Solution of Some Homework Problems
(3.1) Prove that a commutative ring R has a unique 1.
Proof: Let 1R and 10R be two multiplicative identities of R. Then since 10R is an identity,
1R = 1R 10R . Since 1R is an identity, 1R 10R = 10R . Thus 1R = 1R 10R = 10R , and so the
multiplicative identity is unique.
(3.4) (i) If X is a set, prove that the Boolean group B(X) in Example 2.18 with elements
the subsets of X and with addition given by
U + V = (U − V ) ∪ (V − U ),
where U − V = {x ∈ U : x 6∈ V }, is a commutative ring if one defines multiplication
U V = U ∩ V.
We call B(X) a boolean ring.
(ii) Prove that B(X) contains exactly one unit.
(iii) If Y is a proper subset of X, show that the unit in B(Y ) is distinct from the unit in
B(X). Conclude that B(Y ) is not a subring of B(X).
Proof:
(i) By the definition of U + V , we have U + V ∈ B(X) and U + V = (U − V ) ∪
(V − U ) = (V − U ) ∪ (U − V ) = V + U , and so + is a commutative binary operation in
B(X). Note that
U + V = (U − V ) ∪ (V − U ) = (U ∪ V ) − (U ∩ V ).
Therefore, ∀U, V, W ∈ B(X), the set (U + V ) + W = [((U ∪ V ) − (U ∩ V )) ∪ W ] − [((U ∪ V ) −
(U ∩ V )) ∩ W ] consists of the elements in X that occurs in an odd number of sets in U, V
and W (in either one of them or in all of them). This is the same as the set of U + (V + W ),
and so
(U + V ) + W = U + (V + W ).
Thus + is associative.
Since the empty set ∅ ∈ B(X), and since ∀U ∈ B(X), U + ∅ = (U − ∅) ∪ (∅ − U ) = U ,
the empty set ∅ is the additive identity in B(X). For any U ∈ B(X), U + U = ∅ and so
every element in B(X) has an additive inverse. This indicates that B(X) with addition is
an abelian group.
1
By the definition of multiplication, ∀U, V ∈ B(X), U V = U ∩U ∈ B(X) is a commutative
binary operation of B(X). For U, V, W ∈ B(X),
(U V )W = (U ∩ V ) ∩ W = U ∩ (V ∩ W ) = U (V W ),
and so multiplication is associative in B(X). As for any U ∈ B(X), U X = U ∩ X = U , X
is the multiplicative identity of B(X).
We also need to show that distributive law holds. ∀U, V, W ∈ B(X),
U (V + W ) = U ∩ [(V ∪ W ) − (V ∩ W )] = (U ∩ (V ∪ W )) − (U ∩ (V ∩ W ))
= [(U ∩ V ) ∪ (U ∩ W )] − [(U ∩ V ) ∩ (U ∩ W )] = U V + U W.
Therefore, B(X) is a commutative ring.
(ii). By (i), we know that X is a unit. Suppose that U is a unit of B(X). Then for some
V inB(X), U V = U ∩ V = X. Since U ∈ B(X), U ⊆ X. Since X = U V = U ∩ V ⊆ U , we
have U = X and so X is the only unit of B(X).
(iii) Let Y ⊂ X be a proper subset of X. Then Y 6= X. By (i), B(Y ) is a commutative
ring. By (ii), Y is the only unit in B(Y ), and Y is not a unit in B(X). Therefore, B(Y ) is
not a subring of B(X).
(3.5) Show that U (Zm ) = {[k] ∈ Zm : (k, m) = 1}.
Proof: If (k, m) = 1 (as integers), then there exist integers s and t such that sk + tm = 1
(as integers). Therefore,
sk ≡ sk + tm ≡ 1 (mod m).
It follows that [s][k] = [1] in Zm , and so {[k] ∈ Im : (k, m) = 1} ⊆ U (Zm ).
Conversely, let [k] ∈ U (Zm ). Then ∃[s] ∈ Zm such that [sk] = [s][k] = [1], and so sk ≡ 1
(mod m). Hence there exists an integer t such that sk = 1 + tm, or sk + (−t)m = 1 (as
integers). This implies that (m, k) = 1 and so {[k] ∈ Zm : (k, m) = 1} ⊇ U (Zm ).
√
(3.10) (i) Prove that R = {a + b 2 : a, b ∈ Z} is a domain.
√
(ii) Prove that R = { 21 (a + b 2) : a, b ∈ Z} is not a domain.
√
(iii) Using the fact that α = 12 (1 + −19) is a root of x2 − x + 5, prove that R = {a + bα :
a, b ∈ Z} is a domain.
