Math 456
February 23, 2016
QUIZ SOLUTIONS
QUIZ 1
Let R be a commutative ring with 1 6= 0 and a, b ∈ R. Mark the following as True or False:
(1) ab = 0 implies ba = 0. Answer: True, since we assumed that R is commutative.
(2) If a is a unit of R, then so is −a. Answer: True. We have a theorem that says (−a)(−b) =
ab for all a, b ∈ R. So if a is a unit, then aa−1 = a−1 a = 1 implies that (−a)(−a−1 ) =
(−a−1 )(−a) = 1 and so the multiplicative inverse of −a is −a−1 .
(3) a has an additive inverse. Answer: True. Since hR, +i is an abelian group.
(4) If a and b are units, then so is a + b. Answer: False. Counterexample: a = 1 and b = −1
in R = Z.
(5) Subtraction is an associative operation on R. Answer: False. Counterexample: (1−1)−1 6=
1 − (1 − 1) in R = Z.
(6) If 2 = 0 in R, then a = −a. Answer: True. a + a = 2a = 0a = 0, so −a = a.
(7) 2 = 3 is impossible in R. Answer: True. If 2 = 3 then, by cancellation, 1 = 0, contradicting our assumption that 1 6= 0.
(8) If 3a = 0 and a 6= 0, then 3 = 0. Answer: False. Counterexample: In R = Z6 we have
3 · 2 = 0 with 2 6= 0 and 3 6= 0.
(9) If 3a = 0 then −a = 2a. Answer: True. Since a + (a + a) = 0, the additive inverse of a is
a + a = 2a.
(10) 3 is a unit in Z13 . Answer: True. gcd(3, 13) = 1 so 3 is a unit OR since 3 · 10 = 1 in Z13 ,
the multiplicative inverse of 3 is 10 and 3 is a unit.
QUIZ 2
(1) A domain D has 8 elements. How many of these elements are units? How many of these
elements are zero divisors? Answer: By Theorem 19.11, a finite domain is a field. So all
nonzero elements of D are units. Thus there are 7 units. Units cannot be zero divisors, and
0 is not a zero divisor by definition, so there are no zero divisors in D.
(2) An element a of a ring is idempotent if a2 = a. Show that a domain contains exactly 2
idempotents. Answer: Suppose that R is a domain. If a ∈ R and a2 = a then a2 − a = 0
or a(a − 1) = 0. In a domain this implies that a = 0 or a = 1. Thus R contains at most two
idempotents, 0 and 1. By the definition of a domain, 1 6= 0 and so R contains exactly two
idempotents.
(3) Define a function φ : C → C by φ(a+ib) = b+ia for all a, b ∈ R. Is φ a ring homomorphism?
Explain. Answer: No. We proved that any nontrivial homomorphism between fields sends
1 to 1. Since φ(1) = i 6= 1, φ is not a homomorphism. OR φ(i · i) = φ(−1) = −i whereas
φ(i)φ(i) = 1 · 1 = 1. Thus φ(i · i) 6= φ(i)φ(i) and φ is not a homomorphism.
QUIZ 3
Suppose that F is a field and σ ∈ F is a zero of f (x) = x4 + x3 + x2 + x + 1 ∈ F [x].
(1) Show that σ is a zero of g(x) = x5 − 1.
Answer: g(σ) = σ 5 − 1 = (σ − 1)(σ 4 + σ 3 + σ 2 + σ + 1) = (σ − 1)f (σ) = 0.
(2) Show that α = σ + σ 4 is a zero of h(x) = x2 + x − 1.
Answer:
h(α) = α2 + α − 1 = (σ + σ 4 )2 + (σ + σ 4 ) − 1
= σ 2 + 2σ 5 + σ 8 + σ + σ 4 − 1
= σ 2 + 2 + σ 3 + σ + σ 4 − 1 = f (σ) = 0
We are using the equation σ 5 = 1 obtained from (1).
1
2
(3) Show that 2α + 1 is a square root of 5.
