Math 403 - Solutions for problem set 1
Page 230, problem 3. Suppose that R is a ring with identity 1R , that S is a subring of
R, and that 1R ∈ S. Then 1R r = r1R for all r ∈ R. In particular, 1R s = s1R for all s ∈ S.
Since 1R ∈ S, we can therefore consider 1R as an identity element in S. Thus, S is a ring
with identity 1S , namely 1S = 1R .
Now suppose that u is a unit in S. Therefore, there exists an element v ∈ S such that
uv = 1S . Hence, uv = 1R . Since S ⊆ R, we have u, v ∈ R. The equation uv = 1R therefore
implies that u is a unit in R, which is what we wanted to prove.
To see that the converse is not true in general, consider S = Z and R = Q. Let
u = 2. Then u is a unit in R. But u is not a unit in S since S × = {1, −1}. As another
counterexample, take S = Z and R = Q again. Then s = 1/2 is a unit in R, but is not even
an element in S.
Page 230, problems 5(a). Let R denote the set of rational numbers with odd denominator. Then R ⊂ Q. Also, R is closed under addition. To see this, consider α, β ∈ R. By
definition, we can write α = a/b and β = c/d, where a, b, c, d ∈ Z and both b and d are odd.
Then
α + β = a/b + c/d = (ad + bc)/(bd) .
Now ad + bc and bd are both in Z, although not necessarily relatively prime. However, the
denominator of (ad + bc)/(bd) must divide bd. Since b and d are odd, it follows that bd is also
odd. Every divisor of bd will be odd too. In particular, the denominator of (ad + bc)/(bd)
must be odd. Therefore, α + β ∈ R. Hence R is closed under the addition operation for Q.
Note that the additive identity 0 of Q can be written in reduced form as 0/1 and so
0 ∈ R. Finally, if α ∈ R, then we can write α in reduced form: α = a/b, where a, b ∈ Z,
b > 0, and gcd(a, b) = 1. Since α ∈ R, b is odd. It follows that −α = (−a)/b, which is the
reduced form for α since −a, b ∈ Z, b > 0, and gcd(−a, b) = 1. Therefore, −α ∈ R.
We have proved that R is a subgroup of the underlying additive group of Q. It remains
to prove that R is closed under the multiplication operation in Q. To verify this, suppose
that α, β ∈ R. As before, we write α = a/b and β = c/d, where a, b, c, d ∈ Z and both b and
d are odd. Then
αβ = (ac)/(bd) .
Since b and d are odd, so is bd. Any divisor of bd will also be odd. The denominator of αβ
(when it is written in reduced form) is a divisor of bd and therefore must be odd. It follows
that αβ ∈ R.
We have proved that R is a subring of Q.
Page 230, problem 5 (b). Let R be the set of rational numbers with even denominator.
The additive identity of Q is 0. The reduced form of 0 is 0/1. The denominator is 1, which
is not even. Therefore, 0 6∈ R. Therefore, R is not a subgroup of the underlying additive
group of Q. Hence R is not a subring of Q.
Page 231, problem 7.
Suppose that R is a ring. Let
S = { z ∈ R | zr = rz for all r ∈ R } .
We will prove that S is a subring of R. First of all, we will verify that S is a subgroup of
the underlying additive group of R. For this purpose, suppose that z1 , z2 ∈ S. Then, for all
r ∈ R, we have z1 r = rz1 and z2 r = rz2 . Therefore, using the distributive laws for R, we
have
(z1 + z2 )r = z1 r + z2 r = rz1 + rz2 = r(z1 + z2 )
for all r ∈ R. Therefore, z1 + z2 ∈ S. Furthermore, letting 0 denote the additive identity of
R, we have 0 · r = 0 and r · 0 = 0. Hence 0 · r = r · 0. Therefore, 0 ∈ S.
Finally, suppose that z ∈ S. Let w = −z, the additive inverse of z in R. We have
z + w = 0. Thus, z + w ∈ S. Since z is in S and z + w is in S, it follows that, for all r ∈ R,
we have zr = rz and (z + w)r = r(z + w). Therefore, we have
zr + wr = rz + rw = zr + rw
Thus, we have the equation zr + wr = zr + rw. Applying the cancellation law for the
underlying additive group of R to that equation, it follows that wr = rw for all r ∈ R.
