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MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
D&F 7.6.5:
(a) It suffices to show that if (m, n) = 1 then the ideals (m) and (n) are comaximal. Ideed, if we
knew that this is the case then we’d know that (ni ) and (nj ) were comaximal, and thus that
the Chinese Remainder Theorem applies to the ideals (n1 ), . . . , (nk ). The intersection of
these ideals is exactly (n1 · · · nk ). The equivalences specify an element in Z/n1 × · · · × Z/nk ;
the fact that there is a unique solutions follows from the fact that this is isomorphic to
Z/n1 · · · nk .
If (m, n) = 1 then there exist integers a, b such that am+bn = 1; thus the ideal (m, n) = Z,
and (m) and (n) are comaximal, as desired.
(b) Since x is unique it suffices to show that this x satisfies the above equivalences. For any i,
ni |n0j for j 6= i, so
x ≡ ai ti n0i ≡ ai (mod ni ),
so we are done.
(c) We have
n01 = 2025 ≡8 1
n02 = 648 ≡25 −2
n03 = 200 ≡81 38,
and
t1 = 1
t2 = 12
t3 = 32.
Thus
x ≡ 1 · 1 · 1 + 2 · 12 · (−2) + 3 · 32 · 38 ≡ 3601
(mod 16200),
and
y ≡ 5 · 1 · 1 + 12 · 12 · (−2) + 47 · 32 · 38 ≡ 8269
(mod 16200).
D&F 7.6.6: Suppose that
fi (x) = adi xd + · · · + a0i .
Let aj be the integer such that aj ≡ aji (mod ni ) for each i, obtained from the previous problem.
Then
f (x) = ad xd + · · · + a0 ≡ fi (x) (mod ni ).
If each fi is monic then adi = 1 for all i, and thus we can choose ad = 1.
D&F 8.1.8:
(a) For D
= −1 the proof is in the text. First, suppose that D = −3, −7, −11; then O =
√
√
1+ D
Z[ 2 ]; we can write this as the set of numbers a2 + 2b D, where a, b ∈ Z and a ≡ b
√
√
√
(mod 2). Let α = a + b D and let β = c + d D. Write γ = α/β = r + s D. Let n be an
integer which minimizes
|s − n/2|, and let m be an integer that minimizes |r − m − n/2|.
√
1+ D
We let δ = m + n 2 ∈ O. We claim that
N (α − βδ) < N (β),
which gives us the Euclidean algorithm.
N (α − βδ) = N (β)N (γ − δ),
1
2
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
so it suffices to check that N (γ − δ) < 1. We have
n
n √
N (γ − δ) = N ((r − m − ) + (s − ) D)
2
2
n
1 |D|
n 2
= (r − m − ) + |D|(s − )2 ≤ +
2
2
4
16
4 + |D|
=
< 1.
16
Now suppose that D = −2. We do the same process as above to define γ, but then we
choose m and n so that |s − n| and |r − m| are minimized. Then again we just need to check
that N (γ − δ) < 1. But
N (γ − δ) = (r − m)2 + 2(s − n)2 ≤
1 2
3
+ = < 1.
4 4
4
So again this works.
(b) We need to check that O contains no universal side divisors. The only units in O are ±1.
Thus if we had a universal side divisor u then for every x, either x + 1, x − 1 or x would be
divisible by u.
p
√
The field norm of a + b 1 + D2 is
b
b2
1 + |D| 2
(a + )2 − D = a2 + ab +
b .
2
4
4
Note that for b 6= 0 this is at least |D|/4, which is at least 10. Let x = 2; if we have a u
then u must divide 2 ± 1 or 2, so it must divide either 2 or 3. But N (2) = 4, and the only
elements with norm at most 4 are 0, ±1 or ±2, so 2 has no nonunit divisors. This works
analogously for
√ 3, so the only possible universal side divisors are ±2 and ±3.
1+ D
Let x = 2 . The norm of this is 1+|D|
4 ; in our cases, this is 11, 17, 41. Each of these is
prime, an in particular are not divisible by 4 or 9. The norm of x + 1 is 13, 19, 43, which are
also not divisible by 4 or 9. The norm of x − 1 is again 11, 17, 41, again not divisible by 4
or 9. So none of x, x ± 1 is divisible by ±2 or ±3, so we can’t have a universal side divisor.
D&F 8.1.10: We know that I = (α), and let β ∈ Z[i]. We can use the Euclidean algorithm to see
that
β = qα + r
where N (r) < N (α). Thus modulo I, β is equivalent to r, and every coset of I is represented by
an element of norm at most N (α). It thus remains to check that for all c ∈ Z there exist only
2
2
finitely many
√ β ∈ Z[i] with N (β) < c. For β = a + bi, N (β) = a + b ; in particular, we see that
|a|, |b| < c; thus there are at most 4c such elements.
