# Mathematics 2215: Rings, fields and modules

```Mathematics 2215: Rings, fields and modules
Solutions to homework exercise sheet 1
1. Show that Z[i] = {a + bi : a, b ∈ Z} is a unital commutative ring under addition and multiplication of complex numbers. [This is called the ring of Gaussian integers]. Does it have
zero-divisors? Is it a division ring? Is it a field?
Solution We’ll show that Z[i] is a subring of C. Clearly Z[i] ⊆ C and Z[i] 6= ∅. So it suffices
to show that x, y ∈ Z[i] =⇒ x − y ∈ Z and xy ∈ Z[i]. Let x, y ∈ Z[i] and write x = a + bi and
y = c + di where a, b, c, d ∈ Z. Then x − y = e + f i where e = a − c ∈ Z and f = b − d ∈ Z, so
x − y ∈ Z[i], and xy = (a + bi)(c + di) = g + hi where g = ac − bd ∈ Z and h = bc + ad ∈ Z,
so xy ∈ Z[i].
So Z[i] is a subring of C, so it’s a ring. Since C is a field with unity 1, and 1 = 1 + 0i ∈ Z[i],
the ring Z[i] is unital (with unity 1). Since C is a field, C contains no zero-divisors, so the
subring Z[i] contains no zero-divisors. Z[i] is not a division ring, since 2 ∈ Z[i] but 2(a + bi) =
2a + 2bi 6= 1 for any a, b ∈ Z, so 2 ∈ Z[i]&times; but 2 is not invertible. Since Z[i] isn’t a division
ring, it isn’t a field.
2. Which of the following is a subring of M (2, Z)? For each subring R, determine whether R is
unital, whether R is commutative, whether R is a division ring, whether R is a zero ring and
whether R has zero-divisors.
(a) R = a0 0b : a, b ∈ Z
(b) R = 10 1b : b ∈ Z
0 (c) R = a0 1−a
:a∈Z
(d) R = 00 0b : b ∈ Z
(e) R = a0 a0 : a ∈ Z
(f) R = 00 00
Solution
(a) This is a subring, since it’s a subsetof M (2,Z), and if x = a00b ∈ R and y = 0c d0 ∈ R
a−b c−d
∈ R. The element
where
0 0
0
0 ∈ R and xy =
0 1 a, b,&times; c, d ∈ Z, then x − y = 0 1 0
1
∈
R
is
a
zero-divisor,
since
=
0.
So
R
has
zero-divisors
0 0
0 0
0 0
1 0 1 0 and
1 0 so
R cannot
be a division ring. R is not a zero ring,
0 0
0 0 = 0 0 6=
1 0 since
for example
0. 0R0 is
0
1
1
0
0
1
0
1
not commutative, since for example 0 0 0 0 = 0 0 is not equal to 0 0 0 0 = 0 0 .
(b) This is not a subring because, for example, 10 11 ∈ R but 10 11 − 10 11 = 10 01 6∈ R.
(c) This is not a subring, since for example 00 01 ∈ R but 00 01 − 00 01 = 00 00 6∈ R.
0 c
0 b
(d) This is a subring, since it’s a subset
0 b−c of M (2, Z), and if 0x0 =
0 0 ∈ R and y = 0 0 ∈ R
where b, c ∈ Z, then x − y = 0 0 ∈ R and xy = 0 0 ∈ R. This also shows that R
is a zero ring, since xy = 00 00 = 0R for every x, y ∈ R. So R has plenty of zero-divisors
(every non-zero element is a zero-divisor) so R is not a division ring. R is not unital, since
a zero ring never is. Since R is a zero ring, R is commutative.
a 0
b 0
(e) This is a subring, since it’s a subset
of
M
(2,
Z),
and
if
x
=
∈
R
and
y
=
0
a
0 b ∈ R
a−b 0 ab 0 where
Z, then x − y = 0 a−b ∈ R and xy = 0 ab ∈ R. Moreover, yx =
ba 0 a, b ∈
ab
0
0 0
0 ba =
0 ab = xy, so R is commutative. If xy = 0 0 = 0R then ab = 0, so a = 0
or b = 0 (since Z has no zero-divisors),
so x = 0 or y = 0. So R has no zero-divisors,
1 0
1and
so R is not a zero ring. Since 0 1 = 1M (2,Z) ∈ R, the
R is unital, with unity 0 01 .
2 ring
However, R is not a division ring since, for example 0 02 is not invertible.
