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43. Here is the picture:
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Black dots form I (equivalently the coset 0 + I), white dots the coset 1 + I, and all the dots
together Z[i]. There are two cosets, so the quotient ring Z[i]/I has two elements, 0 and 1, the
images of 0 ∈ Z[i] and 1 ∈ Z[i] under the quotient homomorphism Z[i] Z[i]/I.
To see that S = R/I is ∼
= Z/2Z, note that 0 + a = a, 0 × a = 0 and 1 × a = a for every a in
any ring, and this determines all the sums and products except 1 + 1. Showing that 1 + 1 = 0
(as it is in Z/2Z) can be done in (countably) infinitely many ways, of which here are four:
• Adding 1 maps the coset 0 + I to 1 + I, and so it must map 1 + I to 0 + I.
• If 1 + 1 were 1, subtracting 1 from both sides would give 1 = 0.
• Because 2 = (1 + i)(1 − i) is divisible by 1 + i in R, its image in S is zero.
• Every non-zero ring contains either Z or Z/nZ for some n ≥ 2 as a subring, and size
considerations imply that S cannot contain anything except Z/2Z.
44. Because the polynomial is primitive, it is irreducible over Z if and only if it is irreducible
over Q. As it has degree 3, it is irreducible over Q if and only if it has no roots in Q. Moreover,
if a = n/d ∈ Q is a root, then n|4 and d|1, so a ∈ {±1, ±2, ±4}. Checking these one by one, we
find that none of them is a root.
On the other hand, over Z/3Z it has a root a = 2. In fact the polynomial is x3 + 1 = (x + 1)3 ;
the coefficients 3x and 3x2 that one normally sees in the binomial expansion of (x+1)3 disappear
because 3 = 0 in Z/3Z.
45. The answer is (unique up to units ±1, ±i)
x4 − 3x3 + 2ix2 − (5i + 1)x − 3i + 3 = (x − 3)(x3 + (1 + i)2 x + i(1 + i))
To find the first factor, observe that any root of the given polynomial a = n/d ∈ Q[i] written in
lowest terms satisfies d|1 and n|(−3i + 3) = 3 · (1 + i) · unit. Trying all the possibilities results in
locating the root a = 3. The remaining factor of degree 3 is irreducible by Eisenstein’s criterion
for the prime 1 + i.
47. (1) It is a homomorphism, because complex conjugation preserves addition and multiplication, and it is both injective and surjective because its composition with itself is identity.
(2) f (z) = 0 =⇒ f (z) = 0 =⇒ f (z̄) = 0. For the last equality, observe that hitting the
monomials in f (z) by complex conjugation does not affect the coefficients of f (as they are real)
and only changes all z’s to z̄’s.
(3) Assume f ∈ R[x] is of degree > 1 and irreducible. By (2), the roots f in C (that always
exist) are either real, or come in complex conjugate pairs. If a is a real root, then x − a divides
f in R[x], so f cannot be irreducible. Otherwise, say a1 , ā1 , a2 , ā2 , . . . an , ān ∈ C are the roots of
f in C. Then f = g1 · · · gn where gi = (x − ai )(x − āi ) = x2 − (ai + āi )x + ai āi for i = 1, . . . , n.
But it is clear that each gi has real coefficients, so f is reducible over the reals, unless deg f = 2
and f is one of the gi .
48. Over R every polynomial of odd degree is reducible (see previous exercise), and over C
every polynomial of degree > 1 has roots and is therefore reducible. Finally, note that Z and
Z[ 12 ] are both UFDs, so for primitive polynomials irreducibility in each one of them is equivalent
to irreducibility over Q, their common field of fractions. This solves the question for f and h.
For g the answer is different, because 2 is a unit in Z[ 21 ].
49. (1) First of all, if a ∈ K × = K \ {0}, then so is a2 , and so s does map K × to itself.
It also preserves multiplication, 1, and multiplicative inverses (uv = 1 =⇒ u2 v 2 = 1), and is
therefore a group homomorphism.
(2) The kernel of s is the set of a ∈ K for which a2 = 1. The polynomial x2 −1 = (x−1)(x+1)
has only ±1 as roots, so there are two of them if −1 6= 1 (equivalently 2 6= 0 in K) and one if
−1 = 1 (equivalently 2 = 0 in K).
(3) We proved that s : (Z/pZ)× → (Z/pZ)× has kernel of size 2, so it is a 2-to-1 map. In other
words, the p − 1 non-zero elements group in pairs (namely, {a, −a}) that square to the same
element (namely a2 ), and there are therefore (p − 1)/2 squares and equally many non-squares. If
a is any non-square, then the map x 7→ ax from (Z/pZ)× to itself sends squares to non-squares
(if an element of the form ab2 is a square, then so is a), and, again by counting, must send
non-squares to squares as a result.
(4) Apart from zero, 1 = (±1)2 , 4 = (±2)2 and 2 = (±3)2 are squares, and 3, 5 and 6 are not.
50. If p = 2 then x2 − 2 = x2 has zero as a root. Suppose p is odd. If −1 or 2 is a square in
Z/pZ, we are done. Otherwise −1 · 2 must be a square by the previous exercise, being a product
√
2 + 1)(x2 − 2)(x2 + 2) has a root for every p. But ±1, ± 2,
of two
non-squares.
It
follows
that
(x
√
±i 2 are not in Q, and so it has no rational roots.