Zachary Scherr Math 371 HW 4 Solutions - Bucknell … Scherr Math 371 HW 4 Solutions 1. 8.1.10...

Zachary Scherr Math 371 HW 4 Solutions

1. 8.1.10

Solution: Let I be an ideal in Z[i]. Then since Z[i] is a Euclidean domain, we know there existsα ∈ Z[i] so that I = (α). By the division algorithm, we know that for any β ∈ Z[i] we can findθ, γ ∈ Z[i] so that

β = αθ + γ

where γ = 0 or N(γ) < N(β). This tells us that modulo (α), every element of Z[i] is equivalent tosome element of norm less than N(α). The problem will follow if we can show that there are onlyfinitely many elements in Z[i] of norm less than N(α). Of course, the equation

a2 + b2 < N(α)

has only finitely many integer solutions and so we are done.

2. 8.2.3

Solution: Let P be a prime ideal of the PID R. We know that R/P is an integral domain, and thefourth isomorphism theorem guarantees that any ideal I in R/P equals I/P for some ideal I in Rcontaining P . But then since R is a PID we get I = (α) for some α ∈ R and hence I = I/P = (α)proving that every ideal in R/P is principal.

3. 8.2.5

Solution:

(a) If you look at my notes on Z[√D], I carefully proved that 2, 3 and 1 +

√−5 are irreducible in

Z[√−5]. A similar calculation shows that 2 +

√−5 and 2 −

√−5 are irreducible since both

elements have norm 9 and Z[√−5] has no element of norm 3. Since none of the generators of

the given ideals are associates, if any ideal is principal then necessarily that ideal must equalall of Z[

√−5]. The assumption that any of these ideals is principal will immediately lead to a

contradiction. For example, if I2 is principal then we have (2, 1 +√−5) = (1) by the preceding

remarks. But then there exist α, β ∈ Z[√−5] such that

2α+ (1 +√−5)β = 1.

Multiplying both sides by (1−√−5) gives

2(1−√−5)α+ 6β = 1−

√−5

showing that 2 divides 1−√−5. This is absurd, and the same argument applies to each ideal

to show they they are non principal.

(b) We know that

I22 = (2, 1 +√−5)(2, 1 +

√−5) = (22, 2(1 +

√−5), (1 +

√−5)2) = (4, 2 + 2

√−5,−4 + 2

√−5).

It is clear that I22 ⊆ (2) since all of its generators are divisible by 2. Conversely,

2 = −4 + (2 + 2√−5)− (−4 + 2

√−5)

showing that 2 ∈ I22 and hence (2) ⊆ I22 .


(c) We have that

I2I3 = (2, 1 +√−5)(3, 2 +

√−5) = (6, 3 + 3

√−5, 4 + 2

√−5,−3 + 3

√−5).

This ideal contains 1−√−5 since

1−√−5 = (4 + 2

√−5)− (3 + 3

√−5).

Conversely, we have

6 = (1−√−5)(1 +

√−5)

3 + 3√−5 = (1−

√−5)3

4 + 2√−5 = (1−

√−5)(−1 +

√−5)

−3 + 3√−5 = (1−

√−5)(−3)

proving that I2I3 ⊆ (1 −√−5). Thus we get equality, and a similar argument shows that

I2I′3 = (1 +

√−5). Thus

I22I3I′3 = (I2I3)(I2I

′3) = (1−

√−5)(1 +

√−5) = (6).

4. 8.2.7

Solution:

(a) Suppose first that the integral domain R. Then for any a, b ∈ R, with at least one of a, bnon-zero, we know that (a, b) = (d) for some d ∈ R. Then clearly d divides both a and b. If d′

is any other element of R dividing a and b, then (d) = (a, b) ⊆ (d′) showing that d′ divides d.Thus d is a gcd for a and b, and clearly ax+ by = d has a solution in R since d ∈ (a, b).

Conversely, assume that every a, b, with at least one of a, b non-zero, has a gcd d for whichax+ by = d has a solution in R. Then certainly d ∈ (a, b), and since d divides both a and b wehave (a, b) ⊆ (d). This shows that (a, b) = (d) and so R is a Bezout domain.

