SOLUTIONS TO THE THIRD PROBLEM SHEET FOR
ALGEBRAIC NUMBER THEORY M3P15
AMBRUS PÁL
1. These are:
±1, ±i, ±1 ± i, ±2, ±2i, ±1 ± 2i, ±2 ± 1i, ±2 ± 2i, ±3, ±3i, ±1 ± 3i, ±3 ± i.
2. We have:
N (3 + 4i) = 25 = 52 , N (4 + 5i) = 41, N (5 + 6i) = 61,
N (6 + 7i) = 85 = 5 · 17, N (7 + 8i) = 113.
So 6 + 7i is not a prime, but as 41, 61, 113 are prime numbers, 4 + 5i, 5 + 6i, 7 + 8i
are primes. As 5 ≡ 1(4), the Gaussian integer 3 + 4i is not a prime, in fact
(2 + i)2 = 3 + 4i.
3. We have:
N (1+5i) = 26 = 2·13, N (1+6i) = 37, N (1+7i) = 50 = 2·52 , N (1+8i) = 65 = 5·13.
We immediately get that 1 + 6i is prime. In order to factorise the other Gaussian
integers, using the prime factorisation of the norms, we know which primes we
should try to divide these elements by; primes of norm 2, 5, 13, which are 1 ± i, 1 ±
2i, 2 ± 3i, as well as these multiplied by ±i. After some trial and error we get:
1 + 5i = (1 − i)(−2 + 3i), 1 + 7i = i(1 + i)(2 − i)2 , 1 + 8i = (1 − 2i)(−3 + 2i).
4. Note that
a
ab
=
.
b
N (b)
Using this we get that
20 21 7 + 3i
27
5 4 + 7i
27 31
2 + 5i
=
+ i,
=
− i,
=
+ i.
5 + 2i
29 29 3 + 2i
13 13 5 + i
26 26
Approximating these complex numbers with Gaussian integers, we get:
2 + 5i = (5 + 2i)(1 + i) − 1 − 2i, 7 + 3i = (3 + 2i)2 + 1 − i,
4 + 7i = (5 + i)(1 + i) + i.
5. We have:
N (−9 + 7i) = 130 = 2 · 5 · 13, N (7 − 4i) = 65 = 5 · 13.
Let’s divide the first by the second and compute the residue:
−9 + 7i
91 13
7 1
= − + i = − + i.
7 − 4i
65 65
5 5
We get:
−9 + 7i = (−1)(7 − 4i) + (−2 + 3i).
Date: February 22, 2017.
1
2
AMBRUS PÁL
Doing the same with the pair 7 − 4i, 2 + 3i we get:
7 − 4i = (−2 − i)(−2 + 3i),
so the greatest common divisor is −2 + 3i.
6. Since an element is a unit if and only if its norm is ±1, we need to compute the
norms.
7. (a) This element is clearly non-zero, and not a unit. If it is a product of two
non-units, then its norm is a product of two integers of absolute value greater than
1, a contradiction.
(b) For n = 0 and n = 6 we can use (a). An easy calculation shows that OK
has no elements of norm 2, 3 or 7. Since the norm of an element of OK which is
not a unit and not an irreducible, is a product √of two integers which are norms,
that
we see
for n = √
1, 2, 3, 4 the element n + −5 irreducible. It is clear √
√ that √
5 + −5 = −5(1 − −5) is not irreducible. It remains to understand 7 + −5
whose norm is 54. So lets√find all elements
√ of norm 6 and 9. Apart from ±3, which
−5
and
±2
±
−5. A little experimentation shows that
is useless,
we
have
±1
±
√
√
√
−5)(2 − −5),√so this one is
7 + −5 = (1 + √
√ certainly√not irreducible.
√
8. Write a = x + y d, then x − y d = (z1 + z2 d)(x + y d) (where z = z1 + z2 d)
gives a system of two linear equations in x and y:
(z1 − 1)x + z2 dy,
0
=
0
= z2 x + (z1 + 1)y,
which has zero determinant:
z1 − 1
z2 d
det
= z12 − dz22 − 1 = N (a) − 1 = 0,
z2
z1 + 1
because a has norm 1. So by linear algebra√it is has a non-zero
√ solution.
−5].
So
N
(2,
1
+
−5) = 2 ,√N (3, 1 +
9.
Since
−5
is
3
mod
4,
we
have
O
=
Z[
K
√
√
√
−5) = 3, N (3, 2 + −5) = 3 and N (7, 3 + −5) = 7, since 1, 1 + −5 is a
Z-basis for OK . These are all primes. Alternatively
their norm
√ we can compute
√
√ by
multiplying√them with their conjugates (2, 1 − −5) , (3, 1 − −5), (3, 2 − −5)
and (7, 3 − −5), respectively.
10. Note that I has norm 2 by the above. Since there is no element of OK of norm
2 we get that I is not principal. We have I 2 = (2) which is clearly principal.
11. Clearly (S) is contained in any ideal containing S. So we only need to show
that S is an ideal. However (S) is trivially closed under addition and multiplication
by elements of R, and it contains zero if S is non-empty, so it is an ideal indeed.
12. These are all very easy, and quite similar, so I will only write down a proof
for (S) · (S 0 ) = (S · S 0 ). Clearly S · S 0 ⊆ (S) · (S 0 ), and the right hand side is an
ideal, so (S · S 0 ) ⊆ (S) · (S 0 ) by problem 11 above. If a1 s1 + a2 s2 + · · · ∈ S and
a01 s01 + a02 s02 + · · · ∈ S 0 , with si ∈ S, si ∈ S 0 , ai , a0i ∈ R, then
X
(a1 s1 + a2 s2 + · · · )(a01 s01 + a02 s02 + · · · ) =
ai a0j si s0j ∈ (S · S 0 ),
i,j
so the ideal generated by the product of (S) and (S 0 ) as sets is in (S · S 0 ). But the
former is (S) · (S 0 ), so (S) · (S 0 ) ⊆ (S · S 0 ).