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Math 236
Fall 2006
Dr. Seelinger
Solution for §3.1
Problem 5: Consider the following sets and determine whether each set is a subring of M2 (R).
If a set is a subring of M2 (R), determine whether it has an identity.
0 r
(a) Let S be the set of all matrices of the form
where r is a rational number. We claim
0 0
that S is a subring. 0 r1
0 r2
In particular, let
,
∈ S. Then
0 0
0 0
0 r1
0 r2
0 r1 + r 2
+
=
∈S
0 0
0 0
0
0
so S is closed under addition. Next, note that
0 0
∈S
0 0
Since 0 is a rational number. Furthermore,
0 r1
0 r2
0 0
=
∈S
0 0
0 0
0 0
so S is closed under multiplication. Finally, note the additive inverse of
0 r1
0 0
is
0 −r1
0 0
which is in S since −r1 is rational. Therefore, S is a subring of M2 (R) by Theorem 3.2.
Next,
note that S does not have an identity element since the product of any two elements
0 0
in S is
.
0 0
a b
(b) Let T be the set of all matrices of the form
where a, b, c ∈ Z. We claim that T is a
0 c
subring.
a1 b 1
a2 b 2
In particular, let
,
∈ T . Then
0 c1
0 c2
a1 b 1
a2 b 2
a1 + a2 b 1 + b 2
+
=
∈T
0 c1
0 c2
0
c1 + c2
1
so T is closed under addition. Next,
a1 b 1
a2 b 2
a1 a2 a1 b2 + b1 c2
=
∈T
0 c1
0 c2
0
c1 c2
so T is closed under multiplication. Furthermore, note that
0 0
∈T
0 0
since we can consider the case when a1 = b1 = c1 = 0. Finally, note the additive inverse of
a1 b 1
0 c1
is
−a1 −b1
0 −c1
which is in T since −a1 , −b1 , −c1 ∈ Z. Therefore,
T is a subring of M2 (R) by Theorem 3.2.
1 0
Next, note that the identity element of T is
.
0 1
a a
(c) Let W be the set of all matrices of the form
where a, b ∈ R. We claim that W is a
b b
subring.
a2 a2
a1 a1
∈ W . Then
,
In particular, let
b2 b2
b1 b1
a1 + a2 a1 + a2
a2 a2
a1 a1
∈W
=
+
b1 + b2 b1 + b2
b2 b2
b1 b1
so W is closed under addition. Next,
a1 a2 + a1 b 2 a1 a2 + a1 b 2
a1 a1
a2 a2
=
∈W
b1 b1
b2 b2
b 1 a2 + b 1 b 2 b 1 a2 + b 1 b 2
so W is closed under multiplication. Furthermore, note that
0 0
∈T
0 0
since we can consider the case when a1 = b1 = 0. Finally, note the additive inverse of
a1 a1
b1 b1
is
−a1 −a1
−b1 −b1
2
which is in W . Therefore, W is a subring of M2 (R) by Theorem 3.2.
Note that W does not have an identity element.
a 0
(d) Let X be the set of all matrices of the form
where a ∈ R. We claim that X is a
a 0
subring.
a1 0
a2 0
In particular, let
,
∈ X. Then
a1 0
a2 0
a1 0
a2 0
a1 + a2 0
+
=
∈X
a1 0
a2 0
a1 + a2 0
so X is closed under addition. Next,
a1 0
a2 0
a1 a2 0
=
∈X
a1 0
a2 0
a1 a2 0
so X is closed under multiplication. Furthermore, note that
0 0
∈X
0 0
since we can consider the case when a1 = 0. Finally, note the additive inverse of
a1 0
a1 0
is
−a1 0
−a1 0
which is in X. Therefore, X is a subring of M2 (R) by Theorem 3.2.
1 0
Note that the identity element of X is
.
1 0
a 0
(e) Let D be the set of all matrices of the form
where a, b ∈ R. We claim that D is a
0 b
subring.
a1 0
a2 0
In particular, let
,
∈ D. Then
0 b1
0 b2
a1 0
a2 0
a1 + a2
0
+
=
∈D
0 b1
0 b2
0
b1 + b2
so D is closed under addition. Next,
a1 0
a2 0
a1 a2 0
=
∈D
0 b1
b2 0
0 b1 b2
3
so D is closed under multiplication. Furthermore, note that
0 0
∈D
0 0
since we can consider the case when a1 = b1 = 0. Finally, note the additive inverse of
a1 0
0 b1
is
−a1 0
0 −b1
which is in D. Therefore, D is a subring of M2 (R) by Theorem 3.2.