(b) We proceed by induction on the number of generators in an ideal in R. The base case is idealsof the form (a, b), which we know to be principal. Suppose we know that every ideal generatedby n elements is principal. Then if

I = (a1, a2, . . . , an+1)

we haveI = (a1, a2, . . . , an−1) + (an, an+1).

The ideal (an, an+1) = (b) for some b so in fact

I = (a1, a2, . . . , an−1) + (b) = (a1, a2, . . . , an−1, b)

and I is therefore principal by the induction hypothesis.

(c) Let F be the fraction field of a Bezout domain R. Consider a fraction ab where a, b ∈ R and

b 6= 0. Let d = gcd(a, b). Then there exist x, y ∈ R so that a = dx and b = dy. Then clearlyab = xy, and we claim that x and y are relatively prime. To prove this let e = gcd(x, y). Thenclearly de divides both a and b, so by definition of gcd we must have de dividing d. Thusd ∈ (de), and clearly de ∈ (d) proving that (d) = (de). Thus there exists u ∈ R so that d = deu,but then since R is an integral domain we have eu = 1 proving that e is a unit.


5. 8.3.5

Solution:

(a) Let 2 = αβ for αβ ∈ Z[√−n]. Then taking norms shows 4 = N(α)N(β). There is no element

in Z[√−n] of norm 2 since n > 3 and there is no integral solution to a2 + nb2 = 2. Thus we

see that at least one of N(α) and N(β) is 1 so indeed 2 is irreducible.

Let√−n = αβ. Taking norms shows n = N(α)N(β). Suppose 1 < N(α). If β = a + b

√−n

then we haven > N(β) = a2 + nb2.

This shows that b = 0, and since a2 divides n, which is squarefree, we must have a = ±1. Thisshows that β is a unit and thus

√−n is irreducible.

Lastly, let 1 +√−n = αβ. Taking norms shows 1 + n = N(α)N(β). Suppose that N(α) > 1.

Then if β = a+ b√−n, we have

N(β) = a2 + nb2 < 1 + n

and N(β) divides 1 +n. We cannot have b 6= 0 since then otherwise either a2 +nb2 is too largeor else a2 + nb2 = n which does not divide 1 + n. Thus β = a, but if a divides 1 +

√−n then

necessarily a must be a unit in Z since gcd(1, 1) = 1. This shows that 1 +√−n is irreducible.

(b) Consider first the case that n is even. Then −n = (√−n)2 ∈ (2), but

√−n 6∈ (2). This shows

that the ideal (2) is not prime, but since 2 is irreducible we cannot have that R is a UFD.Similarly, if n is odd then 1 + n = (1 +

√−n)(1 −

√−n) ∈ (2), but neither 1 +

√−n nor

1−√−n is in (2). This proves that (2) is not prime and hence again R is not a UFD.

(c) Consider the ideal I = (2,√−n) ⊆ R. It is clear that I is maximal since Z[

√−n]/I ∼= Z/2Z

which is a field. I claim that I cannot be principal. To see this, suppose I = (α) for someα ∈ R. Then 2 = αx and

√−n = αy for some x, y ∈ R. Since both 2 and

√−n are irreducible,

we must either have α a unit or x and y units. The elements x and y cannot be units since ifthey were then 2 and

√−n would be associates, but this is clearly not the case. Thus we must

have α a unit, in which case I = (α) = R. This is a contradiction since we saw I is maximaland hence proper.

6. 8.3.6

Solution:

(a) Since N(1 + i) = 2, the division algorithm gives us that every element in Z[i] is equivalent tosome element of norm at most 1 modulo 1 + i. Thus every element in Z[i] is equivalent to oneof 0, 1,−1, i,−i. Of course 1 ≡ −i (mod 1 + i) and −1 ≡ i (mod 1 + i), and less obviously1 ≡ −1 (mod 1 + i) since 1 + 1 = 2 = (1 + i)(1− i). Since 1 + i is not a unit, we know that 0and 1 are not equivalent, and indeed

Z/(1 + i) = {0, 1}.