D&F 8.1.11:
(a) The set of all x such that a|x and b|x is exactly the set (a) ∩ (b). This is the set of all
elements that satisfy (i). If this ideal is principal and generated by e then by definition of
principal domain, this satisfies (ii). Alternately, if e satisfies (ii) then any element in (a)∩(b)
is a multiple of it, and thus the ideal is principal and generated by e.
(b) Since a Euclidean domain is a principal ideal domain, least common multiples exist.
(c) Clearly, x = ab/(a, b) is a multiple of both a and b; thus we just need to show that it satisfies
(ii). Let e be the least common multiple of a and b, so that it divides x. Write e = ya for
a
b
a
y ∈ R. We can write a = (a,b)
(a, b), and then we know that (a,b)
|y (a,b)
. We claim that this
means that
b
(a,b) |y,
so that x|e and we are done.
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
3
More generally, we will show for any x, y, z ∈ R that if (x, y) = 1 and x|yz then x|z.
Suppose not. We can write
1 = αx + βy.
Multiplying both sides by z we get
yz
z = αxz + βyz = x(αz + β );
x
since x divides the right-hand side it must divide the left-hand side.
D&F 8.2.5:
√
(a) If I2 were principal then
both 2 and 1 + −5.
√ there would exist a nonunit a which divides √
N (2) = 4 and N (1 + −5) = 6; thus we must have N (a) = 2. But Z[ −5] only has norms
of the form a2 + 5b2 , none of which are equal to 2, so this is impossible.
√
If I3 √
or I30 were principal then there would be some a dividing 3√and 2 ± −5. N (3) =
N (2 ± −5) = 9, so N (a) = 1, 3, 9. There are no elements in Z[ −5] of norm 3, so it’s
either 1 or 9. If it’s 1 then the ideal would be the whole ring, so there would exist x and y
such that
√
3x + (2 ± −5)y = 1.
But then
(3x + 2y)2 + 5y 2 = 1;
in particulra,
y = 0 and 3x = 1; impossible, since x is an integer. Thus N (a) = 9. Thus
√
2 ± 5 = 3u for some unit u, which is clearly impossible since 3 6 |2. Thus these ideals aren’t
principal.
√
√
√
√
(b) Note that 1 − −5 = 2 − (1 + −5) ∈ I2 . Thus 6 = (1 + −5)(1 − −5) ∈ I22 . Thus
2 = 6 − 22 ∈ I22 and (2) ⊆ I22 . On the other hand,
√
√
√
√
I22 = (22 , (1 + −5)2 , 2(1 + −5)) = (4, −4 + 2 −5, 2 + 2 −5) ⊆ (2).
(c)
√
√
√
I2 I3 = (6, 4 + 2 −5, −3 + 3 −5, 3 + 3 −5 )
| {z }
√
√
−(3+3 −5)+6
√
= (6, 4 + 2 −5, 3 − 3 −5)
√
√
√
= (6, 2(1 − −5) , 3(1 − −5)) ⊆ (1 − −5).
|
{z
}
√
−(4+2 −5)+6
In addition
√
√
√
−5 = (4 + 2 −5) − (3 + 3 −5),
√
√
√
so (1 − −5) ⊆ I2 I3 . That I2 I30 = (1 + −5) works analogously with the signs on the −5
reversed.
Thus
√
√
(6) = (1 − −5)(1 + −5) = I2 I3 I2 I30 = I22 I3 I30 .
1−
D&F 8.2.6:
(a) In order to apply Zorn’s lemma we need to show that the union of nonprincipal ideals is
nonprincipal. So suppose that we are given a chain
I1 ⊆ I2 ⊆ · · ·
S
of nonprincipal ideals, and consider I = ∞
n=1 In . We want to show that this ideal is
nonprincipal. Suppose for contradiction that it were. Then I = (a) for some a, and this
4
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
means that In ⊆ (a) for all n. However, a ∈ Im for some m by the definition of I; thus
Im = (a) and is principal, a contradiction.
(b) Since I is maximal among non-principal ideals, and Ia 6= I since a ∈
/ I, we must have Ia be
principal. We also know I ⊂ Ib , so all we need to check now is that Ib ⊆ J; once we have
this we know that I ⊂ J and thus (by the same logic as above) J must be principal.
Note that for i ∈ I
iIa ⊆ I ⊆ Ia ,
so I ⊆ J. Similarly,
bIa = (bI, ba) ⊆ I,
so Ib ⊆ J, as desired.
Ia J ⊆ I by the definition of J, and it is generated by αβ by the definition of multiplication
of principal ideals.