1 0
Note that the identity element of D is
.
0 1
a 0
(f ) Let R be the set of all matrices of the form
where a ∈ R. We claim that R is a
0 0
subring.
a1 0
a2 0
In particular, let
,
∈ D. Then
0 0
0 0
a1 0
a2 0
a1 + a2 0
+
=
∈R
0 0
0 0
0
0
so R is closed under addition. Next,
a1 0
a2 0
a1 a2 0
=
∈R
0 0
0 0
0 0
so R is closed under multiplication. Furthermore, note that
0 0
∈R
0 0
since we can consider the case when a1 = 0. Finally, note the additive inverse of
a1 0
0 0
is
−a1 0
0 0
which is in R. Therefore, R is a subring of M2 (R) by Theorem 3.2.
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Note that the identity element of R is
1 0
.
0 0
√
√
√
Problem 9 Let Z[ 2] = a + b 2 : a, b ∈ Z ⊆ R. Prove Z[ 2] is a subring of R.
√
√
√
√
√
Proof: Consider a√+ b 2,√
c + d 2 ∈√Z[ 2] where a, b, c, d ∈ Z. Then (a + b 2)
√+ (c + d√ 2) =
(a + c) + (b + d) √
2 ∈ Z[ √
2], so Z[ 2] is closed under addition.
√ Also, (a + b 2)(c + d 2) =
2
∈
Z[
2]
since
ac+2bd,
ad+bc
∈
Z.
So
Z[
2]√is closed under multiplication.
(ac+2bd)+(ad+bc)
√
√
√
√
Next, 0 = 0 + 0 2 ∈ Z[ 2]. Finally,
the
additive
inverse
of
a
+
b
2
is
(−a)
+
(−b)
2
∈
Z[
2].
√
Q.E.D.
Therefore, by Theorem 3.2, Z[ 2] is a subring of R.
Problem 10 Let Z[i] = {a + bi : a, b ∈ Z} ⊆ C. Prove Z[i] is a subring of C.
Proof: Consider a+bi, c+di ∈ Z[i] where a, b, c, d ∈ Z. Then (a+bi)+(c+di) = (a+c)+(b+d)i ∈
Z[i], so Z[i] is closed under addition. Also, (a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ Z[i] since
ac − bd, ad + bc ∈ Z. So Z[i] is closed under multiplication. Next, 0 = 0 + 0i ∈ Z[i]. Finally, the
additive inverse of a + bi is (−a) + (−b)i ∈ Z[i]. Therefore, by Theorem 3.2, Z[i] is a subring of
C.
Q.E.D.
Problem 18 Define a new addition ⊕ and a new multiplication ⊗ on Z by
a⊕b=a+b−1
and
a ⊗ b = a + b − ab,
where the operations on the right-hand side of the equal signs are ordinary addition, subtraction,
and multiplication. Prove that, with the new operations ⊕ and ⊗, Z is an integral domain.
To prove this, we need to check the eight conditions in the definition of a ring, then check
the additional conditions on being an integral domain. So let a, b, c ∈ Z.
(1) Since a, b ∈ Z, we have a ⊕ b = a + b − 1 ∈ Z.
(2) Note a⊕(b⊕c) = a⊕(b+c−1) = a+(b+c−1)−1 = (a+b−1)+c−1 = (a+b−1)⊕c = (a⊕b)⊕c.
So associativity of addition holds.
(3) We also see a ⊕ b = a + b − 1 = b + a − 1 = b ⊕ a, so commutativity of addition holds.
(4) Note that if we set O = 1 we see that a ⊕ O = a ⊕ 1 = a + 1 − 1 = a, so O = 1 is the zero
element.
(5) Consider the equation 1 = O = a ⊕ x = a + x − 1. Solving this equation for x gives
x = 2 − a ∈ Z, so this property holds.
(6) Note that a ⊗ b = a + b − ab ∈ Z since a, b ∈ Z.
(7) Consider a⊗(b⊗c) = a⊗(b+c−bc) = a+(b+c−bc)−a(b+c−bc) = a+b+c−ab−bc−ac+abc
and (a ⊗ b) ⊗ c = (a + b − ab) ⊗ c = (a + b − ab) + c − (a + b − ab)c = a + b + c − ab − ac − bc + abc.
So a ⊗ (b ⊗ c) = (a ⊗ b) ⊗ c and associativity of multiplication holds.