(b) We’ve seen that q is irreducible in the Euclidean domain Z[i], so indeed (q) is maximal. It’sclear that any a + bi ∈ Z[i] is equivalent to some r + si modulo q where 0 ≤ r, s < q since wecan add or subtract multiples of q and qi from a + bi without changing its value in Z[i]/(q).This shows that Z[i]/(q) contains at most q2 elements, and it only remains to check that notwo distinct elements whose real and imaginary parts are in the interval [0, q) are equivalent.


This is clear though. Suppose r + si and r′ + s′i are distinct and 0 ≤ r, r′, s, s′ < q. Then thereal and imaginary parts of (r + si)− (r′ + s′i) = (r − r′) + (s− s′)i cannot be multiples of qthus (r− r′) + (s− s′)i cannot be a multiple of q in Z[i]. This proves that Z[i]/(q) has at leastq2 elements and so we are done.

(c) As suggested, write p = ππ. Then both π and π are irreducible so Z[i]/(π) and Z[i]/(π) arefields. Since π and pi are non-associate irreducible elements of Z[i], we have gcd(π, π) = 1 andhence Z[i] = (π, π). Thus the Chinese remainder theorem tells us that

Z[i]/(p) = Z[i]/(π)(π) ∼= Z[i]/(π)× Z[i]/(π).

The fact that Z[i]/(p) has order p2 follows easily from the argument in part b). Check thatZ[i]/(π) ∼= Z[i]/(π) via the isomorphism

a+ bi 7→ a− bi

in which case both fields must have order equal to p.

Without actually recognizing that both fields are isomorphic, you could just argue by uniquefactorization. Since the only factors of p2 are 1, p, p2, it suffices to prove that neither ring hassize 1. This is equivalent to proving that neither ideal (π) nor (π) equals Z[i], but this is clearsince neither π nor π is a unit.

7. 8.3.7

Solution:

(a) Consider a function ϕ : Z[i]→ (πn)/(πn+1) given by ϕ(α) = απn. The function ϕ is well-definedsince απn ∈ (πn), and it’s clearly a group homomorphism since

ϕ(α+ β) = (α+ β)πn = απn + βπn = απn + βπn = ϕ(α) + ϕ(β).

Since Z[i] is a UFD and π is irreducible, an element α ∈ ker(ϕ) if and only if απn ∈ (πn+1),which is the case if and only if α ∈ (π). Thus ker(ϕ) = (π), so it only remains to show that ϕis surjective. This is clear, however, since the map Z[i] → (πn) given by α 7→ απn is alreadysurjective. Thus by the first isomorphism theorem, ϕ gives an isomorphism of groups,

Z[i]/(π) ∼= (πn)/(πn+1).

(b) We prove this statement by induction on n. There is nothing to prove when n = 1, so assumewe know the statement for some k ≥ 1. When n = k+ 1, the third isomorphism theorem gives

Z[i]/(πk) ∼=(Z[i]/(πk+1)

)/((πk)/(πk+1)

).

Taking sizes, we get: ∣∣Z[i]/(πk+1)∣∣ =

∣∣Z[i]/(πk)∣∣ · ∣∣(πk)/(πk+1)

∣∣= |Z[i]/(π)|k · |Z[i]/(π)|= |Z[i]/(π)|k+1

proving the claim.


(c) Using unique factorization, we can write any α ∈ Z[i] uniquely up to associates as

α = πe11 πe22 · · ·π

ekk

where πi is irreducible. Since πi is relatively prime to πj for i 6= j, the Chinese remaindertheorem gives an isomorphism

Z[i]/(α) ∼= Z[i]/(πe11 )× Z[i]/(πe22 )× . . .× Z[i]/(πekk ).

Taking sizes, we see

|Z[i]/(α)| = |Z[i]/(πe11 )| · |Z[i]/(πe22 )| · · · |Z[i]/(πekk )| .= |Z[i]/(π1)|e1 · |Z[i]/(π2)|e2 · · · |Z[i]/(πk)|ek

= N(π1)e1N(π2)e2 · · ·N(πk)ek

= N(α)

where the second line follows from part b), the third line follows from problem 6, and the lastline follows from the multiplicativity of the norm.