2
Proof:
The binary operations of these sets are assumed to be the addition and multipli-
cation of the real or the complex numbers.
(i) Let R denote the field of real numbers with real number addition and multiplication.
Then R is a subset of R. To see that R is a subring of R, it suffices to show that
√
(1) 1 ∈ R. This is true as 1 = 1 + 0 · 2 ∈ R.
√
√
(2) For any a + b 2, a0 + b0 2 ∈ R, we have
√
√
√
(a + b 2) − (a0 + b0 2) = (a − a0 ) + (b − b0 ) 2 ∈ R.
√
√
(3) For any a + b 2, a0 + b0 2 ∈ R, we have
√
√
√
(a + b 2)(a0 + b0 2) = (aa0 + 2bb0 ) + (ab0 + a0 b) 2 ∈ R.
As R is a subring of a domain R, R itself is also a domain.
(ii) Suppose that this R is a ring. Then multiplication must be a binary operation of
R. Note that
1
2
∈ R but
1
2
·
1
2
=
1
4
6∈ R. Thus multiplication is not a binary operation of R,
and so R is not even a ring, not to mention a domain.
(iii) Let C denote the set of all complex numbers with number addition and multiplication. Then C is a field, and this R is a subset of C. To see that R is a subring of C, it
suffices to show that
(1) 1 ∈ R. This is true as 1 = (2 + 0 ·
(2) For any a +
bα, a0
+
b0 α
√
−19)/2 ∈ R.
∈ R, we have
a + bα − (a0 + b0 α) = (a − a0 ) + (b − b0 )α ∈ R.
(3) For any a + bα, a0 + b0 α ∈ R, we have
(a + bα)(a0 + b0 α) = aa0 + (ab0 + a0 b)α + bb0 α2 .
Since α2 = α − 5, we can rewrite the product as
(a + bα)(a0 + b0 α) = aa0 + (ab0 + a0 b)α + bb0 (α − 5) = (aa0 − 5bb0 ) + (ab0 + a0 b + bb0 )α ∈ R.
Therefore, R is a subring of the field C. Since a field is a domain, R itself is also a domain.
(3.12) (i) If R is a commutative ring, define the circle operation a ◦ b by a ◦ b = a + b − ab.
Prove that the circle operation is associative and that 0 ◦ a = a for all a ∈ R.
3
(ii) Prove that a commutative ring R is a field if and only if {r ∈ R : r 6= 1} is an abelian
group under the circle operation.
Proof: (i) ∀a, b, c ∈ R,
(a ◦ b) ◦ c = (a + b − ab) ◦ c = (a + b − ab) + c − c(a + b − ab) = a + b + c − ab − ac − bc + abc
= a + (b + c − bc) − a(b + c − bc) = a ◦ (b ◦ c).
Thus the operation ◦ is associative. Moreover,
∀a ∈ R, 0 ◦ a = (0 + a) − 0 · a = a.
(ii) Let A = {r ∈ R : r 6= 1}. Suppose that R is a field. Then R 6= {0}, and so 0 ∈ A.
For a1 , a2 ∈ A, if 1 = a1 ◦ a2 = a1 + a2 − a1 a2 , then 1 − a2 = a1 − a1 a2 = a1 (1 − a2 ), or
(1 − a1 )(1 − a2 ) = 0. Since R has no zero divisor, either a1 = 1 or a2 = 1, contrary to the
assumption that a1 , a2 ∈ A. Therefore, a1 ◦a2 ∈ A also. Moreover, a1 ◦a2 = a1 +a2 −a1 a2 =
a2 + a1 = a2 ◦ a1 . Thus (also by (i)), ◦ is an associative and commutative binary operation
of A, with 0 serves as the identity (see (i)).
Let a ∈ A. To see that a has an inverse in A with respect to the ◦ operation, we solve
the ”equation” a + b − ab = 0 for the unknown b in R. Since R is a field, the inverse of a
in A is b = a(1 − a)−1 (it exists as a 6= 1). Thus when R is a field, A with ◦ is an abelian
group.
Conversely, we assume that A with ◦ is an abelian group. Since R is a commutative
ring, it suffices to show that any r ∈ R −{0} has a multiplicative inverse. Let r ∈ R −{0, 1}.
Then r ∈ A. Since r 6= 0, r + 1 ∈ A. Since (A, ◦) is a group, there exists a b ∈ A such that
0 = (r + 1) ◦ b = r + 1 + b − (r + 1)b = r + 1 + b − rb − b = r(1 − b) + 1.
Thus r−1 = b − 1 ∈ R, and so R is a field.