(c) Let x ∈ I. Ia ⊇ I, so x ∈ Ia ; thus in particular, x = sα for some s. But since sα ∈ I this
means that sIa ⊆ I, and s ∈ J, as desired.
We have shown that I = Ia J = (αβ), so it is principal. However, this contradicts part
(a), where we checked that it wasn’t. The only way this can happen is if in our use of Zorn’s
lemma the partial order were empty and there were no nonprincipal ideals.
D&F 8.2.7:
(a) If we know that (a, b) = (d) then d|a and d|b. On the other hand, if (a, b) = d then
d = xa + yb for some x and y, so that any other e such that e|a and e|b also divides d. Thus
if R is a Bezout domain then we have a GCD for a and b that can be written as a linear
combination of a and b.
On the other hand, if a and b have a GCD that can be written as a linear combination of
a and b then d ∈ (a, b) so (a, b) ⊇ (d). If d|a and d|b then (a, b) ⊆ (d), and thus (a, b) = (d).
(b) Suppose that I = (a1 , . . . , an ), let d2 be the GCD of a1 and a2 , and let di be the GCD of
di−1 and ai . Then
(a1 , . . . , an ) = (d2 , . . . , an ) = (d2 , a3 , . . . , an ) = (d3 , a4 , . . . , an ) = · · · = (dn ).
Thus every ideal is principal.
(c) Let a/b ∈ F , and consider the ideal (a, b) = (d). Then we know that a = da0 and b = db0 .
We know that a/b = a0 /b0 so it remains to check that (a0 , b0 ) = 1. We have x and y such
that d = ax + by. We can write d = d(a0 x + b0 y), and canceling the d’s give sus 1 = a0 x + b0 y,
as desired.
D&F 8.3.3: 101 can be written as the sum of two squares in exactly one way: 102 + 12 . 73 can
only be written in one way: 82 + 32 . Thus the elements whose norm can be 2130797 are
17(8 ± 3i)(10 ± i)
and
(4 ± i)(8 ± 3i)(10 ± i).
These give the possible ways of writing 2130797 as a sum of two squares.
D&F 8.3.4: If n = (r/s)2 + (a/b)2 then ns2 b2 = (rb)2 + (as)2 . Thus every factor of ns2 b2 which
is to an odd power is 1 mod 4 or 2, and therefore every factor of n that is to an odd power is the
same. Thus n is the sum of integer squares.
Problem 5: Prove that for every number n there exists n consecutive positive integers such that
none of them is a power of a prime. (Hint: Not being a power of a prime means that you have at
least two different prime divisors. The Chinese Remainder Theorem will help you.)
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
5
Solution: Pick 2n distinct primes, p1 , . . . , pn and q1 , . . . , qn . We want to find an integer N such
that N ≡ −i (mod pi qi ). Then pi qi |N + i, so none of the n consecutive numbers N + 1, N +
2, . . . , N + n will be prime powers. But such an N exists by the Chinese remainder theorem, since
pi qi and pj qj are relatively prime for distinct i and j.
Problem 6: Let R be a commutative integral domain, and let I be a prime ideal of R. Let
S = R\I.
(a) Show that S is closed under multiplication.
(b) Let R0 = S −1 R; in other words, R0 is the ring of fractions with denominators in S. (See
Section 7.5 of Dummit and Foote.) Show that R0 has exactly one maximal ideal.
(c) Let J = IR0 . Define a topology on R0 by letting the basis of open sets be the cosets
x + Jk
x ∈ R, k ≥ 1.
Prove that this is a well-defined topology. Show that for any two open sets U and V , either
U ⊆ V , V ⊆ U or V ∩ U = ∅. This kind of topology is called “totally disconnected.”
Solution:
(a) Let a, b ∈ S. If ab ∈ I then because I is prime, a ∈ I or b ∈ I. However, by definition this
can’t be the case, so ab ∈
/ I and S is closed under multiplication.
(b) Let M be the subset of R0 containing all of those fractions si with i ∈ I and s ∈ S. We claim
that M is an ideal and that all other ideals of R0 contain M . First, note that all elements
of the form xs , where x ∈
/ I are units in R0 , so if xs is in an ideal J then J = R. Thus if M
is an ideal it must be maximal.
To check that M is an ideal, first note that
i0
is0 + i0 s
i
+ 0 =
∈ M,
s s
ss0
since is0 , i0 s ∈ I. On the other hand,
xi
xi
= 0 ∈ M,
0
s s
ss
since xi ∈ I because I is an ideal. Thus M is an ideal.
(c) To check that this is a basis for a topology, we need to show that the intersection of any
two of these can be covered by basis elements. Thus we need to show that
(x + I k ) ∩ (y + I ` )
can be covered by cosets of I m for some m. Assume WLOG that k ≤ `. Note that
(x + I k ) ∩ (y + I ` ) = x + (((y − x) + I ` ) ∩ I k ).