(8) For the distributive property, note a ⊗ (b ⊕ c) = a ⊗ (b + c − 1) = a + b + c − 1 − a(b + c − 1) =
2a + b + c − ab − ac − 1 = (a + b − ab) + (a + c − ac) − 1 = (a ⊗ b) + (a ⊗ c) − 1 = (a ⊗ b) ⊕ (a ⊗ c)
and (a ⊕ b) ⊗ c = (a + b − 1) ⊗ c = a + b − 1 + c − (a + b − 1)c = a + b + 2c − ac − bc − 1 =
(a + c − ab) + (b + c − bc) − 1 = (a ⊗ c) + (b ⊗ c) − 1 = (a ⊗ c) ⊕ (b ⊗ c) so the distributive
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properties hold.
(9) Note that a ⊗ b = a + b − ab = b + a − ba = b ⊗ a, so this ring is commutative.
(10) Let IR = 0. Then a ⊗ IR = a ⊗ 0 = a + 0 − a0 = a and IR ⊗ a = 0 ⊗ a = 0 + a − 0a = a, so
IR = 0 is the multiplicative identity.
The above shows that Z with the operations ⊕ and ⊗ is a commutative ring with identity.
Now we need to show that it is also an integral domain. So assume a ⊗ b = O. This equation
translates to a + b − ab = 1. But a + b − ab = 1 ⇒ 0 = ab − a − b + 1 = (a − 1)(b − 1) ⇒ a − 1 =
0 or b − 1 = 0 ⇒ a = 1 = O or b = 1 = O. Hence this ring is an integral domain.
Q.E.D.
Problem 22 Let L be the set of all positive real numbers and for any a, b ∈ L define a ⊕ b = ab
and a⊗b = alog b . (a) Prove that L is a ring under the operations ⊕ and ⊗. (b) Is L a commutative
ring? (c) Is L a field?
(a) We need to demonstrate the eight properties in the definition of a ring. So let a, b, c ∈ L.
(1) Since a, b ∈ L, then a ⊕ b = ab ∈ L as the product of two positive real numbers is a
positive real number.
(2) Note a ⊕ (b ⊕ c) = a ⊕ (bc) = a(bc) = (ab)c = (ab) ⊕ c = (a ⊕ b) ⊕ c.
(3) Next, a ⊕ b = ab = ba = b ⊕ a.
(4) To get a zero element, we need a number OL such that a = a ⊕ OL = aOL . Hence
OL = 1 ∈ L is the zero element.
(5) We need to solve a ⊕ x = OL = 1. This translates to ax = 1, so x = (1/a) ∈ L since a is
a positive (non-zero) real number.
(6) Now a ⊗ b = alog b . But since b is a positive real number, log b is a real number. Therefore,
the positive number a raised to a real exponent log b is still positive, hence alog b ∈ L.
log c
(7) Note a⊗(b⊗c) = a⊗(blog c ) = alog(b ) = a(log b)(log c) = (alog b )log c = (a⊗b)log c = (a⊗b)⊗c,
since log(blog c ) = (log c)(log b) by the properties of logs.
(8) Now a⊗(b⊕c) = a⊗(bc) = alog(bc) = alog b+log c = alog b alog c = (a⊗b)(a⊗c) = (a⊗b)⊕(a⊗c)
and (a ⊕ b) ⊗ c = (ab) ⊗ c = (ab)log c = alog c blog c = (a ⊗ c)(b ⊗ c) = (a ⊗ c) ⊕ (b ⊗ c).
Therefore, since L satisfies the definition of a ring, L is a ring under the operations ⊕ and ⊗.
(b) To show L is commutative, we need to show a⊗b = b⊗a. But a⊗b = alog b = (eloga )log b =
(log a)(log b)
e
= e(log b)(log a) = (elog b )log a = blog a = b ⊗ a. Therefore, L is a commutative ring.
(c) In order for L to be a field, we need to show first that L has an identity, then if a 6= OL ,
−1
a exists. First we show L has an identity, I. So we need to solve a ⊗ I = a. So alog I = a,
which implies log I = 1 when a 6= Ol = 1. If log I = 1, then I = e1 = e. Therefore, e is the
multiplicative identity.
Now let a 6= 1 = OL and set a⊗x = e = I. Then e = alog x = e(log a)(log x) . So (log a)(log x) = 1
which gives us that x = e(1/(log a)) ∈ L when a 6= 1. Therefore, every a 6= 1 = OL in L has a
multiplicative inverse e(1/(log a)) so L is a field.
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