Therefore, w ∈ S. That is, −z ∈ S. This completes the verification that S is a subgroup of
the underlying additive subgroup of the ring R.
To complete the proof that S is a subring of R, we must show that if z1 and z2 are in
S, then so is z1 z2 . So, assume that z1 , z2 ∈ S. Then, for all r ∈ R, we have z1 r = rz1
and z2 r = rz2 . Consider z1 z2 , which is an element of R. Using the associative law for
multiplication in R many times, it follows that
(z1 z2 )r = z1 (z2 r) = z1 (rz2 ) = (z1 r)z2 = (rz1 )z2 = r(z1 z2 )
for all r ∈ R. Therefore, we indeed have z1 z2 ∈ S.
We have shown that S is a subring of R. The subring S of R is often called the “center
of R”.
Now assume that R is a ring with identity. Let S be the center of R, as defined above.
Let 1 denote the multiplicative identity element of R. By definition, 1 · r = r and r · 1 = r
for all r ∈ R. Therefore, 1 · r = r · 1 for all r ∈ R. Therefore, we have 1 ∈ S.
Now assume that R is a division ring. Then, by definition, R is a ring with identity 1,
1 6= 0, and every nonzero element of R is a unit of R. Suppose that S is the center of R.
Then 1 ∈ S and hence S is a ring with identity. Also, 0 is the additive identity of R and is
also the additive identity of the ring S. We have 1 6= 0. We now prove that S is a division
ring. It suffices to prove that S × = S − {0}. Assume that z ∈ S and z 6= 0. Since z ∈ R× ,
there exists an element w ∈ R such that zw = 1 and wz = 1. Since z ∈ S, we have zr = rz
for all r ∈ R. We also have the implications
zr = rz =⇒ w(zr) = w(rz) =⇒ (wz)r = (wr)z =⇒ 1r = (wr)z =⇒ r = (wr)z
=⇒ rw = (wr)z w =⇒ rw = (wr)(zw) =⇒ rw = (wr) · 1 =⇒ rw = wr .
Thus, if we assume that z ∈ S, then wr = rw for all r ∈ R. Therefore, w ∈ S. We have
proved that if z is a nonzero element of S, then there exists an element w ∈ S such that
zw = 1 and wz = 1. Hence S is a division ring.
Finally, if a ∈ S, then ar = ra for all r ∈ R. Since S ⊆ R, we can say that ab = ba for
all b ∈ S. Hence S is a commutative ring. Since S has been proved to be a division ring, it
follows that S is a field. We have proved that if R is a division ring, then the center of R is
a field.
Page 231, problem 8. Let H denote the ring of quaternions. Suppose that a, b, c, d ∈ R
and that α = a + bi + cj + dk is in the center of H. It follows that αβ = βα for all β ∈ H.
We will first take β = i and then we will take β = j. We have
αi = ai + b(−1) + c(−k) + dj = − b + ai + dj + (−c)k,
iα = ai + b(−1) + ck + (−d)j = − b + ai + (−d)j + ck
.
Therefore,
αi = iα ⇐⇒ d = −d and c = −c ⇐⇒ c = d = 0 .
Also,
αj = aj + bk + c(−1) + (−d)i = − c + (−d)i + aj + bk,
jα = aj + (−b)k + c(−1) + di = − c + di + aj + (−b)k
Therefore,
αj = jα
⇐⇒
d = −d and b = −b
⇐⇒
b=d=0 .
.
If α is in the center of H, it follows that αi = iα and αj = jα, and therefore it follows that
b = c = d = 0. Thus, α has the form α = a + 0i + 0j + 0k. In the definition of H, we identify
such a quaternion α with the real number a, and thereby regard R as a subring of H. With
that identification, we have proved that if α is in the center of H, then α ∈ R. Conversely,
if α ∈ R, then α is in the center of H. This is part of the definition of multiplication in H.
Therefore, we have proved that the center of H is the subring R, which is explicitly given
as
R = { a + 0i + 0j + 0k | a ∈ R } .