0 0
(f) R is a subring of M (2, Z), since it’s a subset(the subset containing
only
0 0 ), and if
0 0
0
0
0
0
x, y ∈ R then x = y = 0 0 and so x − y = 0 0 ∈ R and xy = 0 0 ∈ R. It’s a zero
ring, so it’s commutative and non-unital. Since R&times; = ∅, there are no zero-divisors. Since
R is non-unital, it’s not a division ring.
3. Let R be a ring. Show that the following conditions are equivalent:
(a) R is a commutative ring
(b) (a + b)(a − b) = a2 − b2 for every a, b ∈ R
(c) (a + b)2 = a2 + 2ab + b2 for every a, b ∈ R
Solution First, let’s expand these expressions without assuming that R is commutative. By
the distributive laws, we have
(a + b)(a − b) = a(a − b) + b(a − b) = a2 − ab + ba − b2
and
(a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ba + b2 .
(a)⇒(b): Suppose R is a commutative ring. Then ab = ba for every a, b ∈ R, so −ab + ba = 0,
so by the first equation above, (a + b)(a − b) = a2 + 0 − b2 = a2 − b2 .
(b)⇒(a): Suppose that (a + b)(a − b) = a2 − b2 . Then by the first equation above, a2 − ab +
ba + b2 = a2 − b2 , so −ab + ba = 0, so ba = ab, for every a, b ∈ R. So R is commutative.
(a)⇒(c): Suppose R is a commutative ring. Then ab = ba for every a, b ∈ R, so ab + ba =
ab + ab = 2ab, so by the second equation above, (a + b)2 = a2 + 2ab + b2 .
(c)⇒(a): Suppose that (a + b)2 = a2 + 2ab + b2 . Then by the second equation above, a2 + ab +
ba + b2 = a2 + 2ab + b2 , so ab + ba = 2ab, so ba = ab, for every a, b ∈ R. So R is commutative.
4. Let R be a ring.
(a) If x1 , . . . , xm ∈ R, we write
m
X
xi = x1 + &middot; &middot; &middot; + xm .
i=1
Prove by induction that for every y ∈ R we have
y
m
X
i=1
xi =
m
X
yxi
i=1
and
m
X
i=1
2
xi y =
m
X
i=1
xi y.
(b) If a1 , . . . , am ∈ R and b1 , . . . bn ∈ R, show that
m X
n
n X
m
X
X
ai bj =
ai b j .
i=1
j=1
j=1
i=1
[This result is sometimes referred to by saying that “you can interchange the order of
summation”.]
Hint: Pshow that both of the expressions above are equal to ab where
P
a= m
a
and
b= m
i=1 i
j=1 bj .
Solution
(a) If m = 1 then this is trivial, because the sums then have only one term, x1 or yx1 or x1 y,
and we clearly have equality.
P
Pm
P
Pm
Now assume that y( m
yxi and ( m
i )y =
i=1 xi ) =
i=1 P
i=1 xP
i=1 xi y. Then by
Pmthe dism+1
m
tributive law and our assumption, y( i=1 xi ) = y ( i=1 xi ) + xm+1 = y( i=1 xi ) +
P
Pm+1
Pm+1
Pm
yxm+1 = ( m
yx
)
+
yx
=
yx
,
and
(
x
)y
=
(
x
)
+
x
y =
i
m+1
i
i
i
m+1
i=1
i=1
i=1
Pm
Pm
Pm+1i=1
( i=1 xi )y + xm+1 y = ( i=1 xi y) + xm+1y =
i=1 xi y. So the statement is true for
m = 1, and if it holds for m then it also holds for m + 1. By inductions, the statements
is true for all m ∈ N.
Pn
P
(b) Let a = m
j=1 bj . Then
i=1 ai and b =
ab = a
n
X
j=1
bj =
n
X
abj =
n X
m
n X
m
X
X
ai bj =
ai b j ,
j=1
j=1
i=1
j=1
i=1
using (a) twice. Similarly,
m
m
m X
n
m X
n
X
X
X
X
ab =
ai
ai b =
ai b =
bj
=
ai b j ,
i=1
i=1
i=1
j=1
i=1
j=1
again using (a) twice.
5. Let n ∈ N, let R be a ring and let M (n, R) be the set of n &times; n matrices with entries in R.
We define two operations, matrix addition and matrix multiplication, on M (n, R) as follows.
For A, B ∈ M (n, R), if the i, j entry of A is aij and the i, j entry of B is bij , we define
A+B =C
where the i, j entry of C is cij = aij + bij , and
n
X
A &middot; B = D where the i, j entry of D is dij =
aik bkj .
k=1
(a) Explain how the definition of A + B and A &middot; B depend on the definition of addition and
multiplication in the ring R.