√
(3.16) Show that F = {a + b 2 : a, b ∈ Q} is a field, where Q denotes the field of rational numbers.
Proof:
Note that F ⊂ R is a subset of the real number field. We first verify that F is a
subring of R by checking each of the following:
√
2+0 2
∈ F.
2
√
√
a+b 2 a0 +b0 2
,
∈ F , then
2
2
(1) 1 ∈ F , as 1 =
(2) If
√
√
a + b 2 a0 + b0 2
−
=
2
2
4
a−a0
2
0
+ b−b
2
2
√
2
∈ F.
(3) If
√
√
a+b 2 a0 +b0 2
,
2
2
∈ F , then
√
√
a + b 2 a0 + b0 2
·
=
2
2
aa0 +2bb0
2
0
0b √
+ ab +a
2
2
∈ F.
2
Therefore, F is a subring of R.
√
It remains to show that every non zero element in F has an inverse. Let r = a+b2 2 ∈ F −
√
{0}. Since 2 is not a rational number, for any rational numbers a and b with a2 = b2 6= 0,
a2 − 2b2 6= 0. Let a0 =
√
1 0
0 2) ∈ F .
(a
+
b
2
4a
,
a2 −2b2
b0 =
−4b
.
a2 −2b2
Then a0 , b0 are rational numbers and so
√
√
a + b 2 a0 + b0 2
4(a− 2b2 )
·
=
= 1.
2
2
4(a2 − 2b2 )
and so r has a multiplicative inverse in F .
(3.21) (i) If R is a domain, show that if a polynomial in R[x] is a unit, then it is a nonzero
constant (the converse is true if R is a field).
(ii) Show that (2x + 1)2 = 1 in Z4 [x]. Conclude that the hypothesis in part (i) that R be a
domain is necessary.
Proof: (i) Suppose that u(x) ∈ R[x] is a unit. Then ∃v(x) ∈ R[x] such that u(x)v(x) = 1.
Since R is a domain,
deg(u(x)) + deg(v(x)) = deg(1) = 0,
and so deg(u(x)) = 0. It follows that u(x) = u ∈ R − {0}.
If R is a field, then ∀u ∈ R − {0}, there exists r−1 ∈ R, and so r−1 ∈ R[x]. Thus as a
constant polynomial, u ∈ R − {0} is a unit in R[x].
(ii) Note that Z4 is not a domain, as [2] · [2] = [0]. Since (2x + 1)2 = 4x2 + 4x + 1 = 1 in
Z4 [x], thus when R is not a domain, a unit of R[x] does not have to be a non zero constant
in R.
(3.23) If R is a commutative ring and f (x) =
n
P
si xi ∈ R[x] has degree n ≥ 1, define
i=0
its derivative f 0 (x) ∈ R[x] by
f 0 (x) = s1 + 2s2 x + 3s3 x2 + · + nsn xn−1 ;
if f (x) is a constant polynomial, define its derivative to be the zero polynomial. Prove that
the usual rules of calculus hold:
(i) (f + g)0 = f 0 + g 0 .
(ii) (rf )0 = r(f )0 .
5
(iii) (f g)0 = f g 0 + f 0 g.
(iv) (f n )0 = nf n−1 f 0 , for all n ≥ 1.
Proof:
n
P
Let f (x) =
m
P
si xi and g(x) =
i=0
ti xi be two polynomials in R[x]. Without loss
i=0
of generality, we may assume that m ≥ n.
(i). Let aj = 0 for j = n + 1, · · · , m. We have f (x) + g(x) =
m
P
(si + ti )xi . Thus
i=0
(f + g)0 =
m
X
i((si + ti )xi−1 =
i=1
m
X
isi xi−1 +
i=1
(ii). Fix an r ∈ R. Then rf (x) =
n
P
m
X
iti xi−1 = f 0 + g 0 .
i=1
rsi xi . Thus
i=0
(rf (x))0 =
n
X
risi xi−1 = r
i=1
(iii). Let ck =
k
P
si tk−i . Then f (x)g(x) =
m+n
X
ici xi−1 =
m+n−1
X
ci xi . Thus (using k = i − 1)
m+n−1
X
k
X
(k + 1)ck+1 xk

(k + 1)sj tk−j+1  xk
j=0

k
X

k=0
0
m+n−1
X
k=0


k=0
=
nm
P
i=0
i=1
=
isi xi−1 = r(f (x))0 .
i=1
i=0
(f (x)g(x))0 =
n
X

jsj tk−j+1  xk +
j=0
m+n−1
X

k
X

k=0

(k − j + 1)sj tk−j+1  xk
j=0
0
= f (x)g(x) + f (x)g (x).