Now let S = { a + bi + 0j + 0k | a, b ∈ R }. We will prove that S is a subring of H
and that S is isomorphic to C. First of all, note that 0H = 0 + 0i + 0j + 0k is clearly in
S. Suppose that a, b, a′ , b′ ∈ R. Using the definition of addition and multiplication in H, we
have
(a + bi + 0j + 0k) + (a′ + b′ i + 0j + 0k) = (a + a′ ) + (b + b′ )i + 0j + 0k,
(a + bi + 0j + 0k)(a′ + b′ i + 0j + 0k) = (aa′ − bb′ ) + (ab′ + ba′ )i + 0j + 0k .
Both of these elements of H are actually in S. Hence S is closed under the operations of
addition and multiplication for H. Furthermore, the additive inverse of a + bi + 0j + 0k
is (−a) + (−b)i + 0j + 0k, which is clearly in S. It follows that S is a subgroup of the
underlying additive group of H and that S is closed under multiplication. Therefore, S is
indeed a subring of H.
Define a map ϕ : C → S as follows. For all a, b ∈ R, define
ϕ(a + bi) = a + bi + 0j + 0k
.
Suppose that a, b, a′ , b′ ∈ R. Let α = a + bi, α′ = a′ + b′ i.Thus,
α + α′ = (a + a′ ) + (b + b′ )i,
αα′ = (aa′ − bb′ ) + (ab′ + a′ b)i .
Using the above calculations, we see that
ϕ(α+α′ ) = (a+a′ )+(b+b′ )i+0j +0k = (a+bi+0j +0k)+(a′ +b′ i+0j +0k) = ϕ(α)+ϕ(α′ )
and
ϕ(αα′ ) = (aa′ −bb′ )+(ab′ +a′ b)i+0j +0k = (a+bi+0j +0k)(a′ +b′ i+0j +0k) = ϕ(α)ϕ(α′ )
Note also that ϕ is a bijection from C to S. Therefore, ϕ is an isomorphism of the ring C to
the subring S of H.
Finally, if a, b ∈ R and b 6= 0, then a + bi + 0j + 0k is in S, but not in the center of H
(which we determined previously). Therefore, S is not contained in the center of H.
Page 231, problem 17. This problem concerns the direct product R × S of two rings R
and S. As a set, R × S = { (r, s) | r ∈ R, s ∈ S }. We define addition and multiplication
in R × S as follows. If (r, s) and (r′ , s′ ) are in R × S, then we define
(r, s) + (r′ , s′ ) = (r + r′ , s + s′ ) ,
(r, s) · (r′ , s′ ) = (r · r′ , s · s′ ) .
Under addition, R × S is the direct product of the underlying additive groups of R and S.
Thus, R × S is an abelian group under the above defined addition operation. Since r · r′ ∈ R
and s · s′ ∈ S, we do have (r, s) · (r′ , s′ ) ∈ R × S, and so multiplication, as just defined, is
indeed a binary operation on R × S.
We must verify the associative law for multiplication and the distributive laws. Suppose
that (u, v) ∈ R × S. Thus, u ∈ R and v ∈ S. To verify the associative law, note that
(u, v) · (r, s) · (r′ , s′ ) = (u, v) · (r · r′ , s · s′ ) = u · (r · r′ ), v · (s · s′ )
= (u · r) · r′ , (v · s) · s′ = (u · r, v · s) · (r′ , s′ ) = (u, v) · (r, s) · (r′ , s′ ) .
To verify the left distributive law, note that
(u, v) · (r, s) + (r′ , s′ ) = (u, v) · (r + r′ , s + s′ ) = u · (r + r′ ), v · (s + s′ )
= u · r + u · r′ , v · s + v · s′ = (u · r, v · s) + (u · r′ , v · s′ )
= (u, v) · (r, s) + (u, v) · (r′ , s′ ) .
The right associative law is verified in a similar way.
Next we consider commutativity of multiplication. As above, suppose that α and β are
in R × S. We can write α = (r, s) and β = (r′ , s′ ), where r, r′ ∈ R and s, s′ ∈ S. Then
α · β = (r · r′ , s · s′ )
and
β · α = (r′ · r, s′ · s)
Therefore, α · β = β · α if and only if r · r′ = r′ · r and s · s′ = s′ · s.