8. Prove that Z[√−2] and Z[

√2] are Euclidean domains. For Z[

√−2] use N(a + b

√−2) = a2 + 2b2. For

Z[√

2] use N(a+ b√

2) = |a2 − 2b2|. (Hint: Example 3 on page 271 might be very helpful)

Solution: For both of these rings, we follow the same strategy used for Z[i]. Let α, β ∈ Z[√−2]

with β 6= 0. Rationalize αβ = x+ y

√−2, where x, y ∈ Q, and choose integers r, s so that |r− x| ≤ 1

2

and |s− r| ≤ 12 . Let θ = r + s

√−2 ∈ Z[

√−2], then

N

(α

β− θ

)= N((x+ y

√−2)− (r + s

√−2))

= N((x− r) + (y − s)√−2)

= (x− r)2 + 2(y − s)2

≤ 1

4+ 2

1

4

=3

4.

Thus

N(α− βθ) = N(β)N

(α

β− θ

)≤ 3

4N(β).

If we let γ = α− βθ ∈ Z[√−2], then

α = βθ + γ

where we just proved N(γ) ≤ 34N(β) < N(β), so indeed Z[

√−2] is a Euclidean domain.

The same proof, almost verbatim, works for Z[√

2]. Again you take α, β ∈ Z[√

2] with β 6= 0, and


take integers r, s so that if αβ = x+ y√

2 then |x− r| ≤ 12 and |y− s| ≤ 1

2 . Letting θ = r+ s√

2, then

N

(α

β− θ

)= N((x+ y

√2)− (r + s

√2))

= N((x− r) + (y − s)√

2)

= |(x− r)2 − 2(y − s)2|≤ |x− r|2 + 2|y − s|2

=1

4+ 2

1

4

=3

4.

Thus again if γ = α − βθ ∈ Z[√

2] then we have N(γ) < N(β) and indeed Z[√

2] is a Euclideandomain.

1 For Fun

1. In the 1600s, Fermat, a French mathematician, challenged the English mathematicians to find all integersolutions to the equation y3 = x2 + 2. Solve this equation. (Hint: Instead of working in Z, try to thinkabout this equation as y3 = (x+

√−2)(x−

√−2), an equation in Z[

√−2].)

Solution: First off, we argue that if there is a solution then necessarily x is odd. To see this, reducethe equation mod 4 to obtain

y3 ≡ x2 + 2 (mod 4).

If x is even then x2 ≡ 0 (mod 4), but it is easily checked that 2 is not a cube modulo 4. Thus wenecessarily have x odd.

Rewrite the equation asy3 = (x+

√−2)(x−

√−2).

The ring Z[√−2] is a UFD, and I claim that when x is odd we have that x +

√−2 and x −

√−2

are relatively prime. To see this, first note that both y3 and 8 are contained in the ideal I =(x +

√−2, x −

√−2) since y3 = (x +

√−2)(x − sqrt−2) ∈ I and 8 = (2

√−2)2 and 2

√−2 =

(x+√−2)− (x−

√−2) ∈ I. Since y3 is an odd integer, we know that 1 ∈ I and so indeed x+

√−2

and x−√−2 are relatively prime.

Thus, if (x +√−2)(x −

√−2) is a square in Z[

√−2] then necessarily both x +

√−2 and x −

√−2

better be cubes themselves. Hence we can find a, b ∈ Z so that

x+√−2 = (a+ b

√−2)3.

Comparing coefficients, we see that

x = a3 − 6ab2

1 = 3a2b− 2b3.

The second equation rewrites as 1 = b(3a2 − 2b2) which immediately gives b = ±1. We cansolve in both cases to see that b = 1 and a = ±1. Plugging this into the first equation we getx = (±1)3 − 6(±1) which gives x = ±5.

Thus our original equation only has the two solutions (5, 3) and (−5, 3).


It’s worth pointing out that although the equation y3 = x2 + 2 has only two integral solutions,more modern techniques can show that it has infinitely many rational solutions. Besides these twointegral solutions, the next simplest rational solutions are(

± 383

1000,

129

100

).

Zachary Scherr Math 371 HW 4 Solutions - Bucknell … Scherr Math 371 HW 4 Solutions 1. 8.1.10...

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