(iv). To show (iv), we argue by induction on n ≥ 1. When n = 1, this is the definition
of f 0 . Assume that n ≥ 2 and assume that the formula holds for smaller values of n. Define
f 0 (x) = 1 and f n (x) = (f (x))n . Then by (iii) and by induction,
(f n (x))0 = (f (x)f n−1 (x))0 = f (x) · (n − 1)f n−2 (x)f 0 (x) + f n−1 (x)f 0 (x) = nf n−1 (x)f 0 (x).
(3.24) Let R be a commutative ring and let f (x) ∈ R[x].
(i) Prove that if (x − a)2 |f (x), then (x − a)|f 0 (x) in R[x].
(ii) Prove that if (x − a)|f (x) and (x − a)|f 0 (x), then (x − a)2 |f (x).
Proof:
(i) Let g(x) = (x − a)2 . Then by Exercise 3.23(iv), g 0 (x) = 2(x − a). Since
g(x)|f (x), ∃h(x) ∈ R[x] such that f (x) = g(x)h(x). By Exercise 3.23(iii),
f 0 (x) = g 0 (x)h(x) + g(x)h0 (x) = (x − a)(2h(x) + (x − a)h0 (x)).
6
Therefore, (x − a)|f 0 (x).
(ii) Since (x − a)|f (x), we can write f (x) = (x − a)n h(x), where n ≥ 1 in n integer, and where h(x) ∈ R[x] such that (x − a) 6 |h(x). By Exercise 3.25(iii) and (iv),
f 0 (x) = n(x−a)n−1 h(x)+(x−a)n h0 (x). Since (x−a)|f 0 (x), we have (x−a)|((x−a)n−1 h(x)).
Since (x − a) 6 |h(x), we must have n − 1 ≥ 1 and so n ≥ 2.
(3.28) Find the gcd of x2 − x − 2 and x3 − 7x + 6 in Z5 [x], and express it as a linear
combination of them.
Proof: Let f (x) = x2 −x−2 and g(x) = x3 −7x+6. Note that in Z5 [x], g(x) = x3 −2x+1.
Apply division algorithm,
x3 − 2x + 1 = (x2 − x − 2)(x + 1) + (x + 3)
x2 − x − 2 = (x + 3)(x − 4) − 10 = (x + 3)(x − 4).
Therefore, (f (x), g(x)) = x + 3. And
x + 3 = 1 · (x3 − 2x + 1) + (−x − 1)(x2 − x − 2) = g(x) + (−x − 1)f (x).
(3.31) Prove the converse of Euclid’s lemma. Let k be a field and let f (x) ∈ k[x] be a polynomial of degree ≥ 1; if, whenever f (x) divides a product of two polynomials, it necessarily
divides one of the factors, then f (x) is irreducible.
Proof:
Suppose, by contradiction, that f (x) is not irreducible. Then f (x) = a(x)b(x)
such that both a(x) and b(x) has positive degree. It follows that f (x)|a(x)b(x). By the
assumption, f (x)|a(x) or f (x)|b(x). But neither is possible as the comparison of the degrees
both sides would lead to a contradiction.
(3.33) Let k be a field, and let f (x), g(x) ∈ k[x] be relatively prime. If h(x) ∈ k[x],
prove that f (x)|h(x) and g(x)|h(x) imply f (x)g(x)|h(x).
Proof:
Since f (x) and g(x) are relatively prime, ∃s(x), t(x) ∈ k[x] such that s(x)f (x) +
t(x)g(x) = 1. Since f (x)|h(x), there exists a h1 (x) ∈ k[x] such that h(x) = h1 (x)f (x).
Multiply both sides of s(x)f (x) + t(x)g(x) = 1 by h1 (x) to get
s(x)f (x)h1 (x) + t(x)g(x)h1 (x) = h1 (x).
Since g(x)|h(x) and since h(x) = h1 (x)f (x), g(x) divides both functions on the left hand
side of the displayed equality above. Therefore, g(x)|h1 (x), and so g(x)f (x)|h(x).
7
(3.37) (i) Let f (x) = (x − a1 ) · · · (x − an ) ∈ K[x], where k is a field. Show that f (x)
has no repeated roots if and only if the gcd (f, f 0 ) = 1, where f 0 (x) is the derivative of f .
(ii) Prove that if p(x) ∈ Q[x] is an irreducible polynomial, then p(x) has no repeated roots
in C.
Proof:
(i) Suppose that f (x) has a repeated root a. Then (x − a)2 |f (x). By Exercise
3.24(i), (f, f 0 ) 6= 1.