In particular, if R and S are commutative rings, then r · r′ = r′ · r for all r, r′ ∈ R and
s · s′ = s′ · s for all s ∈ S. Therefore, if R and S are commutative rings, it follows that
α · β = β · α for all α, β ∈ R × S. Therefore, R × S is a commutative ring. Conversely,
assume that R × S is a commutative ring. Consider r, r′ ∈ R and s, s′ ∈ S. Let α = (r, s)
and β = (r′ , s′ ), which are elements in R × S. Since R × S is a commutative ring, we have
α · β = β · α. Therefore, r · r′ = r′ · r. Hence R is a commutative ring. Also, s · s′ = s′ · s and
so S is a commutative ring too.
Finally, we consider the existence of a multiplicative identity element. Assume that R
and S are rings with identity. Let 1R and 1S denote the identity elements of R and S,
respectively. Consider the element (1R , 1S ) in R × S. For all r ∈ R and s ∈ S, we have
(1R , 1S )·(r, s) = (1R ·r, 1S ·s) = (r, s) ,
(r, s)·(1R , 1S ) = (r ·1R , s·1S ) = (r, s)
Therefore, R × S has a multiplicative identity element, namely the element (1R , 1S ).
Conversely, assume that R×S has a multiplicative identity element. Denote that element
by ε. We can write ε = (a, b), where a ∈ R and b ∈ S. Suppose that r ∈ R and s ∈ S.
Let α = (r, s). By assumption, we have εα = α and αε = α. Equivalently, these equations
mean that
(a · r, b · s) = (r, s)
and
(r · a, s · b) = (r, s) .
Therefore, we have a · r = r and r · a = r for all r ∈ R. Hence R is a ring with identity,
namely the element a of R. Furthermore, we have b · s = s and s · b = s for all s ∈ S. Hence
S is a ring with identity, namely the element b of S.
√
Page 232, problem
√ 24. For D = 3, 6, and 7, the ring in question is Z[ D]. For D = 5,
the ring is Z[(1 + 5)/2]. In each case, we will give a unit θ which satisfies the inequality
θ > 1. The group of units will then contain all powers of θ. Since θ > 1, we have θn+1 > θn
for all positive integers n. Therefore, the powers of θ provide an infinite number of units in
the ring in question.
√
√
√
which
is
also
in
the
ring
Z[
3]. Note that
For D = 3, let θ = 2 + 3. Let θ′ = 2 − 3,
√
θθ′ = 4 − 3 = 1. Hence θ is indeed a unit in Z[ 3]. Also, we do have θ > 1.
√
√
√
For D = 6, let θ = 5 + 2 6. Let θ′ = 5 − 2 6, which
is also in the ring Z[ 6]. Note that
√
θθ′ = 25 − 4 · 6 = 1. Hence θ is indeed a unit in Z[ 6]. Also, we do have θ > 1.
√
√
√
For D = 7, let θ = 8 + 3 7. Let θ′ = 8 − 3 7, which
is also in the ring Z[ 7]. Note that
√
θθ′ = 64 − 9 · 7 = 1. Hence θ is indeed a unit in Z[ 7]. Also, we do have θ > 1.
√
Finally, consider
D
=
5.
The
ring
in
question
is
Z[ω],
where
ω
=
(1
+
5)/2. Let θ = ω.
√
Let θ′ = (1 − 5)/2. Note that θ′ = 1 − ω and so θ′ is also in the ring Z[ω]. We have
√
√
θθ′ = (1 + 5)/2 (1 − 5)/2 = (1 − 5)/4 = −1 .
Hence θ(−θ′ ) = 1. Note that −θ′ is in Z[ω]. It follows that θ is a unit in Z[ω]. Also, θ > 1.
Additional Problem A. Let R = Z×Z, the direct product of the ring Z with itself. Then
R is a commutative ring with identity and the multiplicative identity element of R is (1, 1).
The additive identity element of R is (0, 0). Suppose that a = (1, 0) and b = (0, 1). Then
a and b are elements of R, and neither is equal to the additive identity element 0R = (0, 0).