(b) Prove that M (n, R) with matrix addition and matrix multiplication is a ring.
(c) Prove that if n &gt; 1 then the ring M (n, R) is commutative if and only if ab = 0 for all
a, b ∈ R.
Hint: for “⇒”, first think about n = 2 and the matrices a0 00 and 00 0b .
(d) If R is a unital ring, prove that M (n, R) is also a unital ring.
(e) Prove that if n &gt; 1 then M (n, R) is not a division ring.
3
Solution
(a)
A+B =C
A&middot;B =D
where the i, j entry of C is cij = aij + bij , and
where the i, j entry of D is dij =
n
X
aik bkj .
k=1
The bits of the definition in boxes use the addition of R (in the first and second boxes)
and the multiplication in R (in the second box only).
(b) Since R is closed under addition and multiplication, it’s easy to see that M (n, R) is closed
under matrix addition and matrix multiplication. So these are operations on M (n, R).
Let’s show that M (n, R) is an abelian group under matrix addition.
Matrix addition is associative, since the i, j entry of A + (B + C) is aij + (bij + cij ) and
the i, j entry of (A + B) + C is (aij + bij ) + cij , and these are equal by associativity of
Matrix addition is commutative, since i, j entry of A + B is aij + bij and the i, j entry of
B + A is bij + aij , and these are equal by commutativity of addition in R.
Let us write 0 for the zero matrix, that is, the matrix with i, j entry 0R for every i, j. The
i, j entry of A + 0 is aij + 0R = aij , so A + 0 = A and also 0 + A = A since matrix addition
is commutative. So 0 is the identity element for matrix addition.
If B is the matrix in M (n, R) with bij = −aij then A + B = 0. So every A ∈ M (n, R) is
invertible with respect to matrix addition.
So we’ve shown that M (n, R) is an abelian group under matrix addition.
Now the i, j entry of A &middot; (B &middot; C) is
n X
n
n
n
X
X
X
aik (bk` c`j )
(by 4(a), taking y = aik for fixed i, k)
aik
bk` c`j =
k=1
`=1
=
k=1
`=1
n
n
X X
(aik bk` )c`j
k=1
(by associativity of multiplication in R)
`=1
n X
n
X
=
(aik bk` )c`j
(since we can swap order of summation)
`=1
=
k=1
n X
n
X
`=1
aik bk` c`j
(by 4(a), taking y = c`j for fixed `, j)
k=1
which is the i, j entry of (A &middot; B) &middot; C. So A &middot; (B &middot; C) = (A &middot; B) &middot; C, so matrix multiplication
is associative.
Remark:
the order
P
P “you
Pn can swap
Pmof summation” means that if R is a ring and xk` ∈ R,
n
then m
x
=
k=1
`=1 k`
`=1
k=1 xk` ). You can prove this by induction. When I set
the exercise sheet, I had thought 4(b) would be enough to show that matrix multiplication
is associative, but that’s wrong since it only justifies swapping the order of summation for
sums of the form xk` = ak b` , meaning that the terms you are adding up have factorisations
which are related in a special way.
4
Pm
Pm
Finally,
the
i,
j
entry
of
A
&middot;
(B
+
C)
is
a
(b
+
c
)
=
ik
kj
kj
k=1
k=1 (aik bkj + aik ckj ) =
Pm
Pn
k=1 aik bkj +
`=1 ai` b`j , which is the i, j entry of A &middot; B + A &middot; C. [Here, we’ve used the
distributive law in R and the fact that you can reorder sums in an abelian group without
changing the answer]. So A &middot; (B + C) = A &middot; B + A &middot; C. Similarly, (B + C) &middot; A = B &middot; A + C &middot; A.
P
Pm
(c) “⇐” If ab = 0 for all a, b ∈ R, then the i, j entry of A &middot; B is m
k=1 aik bkj =
k=1 0 = 0, so
A &middot; B = 0 for every A, B ∈ M (n, R), so A &middot; B = B &middot; A = 0 for every A, B ∈ M (n, R), so
M (n, R) is commutative.
“⇒” If M (n, R) is commutative and a, b ∈ R, consider the matrix A with a11 = 0 and
aij = 0 if (i, j) 6= (1, 1), and the matrix B with b12 = b and bij = 0 if (i, j) 6= (1, 2). Then
B &middot; A = 0 but


ab 0 0 . . . 0
 0 0 0 . . . 0


A &middot; B =  ..