Conversely, suppose that d = (f, f 0 ) 6= 1. Since K is a a field, deg(d) > 0. Let x − a be
a factor of d. Then (x − a)|f and (x − a)|f 0 . By Exercise 3.24(ii), (x − a)2 |f , and so f has
a repeated root.
(ii) By contradiction, suppose that p(x) has a repeated root. Then by (i), (p, p0 ) 6= 1.
Since p(x) is an irreducible polynomial in Q[x], either (p, p0 ) = 1 or (p, p0 ) = p. Therefore
as (p, p0 ) 6= 1, we must have p|p0 . But deg(p0 (x)) = deg(p(x)) − 1, and so p|p0 is not possible.
This contradiction implies that p(x) cannot have a repeated root.
(3.39) (i) let φ : A → R be an isomorphism, and let ψ : R → A be its inverse. Show
that ψ is an isomorphism.
(ii) Show that the composite of two homomorphisms (isomorphisms) is again a homomorphism(isomorphism).
(iii) Show that A ∼
= R defines an equivalence relation on the class of all commutative rings.
Proof:
(i) For any a, b ∈ R, since φ is an isomorphism, ∃a0 , b0 ∈ A such that φ(a0 ) = a,
φ(b0 ) = b, φ(a0 + b0 ) = a + b and φ(a0 b0 ) = ab. Since ψ is the inverse map of φ, we have
ψ(a) = a0 and ψ(b) = b0 , and
ψ(a + b) = a0 + b0 = ψ(a) + ψ(b), ψ(ab) = a0 b0 = ψ(a)ψ(b).
Therefore, ψ is a ring homomorphism. Since ψ is the inverse map of a bijection, ψ is also
a bijection, and soψ is a ring isomorphism.
(ii) Let f : R 7→ R0 and g : R0 7→ R00 be ting homomorphisms. Then for any a, b ∈ R,
(gf )(a + b) = g(f (a) + f (b)) = g(f (a)) + g(f (b)), (gf )(ab) = g(f (a)f (b)) = g(f (a))g(f (b)).
Thus (gf ) is also a ring homomorphism. If both f and g are isomorphisms, then as the
composition of bijections is also a bijection, (gf ) is also a ring isomorphism.
(iii) Let Ω denote the set of all commutative rings. Define a relation ∼
= on Ω as follows:
for R, R0 ∈ Ω, R ∼
= R0 if and only if R is isomorphic to R0 . Then since the identity map is
8
∼ R, ∀R ∈ Ω, and so ∼
an isomorphism, R =
= is reflexive. If R ∼
= R0 , then by (i), R0 ∼
= R also,
0
00
0
0
00
and so ∼
= is symmetric. Now let R, R , R ∈ Ω be such that R ∼
= R and R ∼
= R . Then by
(ii), we also have R ∼
= R00 and so ∼
= is transitive. Thus ∼
= is an equivalence relation.
(3.42) Let R be a commutative ring. Show that the function : R[x] 7→ R, defined by
(a0 + a1 x + a2 x2 + · + an xn ) = a0 , is a homomorphism. Describe ker in terms of roots of
polynomials.
Proof: Let f (x) =
n
P
i=0
m
P
si xi and g(x) =
ti xi be two polynomials in R[x]. Then
i+0
(f + g) = s0 + t0 = (f ) + (g), (f g) = s0 t0 = (f )(g).
Thus is a ring homomorphism. Let E0 denote the kernel of . Then E0 = {f (x) ∈ R[x] :
f (x) =
n
P
si xi } = {f (x) ∈ R[x] : f (0) = 0}.
i=1
(3.45) Suppose that p is a prime integer.
(i) Show that every element a ∈ Zp has a pth root (i.e., there is b ∈ Zp with a = bp ).
(ii) Let K be a field that contains Zp as a subfield. For every positive integer n, show that
n
the function φn : K → K, given by φ(a) = ap , is a ring homomorphism.
Proof: (i) Recall that Fermat’s Theorem (Theorem 1.24) states that if p is a prime, then
n
for any a ∈ Zp , and for any integer n ≥ 1, ap = a in Zp . In particular, a = ap in Zp .
(ii) Then ch(K) = p. That is, when 1 = 1K , then p · 1 = 0. This implies that for any
a, b ∈ K, (a + b)p = ap = +bp . Arguing by induction on n, we have, for any positive integer
n, and for any a, b ∈ K,
n
n
n
(a + b)p = ap + bp .
n
Thus φn (a + b) = φn (a) + φn (b). The law of exponents indicated that φn (ab) = (ab)p =
n
n
ap bp = φn (a)φn (b), and so φn is a ring homomorphism.
9