Answer:
(2α + 1)2 − 5 = 4α2 + 4α + 1 − 5 = 4(α2 + α − 1) = 4h(α) = 0
OR
(2α + 1)2 − 5 = (2(σ + σ 4 ) + 1)2 − 5
= 4σ 8 + 8σ 5 + 4σ 4 + 4σ 2 + 4σ − 4
= 4σ 3 + 8 + 4σ 4 + 4σ 2 + 4σ − 4
= 4(σ 4 + σ 3 + σ 2 + σ + 1)
=0
QUIZ 4
(1) Find the quotient and remainder of f (x) = x5 − 1 divided by g(x) = 2x2 + 1 in Z7 [x].
Answer: Quotient: 4x3 + 5x; Remainder: 2x + 6.
(2) Complete the following definition: A nonconstant polynomial f ∈ F [x] is irreducible if. . .
Answer: f has no nontrivial factorizations, OR f = gh with g, h ∈ F [x] implies deg g = 0
or deg h = 0.
(3) Is f (x) = x3 + 2 reducible over Z5 ? Explain. Answer: Since f (2) = 0, f has a zero in Z5
and so f is reducible. In fact, f (x) = (x − 2)(x2 + 2x + 4).
QUIZ 5
(1) Let α ∈ C be a zero of x2 + 2x√− 4 ∈ Q[x]. √
Find irr(α + 1, Q). Explain. Answer: By the
quadratic formula, α is −1 + 5 or −1 − 5. Either way, α + 1 is a square root of 5 and
a zero of x2 − 5. Since x2 − 5 ∈ Q[x] is monic and irreducible (by Eisenstein, for example),
irr(α + 1, Q) = x2 − 5.
OR
Set β = α + 1. Then α = β − 1 and so
0 = α2 + 2α − 4 = (β − 1)2 + 2(β − 1) − 4 = β 2 − 5.
Hence β p
is a zero of x2 − 5 ∈ Q[x] and irr(α + 1, Q) = x2 − 5 as above.
√
3
(2) Let α =
3 + 1.
(a) Find deg(α, Q). Explain.
√
(b) Show that deg(α, Q( 3)) ≤ 3. Explain.
√
√
√
Answer: We have,√of course, α3 = 3+1. So α is a zero of x3 − 3+1 ∈ Q(√ 3)[x]. This
implies that deg(α, Q(√ 3)) ≤ 3. (Later we will be able to show that deg(α, Q( 3)) = 3.)
Squaring α3 − 1 = 3 on both sides we get (α3 − 1)2 = 3, or x6 − 2x3 − 2 so α is a zero
of x6 − 2x3 − 2 ∈ Q[x]. This polynomial is irreducible over Q by Eisenstein with p = 2. So
irr(α, Q) = x6 − 2x3 − 2 and deg(α, Q) = 6.
3
QUIZ 6
Suppose that α ∈ C has irr(α, Q) = x4 + 1. Then deg(α, Q) = 4 and every element of Q(α) can be
written uniquely in the “standard” form a + bα + cα2 + dα3 with a, b, c, d ∈ Q. Let β = (α2 + 1)/α.
(1) Write β in standard form. Answer: From α4 = −1 we get 1/α = −α3 . So β = (α2 +1)/α =
α + 1/α = α − α3 .
(2) Find irr(β, Q).
Answer: Because α4 = −1 we get β 2 = (α − α3 )2 = α2 − 2α4 + α6 = 2,
and β is a zero of x2 − 2 ∈ Q[x]. Since this polynomial is irreducible (by Eisenstein, for
example), irr(β, Q) = x2 − 2.
(3) Show that α 6∈ Q(β). Answer: The polynomial x4 + 1 has no real zeros since its value
√
positive for all real x. This means that α is not a real number. On the other hand, β = ± 2
is real and so Q(β) is contained in R. Thus α 6∈ Q(β).
OR
If α ∈ Q(β), then the degree four extension Q(α) is contained in the degree two extension
Q(β). Not possible.
OR
Every element in Q(β) can be written uniquely in the form a + bβ = a + b(α − α3 ) for some
a, b ∈ Q. In particular, if α ∈ Q(β) then α = a + b(α − α3 ) for some a, b ∈ Q. But this
violates the uniqueness of the standard forms in Q(α).