MATH470 Winter 2016
Solutions to Assignment 3
Due Friday February 19, 2016
1. Let R be a commutative ring with 1.
(a) (Section 7.3 no 29) Prove that the set of nilpotent elements in R is an ideal,
called the nilradical of R and denoted N(R). [Hint: Use the Binomial
Theorem (Section 7.3 no 25) to show that N(R) is closed under addition.]
(b) (Section 7.4 no 30) Let I be an ideal of R and define
rad(I) = r ∈ R : rn ∈ I for some n ∈ Z+
called the radical of I. Prove that rad(I) is an ideal containing I and that
rad(I)/I is the nilradical of the quotient ring R/I, i.e. rad(I)/I = N(R/I).
Solutions: (a) Let x, y be nilpotent elements and choose m such that xn = 1
and y m = 1. Then,
n+m
X n + m
n+m
(x + y)
=
xi y n+m−i = 0
i
i=0
since in each monomial, either the power of x or the power of y is 0, and then
each monomial is 0. Also, for any r ∈ R, (rx)n = rn xn = 0. This implies that
the set of nilpotent elements is an ideal.
(b) Clearly, I ⊆ rad(I). Let r, s ∈ rad(I) such that rn ∈ I and sm ∈ I. Then, as
in (a), (r + s)n+m ∈ I by the binomial theorem (we have that ri or sm+n−i ∈ I
for each index i, and then each term of the binomial expansion belongs to I
since I is an ideal).
Also, if rn ∈ I, and a ∈ R, then (ar)n = an rn ∈ I since I is an ideal. This
shows that rad(I) is an ideal of R.
Finally, we have that
r ∈ R/I : rn = 0 in R/I for some n ∈ Z+
= r ∈ R/I : rn ∈ I for some n ∈ Z+
N(R/I) =
= rad(I)/I.
2. (Section 7.1 no 23) Let D be a squarefree integer, and let O be the ring of
√
integers in the quadratic field Q( D).
(a) For any positive integer f , prove that Of = Z[f ω] = {a + bf ω : a, b ∈ Z}
is a subring of O containing the identity.
(b) Prove that [O : Of ] = f (the index as an additive abelian group).
(c) Prove conversely that a subring of O containing the identity and having
finite index f in O (as an additive abelian group) is equal to Of .
Solutions: (a) Clearly, 1 ∈ Of . Also, for z1 = a1 + b1 f ω and z2 = a2 + b2 f ω,
we have that
z1 + z2 = (a1 + a2 ) + f ω(b1 + b2 ) ∈ Of
z1 z2 = a1 b1 + b1 b2 f 2 ω 2 + f ω(b1 a2 + a1 b2 )
(
√
a1 b1 + Db1 b2 f 2 + f D(b1 a2 + a1 b2 ) ∈ Of
if D ≡
6 1 (mod 4);
=
D−1
2
a1 b1 + 4 b1 b2 f + f ω(b1 a2 + a1 b2 + f b1 b2 ) ∈ Of if D ≡ 1 (mod 4)
This shows that Of is a subring of O.
(b) We claim that
S = {0, ω, 2ω, . . . , (f − 1)ω}
is a set of representatives for the quotient group O/Of . Indeed, let z = a+bω ∈
O, and write b = f q + r, with 0 ≤ r < f . Then, rω ∈ S, and
z − rω = a + (b − r)ω = a + f qω ∈ Of ⇒ a + bω ∼ rω.
We also have to show that the elements of S are not equivalent, which is true
since iω − jω = (i − j)ω 6∈ Of for 0 ≤ i, j < f , i 6= j.
(c) Let R be a subring of O containing 1 such that the quotient group O/R has
index f , i.e. contains f elements. Since 1 ∈ R which is a subring, we have that
Z ⊆ R. Also, by Lagrange’s theorem, for every x ∈ O/R, f x = 0, which means
that for every a + bω ∈ O, we have that f (a + bω) = f a + f bω ∈ R by definition
of the quotient group. Then
f a + f bω − f a = f bω ∈ R,
for any b ∈ Z, and then Of ⊆ R. Since both quotients have index f , this implies
R = Of .
2
3. (Section 7.3 no 12) Let D be an integer which is not a perfect square in Z and
let
(
!
)
a
b
S=
: a, b ∈ Z .
Db a
(a) Show that S is a subring of M2 (Z).
(b) If D is not a perfect square in Z, then the map
√
ϕ:
Z[ D] −→ S
!
√
a b
a + b D 7→
Db a
is a ring isomorphism.