In particular, since our basis is invariant under translation, we focus on computing
(a + I ` ) ∩ I k
Ik
I`
k ≤ `.
Ik,
If a ∈
then a + ⊆
so when we intersect with I k we still get a + I ` . On the other
hand, if a ∈
/ I k then (a + I ` ) ∩ I k = ∅. Thus the intersection of two basis elements is either
one of the basis elements or empty. This shows both that the topology is well-defined and
proves the property we wanted.
Problem 7: Show that in the rings Z and F [x], for any sequence of ideals
I1 ⊆ I2 ⊆ I3 ⊆ · · ·
there exists an integer N such that In = In+1 for all n ≥ N .
Solution: We know that Z and F [x] are both PIDs. Thus we can write In = (an ) for some an . In
particular, we know that an |an−1 . Note that if an−1 = uan for a unit u then (an−1 ) = (an ); thus in
6
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
particular we want to show that there does not exist a sequence a1 , a2 , . . . in R = Z or F [x] such
that an |an−1 but they do not differ by a unit.
In both Z and F [x], if a|b and a 6= ub with u a unit, N (a) < N (b). Since norms are nonnegative
integers, this means that any sequence a1 , a2 , . . . with an |an−1 must eventually start to differ only
by units; this will mean that at this point (an ) = (an+1 ), as desired.
Problem 8: Let Z be the set of roots of monic polynomials with integer coefficients. You may
assume Z is a ring.
√
(a) Show that the ideals (2) and ( 2) are distinct.
(b) Show that there exists an infinite sequence
I1 ⊂ I2 ⊂ I2 ⊂ · · ·
of ideals in Z.
Solution:
√
(a) Suppose that
√ they were equal. Then there would exist a ∈ Z such that 2a = 2; in other
words, 1/ 2 would be in Z. Thus it√suffices to check that there is no monic polynomial
with integer coefficients such that 1/ 2 is a root of it. Note that if this is possible then
the polynomial has degree at least 2, since the only roots of monic linear polynomials with
integer coefficients are integers.
We prove this by contradiction. Since Z is a ring, we must have 1/2 ∈ Z. We will show
that this is not the case. Suppose that 1/2 is a root of the monic polynomial with integer
coefficients
xn + an−1 x + · · · + a0 .
Plugging in to the polynomial and clearing denominators
1 + 2an−1 + · · · + 2n−1 a1 ∈ 2n Z.
However, this is clearly not possible, since the term on the left is odd and the right-hand
side only contains even integers.
√
Alternate proof: Let I be the ideal of polynomials in Q[x] such that 1/ 2 is a root of
the polynomial. We claim that all such polynomials are divisible by 2x2 − 1. Indeed, let
p(x) ∈ I; by the Euclidean algorithm we can write
p(x) = (2x2 − 1)q(x) + r(x)
√
with deg r ≤ 2. Plugging in 1/ 2 into this equation, we have
√
√
0 = 0q(1/ 2) + r(1/ 2),
√
so in particular we see that 1/ 2 is a root of r(x). Since deg r(x) < 2, r(x) must be zero.
Thus p(x) is divisible by 2x2 + 1, as desired.
2
We claim that if a polynomial in Q[x] has
√integer coefficients and is divisible by 2x − 1
then it is not monic. This will show that 1/ 2 is not in Z. Suppose that such a polynomial
p(x) existed. We know that p(x) = (2x2 − 1)q(x), with
q(x) = an xn + · · · + a0 .
Then an = 1/2, and by induction we can conclude that an−2j = (2kj + 1)/2j+1 for some
kj ∈ Z. We know that the coefficient of x in p(x) is a1 and the constant term is a0 , thus
at least one of these √
is not an integer, contradicting the assumption that p(x) has integer
coefficients. Thus 1/ 2 ∈
/ Z, as desired.
MATH 205 HOMEWORK #8 OFFICIAL SOLUTION
7
(b) We claim that for all n ≥ 0,
n
(21/2 ) ⊂ (21/2
n+1
).
The inclusion is clear, so we just need to show that there does not exist a ∈ Z such that
n
n+1
a21/2 = 21/2
n+1
1/21/2
.
In other words, we want to show that
∈
/ Z. Suppose that it were; then its 2n+1 -st
power would also be in Z, since Z is an integer, and in particular 1/2 ∈ Z. This is not the
case, as we showed in the previous part.
n
n
However 21/2 is a root of the polynomial x2 − 2, so it is in Z. Thus the string of ideals
√
(2) ⊆ ( 2) ⊆ (21/4 ) ⊆ (21/8 ) ⊆ · · ·
satisfies the desired properties.