Math 4400/6400 – Homework #10 solutions
MATH 4400 Exercises.
1. Let p be an odd rational prime. Let k be a positive integer.
(a) Show that the element 1 + p has order pk−1 modulo pk . Hint: Use Exercise 7 on
HW #9, starting with a = 1 + p.
Proof. The assertion is obvious for k = 1, so suppose that k > 1.
We have that (1 + p) − 1 = p is divisible by p but not p2 . So by the referenced
Exercise, (1 + p)p − 1 is divisible by p2 , but not p3 . Continuing in this way, we
eventually find that
(1 + p)p
k−2
− 1 is divisible by pk−1 but not pk ,
(1 + p)p
k−1
− 1 is divisible by pk but not pk+1 .
(1)
and that
k−1
k−1
Since (1 + p)p − 1 is divisible by pk , we have (1 + p)p ≡ 1 (mod pk ), and so
the order of 1 + p divides pk−1 . If the order were less than pk−1 , it would divide
k−2
pk−2 , and so we would have (1 + p)p
≡ 1 (mod pk ), contradicting the second
half of (1).
(b) Show that there is an integer whose order modulo pk is precisely p − 1.
Proof. Let g be a primitive root modulo p. Let ` be the order of g modulo pk .
Then g ` ≡ 1 (mod pk ), and so g ` ≡ 1 (mod p). Hence, p − 1 | `. Writing
` = (p − 1)k, we see that g k has order p − 1 modulo pk .
(c) Using the results of (a) and (c), and an appropriate lemma from our proof that
U (Zp ) is cyclic, show that U (Zpk ) is a cyclic group.
Proof. By (a), 1 + p has order pk−1 modulo pk . By (c), there is an integer a (say)
having order p − 1 modulo pk . Since pk−1 and p − 1 are coprime, the product
(1 + p)a has order pk−1 (p − 1) modulo pk . Since φ(pk ) = pk−1 (p − 1), it follows
that (1 + p)a is a generator for U (Zpk ), and so U (Zpk ) is a cyclic group.
2. Suppose that the nonzero elements κ and γ both qualify as the greatest common divisor
of α and β in Z[i]. Show that κ and γ are associates. Conversely, show that if κ satisfies
the definition of the gcd and γ is an associate of κ, then γ also satisfies the definition
of the gcd.
Proof. Recall our definition of the gcd: It is a common divisor of α and β divisible
by every common divisor. If κ and γ are both gcds, then both are common divisors
of α and β, and so by the second half of the definition, each of κ and γ divides the
other. Hence, κ and γ are associates. (This ‘mutual divisibility’ property was in fact
our definition of being associates.)
Conversely, suppose κ is a common divisor divisible by every common divisor. Let γ be
any associate of κ. Since κ is a common divisor and γ | κ, γ is also a common divisor.
(We use here transitivity of divisibility, which is proved the same way in Z[i] as in
Z.) Similarly, since every common divisor divides κ, and κ divides γ, every common
divisor divides γ. So γ also satisfies the definition of being a gcd.
3. Use the Euclidean algorithm to compute a greatest common divisor of 17 + 4i and
2 + 3i in Z[i]. Express your gcd as a linear combination of the starting numbers.
Solution. Using the Euclidean algorithm, we compute that
17 + 4i = (2 + 3i)(4 − 3i) + (−2i)
2 + 3i = (−2i)(−2 + i) + (−i)
(−2i) = (−i)2 + 0.
Thus, −i is a gcd of α and β, and so α and β are relatively prime.
We also compute that
−i = (2 + 3i)(1) + (−2i)(2 − i)
= (2 + 3i)(1) + ((17 + 4i) + (2 + 3i)(−4 + 3i))(2 − i)
= (17 + 4i)(2 − i) + (2 + 3i)(−4 + 10i).
4. Show that if α and β are nonzero elements of Z[i], and N (α) and N (β) are coprime
ordinary integers, then α and β are coprime Gaussian integers. Give an example to
show that the converse statement is false.
Proof. Let γ be the gcd of α and β. Since γ divides α, we can write α = γδ for some
δ ∈ Z[i]. Taking norms, we get that N (α) = N (γ)N (δ), and so N (γ) divides N (α).
By a similar argument, N (γ) divides N (β). Since N (α) and N (β) are relatively prime,
the only possibility is that N (γ) = 1. But then γ is a unit, as desired.
To show that the converse is false, consider α = 1 − 2i and β = 1 + 2i. Their norms
are both 5 (and so are certainly not relatively prime!), but we can easily check using
the Euclidean algorithm that β and α are relatively prime:
1 + 2i = (1 − 2i)(−1 + i) + (−i)
(1 − 2i) = (−i)(2 + i) + 0.
So a gcd is the unit −i, and hence α and β are coprime.
5. (a) Show that if α is a nonzero Gaussian integer that is not a unit, then α is divisible
by some irreducible π in Z[i].
Proof. Suppose for a contradiction that there is some nonzero, nonunit α which
has no irreducible divisors. Choose α so that the norm of α is minimal, which is
possible by the well-ordering principle.