However, ab = (1, 0)(0, 1) = (0, 0) = 0R . Hence a and b are zero-divisors in the ring R. Thus,
the implication ab = 0R =⇒ a = 0R or b = 0R is not satisfied by the ring R. This implies
that R is not an integral domain.
Additional Problem B. This problem concerns the ring R = Z/10Z, an example of a
commutative ring with unit. Let S be defined as follows:
S = {[a]10 | a is an even integer }
We will show that S is a subring of R and that S is a field.
The fact that S is a subring of R is rather obvious. One just notes that if s1 , s2 ∈ S,
then s1 = [a1 ]10 , s2 = [a2 ]10 , where a1 , a2 are even integers. Then
s1 + s2 = [a1 + a2 ]10 ,
s1 − s2 = [a1 − a2 ]10 ,
s1 s2 = [a1 a2 ]10 ,
are all in S because the integers a1 + a2 , a1 − a2 and a1 a2 are all even.
The ring S is obviously commutative. Also, the ring S has a multiplicative identity, namely
[6]10 . This is verified by checking that
[6]10 [a]10 = [6a]10 = [a]10
for a = 0, 2, 4, 6 and 8. Alternatively, note that, for any integer b, we have 6b ≡ b (mod 5).
This congruence implies that 6(2b) ≡ 2b (mod 10). Therefore, if a = 2b, then we have the
congruence 6a ≡ a (mod 10). This means that [6a]10 = [a]10 for all even integers a.
To see that S is a field, we verify that the four nonzero elements of S are all invertible:
[2]10 [8]10 = [6]10 ,
[4]10 [4]10 = [6]10 ,
[6]10 [6]10 = [6]10 .
Therefore, S is a commutative division ring and hence is a field.
Another subring T of R which is a field is
T = {[a]10 | a is an integer divisible by 5 } = {[0]10 , [5]10 }.
It is easy to verify that T is a subring of R and is a field. The multiplicative identity is [5]10 ,
which is the only nonzero element of T and is clearly invertible.
Additional Problem C. To determine the center of the ring M2 (R), we will first find all
2 × 2 matrices with real entries that commute with the matrix
1 0
.
E11 =
0 0
We have
a b
c d
a 0
1 0
,
=
c 0
0 0
a b
a b
1 0
=
0 0
c d
0 0
A necessary and sufficient condition for these two products to be equal is that b = c = 0.
Thus, the set of 2 × 2 matrices that commute with E11 is
a 0 a, d ∈ R
0 d
Now suppose that A is an element of the center of the ring M2 (R). Then AB = BA for
all B ∈ M2 (R). In particular, we have AE11 = E11 A and AE21 = E21 A, where
0 0
.
E21 =
1 0
As shown above, the fact that AE11 = E11 A implies that A has the form
a 0
A =
0 d
where a, d ∈ R. Now we use the fact that AE21 = E21 A. We have
0 0
a 0
0 0
0 0
0 0
a 0
=
,
E21 A =
=
AE21 =
a 0
0 d
1 0
d 0
1 0
0 d
We have AE21 = E21 A if and only if a = d. Thus,
a 0
= aI2 ,
A =
0 a
1 0
, a scalar multiple of the identity matrix I2 . Note that I2 is the multiwhere I2 =
0 1
plicative identity element in the ring M2 (R). It is obvious that matrices of the form aI2 do
indeed commute with all elements of M2 (R). Thus,
{A ∈ M2 (R) | AB = BA f or all B ∈ M2 (R) } = {aI2 | a ∈ R }
That is, the center of the ring M2 (R) is the subring {aI2 | a ∈ R }.
Additional Problem D. We first prove that the subset
a b S =
a, b ∈ R
.
−b a is a subring of M2 (R). We will then show that S ∼
= C.
0 0
and this is clearly in S. For every
The additive identity element of M2 (R) is
0 0
a b
in S, its additive inverse is
element A =
−b a
−a
−b
,
−A =
−(−b) −a
which isalso in S. Furthermore, suppose that A′ is also in S, and so we can write A′ =
a′ b ′
, where a′ , b′ ∈ R. Then
−b′ a′
′
(a + a′
b + b′
a b′
a b
′
,
=
+
A+A =
−(b + b′ ) (a + a′ )
−b′ a′
−b a
which is in S. We have proved that S is a subgroup of the underlying additive group of
M2 (R).
