1. first note that
0 = ω 3 − 1 = (ω − 1)(ω 2 + ω + 1)
ω 6= 1 ⇒ ω 2 + ω + 1 = 0
a)
N (1 − ω) = (1 − ω)(1 − ω) = (1 − ω)(1 − ω 2 ) = 1 − ω − ω 2 + ω 3 = 3
so 1 − ω is prime because it’s norm is a prime number.
b)
N (a + bω) = (a + bω)(a + bω) = (a + bω)(a + bω 2 ) = a2 + ab(ω + ω 2 ) + b2
= a2 − ab + b2
a + bω
⇔ N (a + bω) = 1
is unit
the solutions of the equation a2 − ab + b2 = 1 are
(a, b) = (±1, 0), (0, ±1), (1, 1), (−1, −1)
because if ab ≥ 0 we can write
a2 − ab + b2 = (a − b)2 + ab = 1
which has solutions
(a, b) = (1, 1), (−1, −1), (±1, 0), (0, ±1)
if ab < 0 then we can write
a2 − ab + b2 = (a + b)2 − 3ab = 1
which has no solution. so we get the following 6 unit elements
±1, ±ω, ±ω 2
c) If we have a factorization q = xy and x, y are not unit then
q 2 = N (x)N (y) ⇒ N (x) = N (y) = q
1
but if x = a+bω then q = a2 −ab+b2 . this equation has no integral solution.
because if
q = a2 − ab + b2 = (a + b)2 − 3ab ≡ (a + b)2
mod 3
we get contradiction because
a + b ≡ 0, 1, 2 mod 3 ⇒ (a + b)2 ≡ 0, 1 mod 3
but the assumption is that q ≡ 2 mod 3.
d) we have
7 = (2 + 3ω)(2 + 3ω 2 )
and
N (2 + 3ω) = N (2 + 3ω 2 ) = 7
as their norm is 7 a prime number they are prime. so this is the prime factorization for 7. for finding such a factorization we try to write 7 as a norm of
some element. so we should solve this equation
7 = a2 − ab + b2
which has a solution (a, b) = (2, 3).
2. we have
(1 −
and so 1 +
√
√
2)(1 +
√
2) = −1
2 is a unit. the set
√
{±(1 + 2)n : n = 0, ±1, ±2, ...}
is an infinite set of unit elements of R. in fact it is the group of units of R.
3. We prove that any non zero and non unit element a ∈ R can be written
as a product of irreducibles by induction on N (a). If N (a) = 1 then a is unit by
the assumption of the problem and so there is nothing to prove. If N (a) = n > 1
and a is irreducible we are done. otherwise a = bc for non units b, c. then by
N (a) = N (b)N (c) we have N (b), N (c) < N (a)(note that b is unit iff N (b) = 1).
by induction we can write
b = p1 ...pm
c = q1 ...qn
for some irreducible elements p1 , ..., pm , q1 , ...qn so we have
a = bc = p1 ...pm q1 ...qn
but it mean that we can write a as a product of irreducibles.
2
4. If d = GCD(a, b) we know from the Theorem 3. on page 38 that d =
ax0 + by0 for some x0 , y0 ∈ Z. Now if d|c then c = nd for some n ∈ Z. so
d = ax0 + by0 ⇒ c = nd = a(nx0 ) + b(ny0 )
x := nx0
y := ny0
on the other hand if c = ax + by for some x, y ∈ Z then
d|a, d|b ⇒ d|ax + by = c
from GCD(7, 12) = 1 we know that the answer is yes. in fact
12(1) + 7(2) = 26
so we have
12x + 7y = 12(1) + 7(2) ⇒ 12(x − 1) = 7(2 − y)
⇒ 7|12(x − 1) ⇒ 7|x − 1 ⇒ x − 1 = 7t
t ∈ Z ⇒ x = 1 + 7t
⇒ 12(7t) = 7(2 − y) ⇒ 2 − y = 12t ⇒ y = 2 − 12t
⇒ (x, y) = (1 + 7t, 2 − 12t) t ∈ Z
on the other hand if for some t ∈ Z (x, y) = (1 + 7t, 2 − 12t) then
12x + 7y = 12(1 + 7t) + 7(2 − 12t) = 26
so they are all the solutions of the equation 12x + 7y = 26.
5. We have X 2 + X + 1 = 0 which implies
2
X +X +1=0
or
X(−X − 1) = 1
so
X
−1
= −X − 1
3