.. 
.
.
0 0 0 ... 0
has i, j entry 0 if (i, j) 6= (1, 1), and the 1, 1 entry is ab. Since M (n, R) is commutative by
assumption, we have A &middot; B = B &middot; A = 0, so ab = 0.
(d) Let 1 be the unity of R. Then the matrix


1 0 ... 0
0 1 . . . 0


I =  ..
.. 
.
.
.
. .
0 0 ... 1
P
is the unity of M (n, R). This because the i, j entry of I &middot; A is m
k=1 Iik akj , and if k 6= i
then Iik akj = 0akj = 0. So this sum is equal to Iii aij = 1aij = aij , so I &middot; A = A. Similarly,
A &middot; I = A.
(e) If R&times; = ∅ then R = {0} and M (n, R) contains only the zero
it can’t be a division ring.
So suppose that R&times; 6= ∅ and let a ∈ R&times; . The matrices



a 0 0 ... 0
0 0
0 0 0 . . . 0
0 a






X = 0 0 0 . . . 0 and Y = 0 0
 ..

 ..
. . .. 
.
.
. .
0 0 0 ... 0
0 0
matrix, so it’s not unital, so

0 ... 0
0 . . . 0

0 . . . 0

. . .. 
. .
0 ... 0
in M (n, R) have X 6= 0 and Y 6= 0 but X &middot; Y = 0, so X is a zero-divisor in M (n, R). So
M (n, R) is not a division ring.
6. Consider the following four matrices in M (2, C):
1 0
i 0
0 1
1=
, i=
, j=
0 1
0 −i
−1 0
0 i
and k =
.
i 0
Let
H = {a1 + bi + cj + dk : a, b, c, d ∈ R}.
5
(a) It’s not hard to see that i2 = j2 = k2 = ijk = −1. Use these equations to calculate ij, jk,
ki, ji, kj, and ik.
(b) Show that H is a unital subring of M (2, C).
(c) Prove that H is not a commutative ring.
(d) Prove that H is a division ring. Hint: for z = a1+bi+cj+dk, consider z = a1−bi−cj−dk
and show that zz = (a2 + b2 + c2 + d2 )1.
(e) Show that the set
Z[i, j, k] = {a1 + bi + cj + dk : a, b, c, d ∈ Z}
is a subring of H which is unital, non-commutative and has no zero-divisors, but is not a
division ring.
Solution
(a) We have ij = ij(−k2 ) = (−ijk)k = 1k = k, so kj = (ij)j = ij2 = −i. Moreover,
jk = (−i2 )jk = −i(ijk) = i, so ik = (jk)k = −j and ji = j(jk) = −k. So ki = (−ji)i =
−ji2 = j.
(b) Clearly, H is a subset of M (2, C). If z, w ∈ H, say z = a1 + bi + cj + dk and w =
e1 + f i + gj + hk, then
z − w = (a − e)1 + (b − f )i + (c − g)j + (d − h)k
and using (a) and the distributive laws (which we know hold for matrix multiplication),
we can calculate that
zw = (ae − bf − cg − dh)1 + (af + be + ch − dg)i + (ag − bh + ce + df )j + (ah + bg − cf + de)k
So z − w and zw ∈ H. So H is a subring of M (2, R). The unity of M (2, C) is 1 which is
in the subring H, so H is a unital subring.
(c) ij = ji since ij = k and ji = −k. So H is a not commutative.
(d) If z ∈ H as in the hint, then we do have zz = (a2 + b2 + c2 + d2 )1. If z ∈ H&times; then
at least one of the real numbers a, b, c, d is non-zero, so a2 + b2 + c2 + d2 &gt; 0. Hence
w = (a2 + b2 + c2 + d2 )−1 z has zw = 1 = wz.
(e) Clearly, Z[i, j, k] is a subset of H. Examining the calculations as in (b), we see that if
a, b, c, d, e, f, g, h are all integers then so are all the coefficients of z − w and zw. So
Z[i, j, k] is a subring of H. Since i, j ∈ Z[i, j, k], the calculation in (c) shows that Z[i, j, k]
is not commutative. The unity of H is 1 which is in Z[i, j, k], so Z[i, j, k] is a unital ring.
The element 21 = 20 02 of Z[i, j, k] is not invertible in Z[i, j, k], so it’s not a division
ring. Since H is a division ring it has no zero-divisors, so the subring Z[i, j, k] has no
zero-divisors either.
6
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