5. Consider the ring R = Z/256Z. Prove that every element of R is either a unit of R or a
nilpotent element. Determine all maximal ideals of R.
Solution. In general, if m ≥ 1, we know that the units in Z/mZ are the elements a + mZ,
where a ∈ Z and gcd(a, m) = 1. Consider m = 256 = 28 . If a ∈ Z, then gcd(a, 256) = 1
if and only if a is odd. Thus, the units in Z/256Z are the elements of the form a + 256Z,
where a is odd.
Now suppose that r ∈ R and that r is not a unit of R. Then r = a + 256Z, where a is
not odd. Thus, a = 2k, where k ∈ Z. Notice that
r8 = (2k + 256Z)8 = (2k)8 + 256Z = 256k 8 + 256Z = 0 + 256Z ,
which is the additive identity 0R in the ring R. Hence r is indeed a nilpotent element in R.
Finally, suppose that I is an ideal of R and that I 6= R. Then I cannot contain any units
of R. To see this, suppose to the contrary that I contains a unit u of R. Since u ∈ R× ,
there exists an element v ∈ R such that uv = 1R . Since u ∈ I and v ∈ R, we have uv ∈ I.
Hence 1R ∈ I. Hence, for every element r ∈ R, we have r = r1R ∈ I. This contradicts the
assumption that I 6= R.
We have shown that if I is an ideal of R and I 6= R, then I contains no units of R.
Thus, the elements of I must be of the form 2k + 256Z, where k ∈ Z. Thus, they must be
of the form (k + 256Z)(2 + 256Z), and hence must be in the principal ideal of R generated
by 2 + 256Z. Therefore, apart from R, every ideal of R is contained in
M = { 2k + 256Z | k ∈ Z } = (2 + 256Z) ,
the principal ideal of R generated by 2 + 256Z. It follows that this ideal M is a maximal
ideal of R and that no other maximal ideals of R exist.
6. Suppose that R is a ring and that I and J are maximal ideals of R.
TRUE OR FALSE: If R/I ∼
= R/J, then I = J.
If true, give a proof. If false, give a specific counterexample.
Solution. The statement is false. As a counterexample, let R be the ring of continuous
functions from the interval (0, 1) to R. Consider the homomorphisms ϕ and ψ from R to R
defined as follows. For every f ∈ R, define
ϕ(f ) = f (1/2) ,
ψ(f ) = f (1/3) .
Let I = Ker(ϕ) and J = Ker(ψ). Both ϕ and ψ are surjecxtive ring homomorphisms.
Therefore, by the first isomorphism theorem, we have
R/I ∼
= R ,
R/J ∼
= R .
Therefore, R/I ∼
= R/J. But, I 6= J. To see this, let f ∈ R be the function defined by
f (x) = x − (1/2) for all x ∈ (0, 1). Then ϕ(f ) = 0, but ψ(f ) 6= 0. Thus, f ∈ I, but f 6∈ J.
Hence, I 6= J.
7. Consider the polynomial ring R = Z[x]. Let I = (x), the principal ideal of R generated
by x. Show that I is a prime ideal of R, but not a maximal ideal of R.
This problem will be part of problem set 4. We have not had time to talk about polynomial rings such as Z[x] in class.
8. Find all of the maximal ideals in the ring R × R.
Solution. We will use the result from problem E in problem set 3. The ring R = R × R has
exactly four ideals. Two of the ideals of R described in the solution to that problem are:
I = { (a, 0) | a ∈ R } ,
J = { (0, b) | b ∈ R } .
The other two ideals of R are R itself and the zero ideal K = { (0, 0) }. By definition, R
is not a maximal ideal. Since K ⊆ I ⊆ R, I 6= K, and I 6= R, it follows that K is not a
maximal ideal of R. However, the only ideals of R containing I (from among the four ideals
in that ring) are I itself and R. Thus, by definition, I is a maximal ideal of R. For a similar
reason, J is also a maximal ideal of R.
9. Suppose that R and S are fields and that ϕ : R → S is a ring homomorphism.
TRUE OR FALSE: If ϕ is surjective, then ϕ is injective.
If true, give a proof. If false, give a specific counterexample.
Solution. The statement is true. Here is a proof. Let K = Ker(ϕ). Thus, K is an ideal in
the ring R. Since R is a field, it has only two ideals, namely R itself or the zero ideal. Thus,
either K = R or K = { 0R }.
We will show that K 6= R. Suppose to the contrary that K = R. Then, we would have
ϕ(r) = 0S for all r ∈ R. Since S is a field, it contains an element 1S such that 1S 6= 0S .
Hence 1S is not in the image of ϕ. Therefore, ϕ cannot be surjective. This contradicts our
assumption that ϕ is surjective.
It follows that K = { 0R }. This implies that ϕ is indeed injective.
10. Suppose that R is a ring. Suppose that I and J are maximal ideals of R and that I 6= J.
Prove that I + J = R.
Solution. Let K = I +J. Then, as shown in class, K is an ideal in the ring R. Furthermore,
K contains I and K contains J. . Thus, we have I ⊆ K and J ⊂ K.
We will prove that K 6= I. Suppose to the contrary that K = I. Since J ⊆ K, it would
then follow that J ⊆ I. By assumption, I 6= J. Also, I 6= R because I is a maximal ideal of
R. Thus, I is an ideal of R, J ⊆ I ⊆ R, I 6= J, and I 6= R. This contradicts the assumption
that J is a maximal ideal of R.
We have proved that I ⊆ K and K 6= I. Since I is a maximal ideal of R, it follows that
K = R. Therefore, we indeed have I + J = R.