(c) If D ≡ 1 mod 4, prove that the set
(
!
)
a
b
S=
: a, b ∈ Z
(D − 1)b/4 a + b
is a subring of M2 (Z) and is isomorphic to the quadratic integer ring O.
Solutions: (a) We have that
!
!
a1
b1
a2
b2
−
=
Db1 a1
Db2 a2
!
!
a1
b1
a2
b2
=
Db1 a1
Db2 a2
(a1 − a2 ) (b1 − b2 )
D(b1 − b2 ) (a1 − a2 )
!
a1 a2 + Db1 b2
a1 b 2 + b 1 a2
D(b1 a2 + a1 b2 ) Db1 b2 + a1 a2
which shows that S is a subring of M2 (Z).
√
√
(b) Let z1 = a1 + Db1 and z2 = a2 + Db2 . By (a), we have that
√
ϕ (z1 + z2 ) = ϕ((a1 + a2 ) + D(b1 + b2 ))
!
(a1 + a2 ) (b1 + b2 )
=
= ϕ(z1 ) + ϕ(Z2 )
D(b1 + b2 ) (a1 + a2 )
√
ϕ(z1 z2 ) = ϕ(a1 a2 + Db1 b2 + D(a1 b2 + b1 a2 ))
!
a1 a2 + Db1 b2
a1 b 2 + b 1 a2
=
= ϕ(z1 )ϕ(z2 )
D(b1 a2 + a1 b2 ) Db1 b2 + a1 a2
3
!
which shows that ϕ is a ring homomorphism. Clearly, ϕ is surjective, and
(
!)
√
0 0
ker(ϕ) =
z = a + Db : ϕ(z) =
0 0
n
o
√
=
z = a + Db : a = 0, b = 0 = {0}
which shows that ϕ is injective, and is then an isomorphism.
(c) Let z1 = a1 + b1 ω and z2 = a2 + b2 ω ∈ O, where ω = (1 +
√
1+2 D+D
D−1
2
ω =
=ω+
,
4
4
√
D)/2. Then,
and
z1 + z2 = (a1 + a2 ) + (b1 + b2 )ω
z1 z2 = a1 a2 + ω 2 b1 b2 + ω (a1 b2 + a2 b1 )
D−1
= a1 a2 +
b1 b2 + (a1 b2 + a2 b1 + b1 b1 ) ω
4
Consider the map
ϕ:
O −→ S
a + bω
7→
a
b
D−1
b a+b
4
!
.
Then,
ϕ(z1 + z2 ) =
ϕ(z1 )ϕ(z2 ) =
!
a1 + a2
b1 + b2
= ϕ(z1 ) + ϕ(z2 )
D−1
(b
+
b
)
(a
+
a
)
+
(b
+
b
)
1
2
1
2
1
2
4
!
!
a1
b1
a2
b2
D−1
D−1
b 1 a1 + b 1
b 2 a2 + b 2
4
4
!
=
a1 a2 + D−1
b1 b2
a1 b 2 + a2 b 1 + b 1 b 1
4
D−1
(a1 b2 + a2 b1 + b1 b2 ) D−1
b 1 b 2 + a1 a2 + b 1 b 2 + a1 b 2 + a2 b 1
4
4
!
ϕ(z1 z2 ) =
b1 b2
a1 b 2 + a2 b 1 + b 1 b 1
a1 a2 + D−1
4
D−1
(a1 b2 + a2 b1 + b1 b1 ) a1 a2 + D−1
b 1 b 2 + a1 b 2 + a2 b 1 + b 1 b 1
4
4
which shows that ϕ is a ring homomorphism. Then, Im(ϕ) = S is a subring of
M2 (Z), and since ϕ is injective, S ' O.
4
√
4. (section 8.1 no 8) Let F = Q( D) be a quadratic field with the usual norm
√
N (a + b D) = a2 − Db2 , and with associated quadratic integer ring O.
(a) Suppose that D = −2, −3, −7 or −11. Prove that O is a Euclidean Domain
with respect to N . [Hint: Apply the same proof as for D = −1 in Section
8.1. For D = −3, −7, −11, show that every element in F differs from an
element in O by an element whose norm is at most (1 + |D|)/16 which is
less than 1 for these values of D. Plotting the points of O in C maybe be
helpful.]
(b) Suppose that D = −43, −67 or −163. Prove that O is not a Euclidean
Domain with respect to any norm. [Hint: Apply the same proof as for
D = −19 in Section 8.1.]
Solutions: We first consider D = −2 i.e.