To complete the verification that S is a subring of M2 (R), it suffices to show that S is
closed under the multiplication operation in M2 (R). Let A and A′ be as in the previous
paragraph. Then
′
aa′ − bb′
ab′ + ba′
aa′ − bb′
ab′ + ba′
a b′
a b
′
,
=
=
AA =
−(ab′ + ba′ ) aa′ − bb′
−ba′ + a(−b′ ) −bb′ + aa′
−b a −b′ a′
which is indeed in the subset S. We have proved that S is a subring of M2 (R).
Now define a map φ from C to S as follows.: For all a, b ∈ R, define
a b
.
φ(a + bi) =
−b a
The map φ is clearly a bijection from C to S. We will prove that φ is a ring homomorphism
and therefore that the subring S of M2 (R) is isomorphic to C.
Consider z = a + bi, w = c + di ∈ C. We have
z + w = (a + c) + (b + d)i,
zw = (ac − bd) + (ad + bc)i
and so
φ(z + w) =
and
a+c
b+d
−(b + d) a + c
=
c
a b
φ(z)φ(w) =
−d
−b a
ac − bd
=
−(ad + bc)
c d
a b
= φ(z) + φ(w)
+
−d c
−b a
d
c
=
ad + bc
ac − bd
ac − bd
ad + bc
−bc − ad −bd + ac
= φ(zw) ,
showing that φ is indeed a ring homomorphism. Since φ is also a bijection, ϕ is an isomorphism of the ring C to the ring S.
Additional Problem E. Suppose that F is any field. Suppose that we define R and its
operations just as stated in the problem. The additive identity 0R of R is 0F + 0F i. The
multiplicative identity 1R of R is 1F + 0F i. Also, it will be convenient to identify an element
a ∈ F with the element a + 0F i in R. Thus, with this identification, we can regard F as a
subring of R.
We distinguish two cases.
Case 1: The equation a2 + b2 = 0F has a solution where a and b are nonzero elements of F .
In this case, consider α = a + bi, which is a nonzero element of R. Let β = a + (−b)i,
which we write more simply as a − bi. Then β is also a nonzero element of R. Furthermore,
we have
αβ = (a + bi)(a − bi) = (a2 + b2 ) + 0F i = 0F + 0F i = 0R
Hence R has zero-divisors and so R is not an integral domain. We proved in class that every
field is an integral domain. Hence R is not a field.
Case 2: The only solution to the equation a2 + b2 = 0F , where a, b ∈ F , is given by
a = b = 0F .
In this case, we will prove that R is a field. First of all, note that R is a commutative
ring with identity 1R (as specified above). Also, 1R 6= 0R . Now suppose that α = a + bi is
any nonzero element of R. This means that a and b are not both zero. Thus, a2 + b2 6= 0F .
Let c = a2 + b2 , which is a nonzero element of F . Hence c is a unit in F . Hence, c has an
inverse in F under multiplication, which we write as c−1 . As stated above, we can regard
c−1 as an element of R.
Let β = a − bi, which is an element of R. Furthermore, we have
αβ = (a2 + b2 ) + 0F i = (c + 0F i)(1F + 0F i) = (c + 0F i)1R = c1R .
Multiplying both sides of the equation by c−1 , we obtain
α(c−1 β) = 1R .
Since c−1 β is an element of R, it follows that α is a unit in R. We have proved that every
nonzero element of R is a unit of R. Since R is a commutative ring with identity and 1R 6= 0R ,
it follows that R is a field.
Now consider F = Z/3Z. There are only two nonzero elements in F , namely 1 + 3Z and
2 + 3Z. Consider a2 + b2 , where a and b are nonzero elements of F . Then, checking the four
possibilities, we see that a2 + b2 is always nonzero. Thus, the only solution to a2 + b2 = 0F
is a = b = 0F . Therefore, we are in case 2. Therefore, R is a field.
Now consider F = Z/5Z. Let a = 1 + 5Z and b = 2 + 5Z, two nonzero elements of F .
We have a2 + b2 = 5 + 5Z = 0 + 5Z = 0F . Hence we are in case 1. Hence R is not a field.