√
√
O = Z[ −2] = a + b −2 : a, b ∈ Z .
√
√
√
√
Let α = a1 + b1 −2 and β = a2 + b2 −2 6= 0. Let x + y −2 ∈ Q( −D) such
that
√
α
= x + y −2.
β
Writing
√
√ √ x + y −2 = m + n −2 + r + s −2 ,
with m, n ∈ Z, r, s ∈ Q and |r|, |s| ≤ 1/2, we have
√ √ α = m + n −2 β + r + s −2 β,
√
√ where m + n −2 and r + s −2 β are in O and
√
√ 1
1
N (r + s −2)β = N r + s −2 N (β) ≤
+2
N (β) < N (β).
4
4
We now consider D = −3, −7, −11 In those 3 cases, D ≡ 1 (mod 4) and
O = Z[ω]. Working as above, let α = a1 + b1 ω and β = a2 + b2 ω 6= 0. Let
√
√
x + y D ∈ Q( D) such that
√
α
= x + y D.
β
5
Let n = k/2 with k ∈ Z such that y = n + s with |s| ≤ 1/4. Now, let m = `/2
where ` has the same parity as k be such that x = m + r with |r| ≤ 1/2. (We
cannot do better than that for r because of the parity condition necessary to
√
have m + n D ∈ O.) Then,
√
√ √ x+y D = m+n D + r+s D ,
√
with m + n D ∈ O, r, s ∈ Q with |s| ≤ 1/4 and |r| ≤ 1/2, and
√ √ α = m + n D β + r + s D β,
√ √
where m + n D and r + s D β are in O and
√
√ 1
1
N (r + s D)β = N r + s D N (β) ≤
+ |D|
4
16
1 11
15
≤
+
N (β) ≤ N (β) < N (β).
4 16
16
(b) We can follow exactly the same proof as in the book showing that O has no
universal side divisor. We recall that for x = a + bω ∈ O, we have that
N (x) = a2 + ab +
1−D 2
b.
4
The only units in O are ±1, so R̃ = {0, ±1}. Suppose that u is a universal side
divisor. Let x ∈ O which is not a unit and of minimal norm. Since
N (x) = a2 + ab +
1−D 2
|D| 2
b = (a + b/2)2 +
b ≥ 11
4
4
when b 6= 0, the elements of minimal norm are x = ±2 with N (x) = 4. Let x =
2. Since u is a universal side divisor, we have that u divides one of x, x−1, x+1,
which implies that N (u) divides either 4 or 9, which means that either u = ±2
or u = ±3. Since u is a universal side divisor, we also have that u divides either
1−D
4
1−D
ω − 1 with N (ω) = 1 − 1 +
4
1−D
ω + 1 with N (ω) = 1 + 1 +
4
ω
with N (ω) = 0 + 0 +
6
If D = −43, this means that N (u) divides 11 or 13, which is impossible; if D =
−67, this means that N (u) divides 17 or 19, which is impossible; If D = −163,
this means that N (u) divides 41 or 43, which is impossible.
5. (section 8.1 no 7) Find a generator for the ideal (85, 1 + 13i) in Z[i], i.e. the
greatest common divisor for 85 and 1 + 3i by the Euclidean Algorithm. Do the
same for the ideal (47 − 13i, 53 + 56i).
Solutions: We have that
(a + bi)(c − di)
a + bi
ac + bd
bc − ad
=
= 2
+i 2
.
2
2
2
c + di
c +d
c +d
c + d2
Then,
85
85
1105
=
−i
⇐⇒ 85 = (−6i)(1 + 13i) + (7 + 6i)
1 + 13i
170
170
Now
97 + 91i
1 + 13i
=
⇐⇒ 1 + 13i = (1 + i)(7 + 6i) + 0
7 + 6i
85
Then, (85, 1 + 13i) = 7 + 6i.
To find (47 − 13i, 53 + 56i), we compute
1763
3321
53 + 56i
=
−i
⇐⇒ 53 + 56i = (47 − 13i)(1 + i) + (−7 + 22i).
47 − 13i
2378
2378
Then,
47 − 13i
−615
−943
=
−i
⇐⇒ 47 − 13i = (−7 + 22i)(−1 − 2i) + (−4 − 5i)
−7 + 22i
533
533
and
−7 + 22i
−82
−123
=
+i
⇐⇒ −7 + 22i = (−4 − 5i)(−2 − 3i) + 0,
−4 − 5i
41
41
which gives (47 − 13i, 53 + 56i) = −4 − 5i.
7