Clearly α cannot be irreducible, since α is a divisor of itself. Hence, we can
decompose α = βγ with neither β nor γ a unit. Since α 6= 0, both β and γ are
also nonzero. It follows that N (β), N (γ) > 1. Consequently, if we write
N (α) = N (β)N (γ),
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then N (β) and N (γ) are positive integers smaller than N (α). Since α was a
counterexample with minimal norm, β is not a counterexample, and so there is
some irreducible π that divides β. But then π | β | α, and so π | α. Hence, α has
an irreducible divisor after all, which is a contradiction.
(b) (continuation) Show that one can factor α in Z[i] as
α = π 1 π2 · · · πk ,
where each of the πi is irreducible.
Proof. Again, if there is a counterexample, there is a counterexample α whose
norm is smallest. As before, α cannot be irreducible itself. We deduce that
α = βγ, where
1 < N (β), N (γ) < N (α).
Neither β nor γ can be counterexamples, and so both β and γ factor as products of irreducibles. Appending the factorizations gives a factorization of α into
irreducibles, which is a contradiction.
6. Show that if p is a rational prime and p ≡ 3 (mod 4), then p is irreducible in Z[i]. For
example, 7 and 19 are irreducible in Z[i].
Proof. Suppose p is not irreducible. Then we can write p = βγ, where neither β nor γ
is a unit in Z[i]. Taking norms gives p2 = N (β)N (γ). Now the only factorizations of p2
into positive integers are 1·p2 , p2 ·1, and p·p. Neither of the first two factorizations can
occur, since then either β or γ would be a unit. But if the third occurs, then N (β) = p.
Writing β = a + bi, we see that p = a2 + b2 , and so p is a sum of two squares. But we
showed in class that no prime p ≡ 3 (mod 4) is a sum of two squares.
MATH 6400 Exercises. Do two of the following.
G1. Show that there are infinitely many irreducible elements in Z[i].
Proof. Suppose that there were only finitely many irreducible elements of Z[i]; then
we could form a complete (finite) list of them, say π1 , . . . , πk . Consider the Gaussian
integer Π = π1 · · · πk + 1. Notice that
|Π| ≥ |π1 · · · πk + 1| ≥ |π1 · · · πk | − 1,
by the triangle inequality. Since 3 must be on our list of irreducibles, |π1 · · · πk | ≥ 3,
and so |Π| ≥ 2 from the above. In particular, N (Π) = |Π|2 ≥ 4, and so Π is not a unit.
Thus, there is some irreducible that divides π. But none of the πi can divide Π, since
then they would also divide Π − π1 · · · πk = 1. This is a contradiction.
G2. For each odd prime p, let n(p) denote the smallest positive quadratic nonresidue modulo
p. For example, n(5) = 2 and n(7) = 3. Below, we abbreviate n(p) to simply n.
(a) Show that p < ndp/ne < p + n. Here the notation dxe (the ‘ceiling’ of x) denotes
the smallest integer not less than x.
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Proof. From the geometry of the number line, we see that p/n ≤ dp/ne < 1+p/n.
Multiplying through by n gives the desired inequality.
(b) Deduce from (a) that dp/ne is a quadratic nonresidue modulo p.
Proof. From part (a), if we reduce ndp/ne modulo p, we get a positive integer less
than n. Since n is the smallest quadratic nonresidue, it must be that ndp/ne is a
quadratic residue modulo p. Hence,
n dp/ne
dp/ne
ndp/ne
=
=−
,
1=
p
p
p
p
and so dp/ne is a quadratic nonresidue.
(c) Using the minimality of n and the result of (b), show that
n<
1 √
+ p.
2
Proof. By the minimality of n, it must be that 1 + p/n > dp/ne ≥ n. Hence,
2
1 √
1
< n2 − n + 1 ≤ p, and so n < + p.
n−
2
2
G3. Let p be a prime number. Let a be an integer not divisible by p. Show that you can
√
always find a pair of integers (x, y) 6= (0, 0) with both |x|, |y| < p and
ax ≡ y
(mod p).
Proof. Consider the integers ax − y, where x and y run independently through the
√
√
integers in the interval [0, p). The number of choices for each of x and y is 1+b pc >
√
p, and so the number of pairs (x, y) exceeds p. Hence, there are at least two such
pairs (x1 , y1 ) and (x2 , y2 ) with ax1 − y1 ≡ ax2 − y2 (mod p), or equivalently,
a(x1 − x2 ) ≡ (y1 − y2 )
(mod p).
The pair (x1 − x2 , y1 − y2 ) 6= (0, 0), and it is easy to see that |x1 − x2 |, |y1 − y2 | <
as desired.
√
p,
G4. Use the result of G3 to give a proof, different from the one presented in class, that
every prime p ≡ 1 (mod 4) can be written as a sum of two squares.
Proof. Choose an integer a with a2 ≡ −1 (mod p), which we know is possible. Choose
a pair (x, y) satisfying the conditions of the previous problem. Since ax ≡ y (mod p),
we have −x2 ≡ (ax)2 ≡ y 2 (mod p), and so p | x2 + y 2 . On the other hand, since
√
(x, y) 6= (0, 0) and both |x|, |y| < p, we also have
0 < x2 + y 2 < 2p.
But the only multiple of p in the interval (0, 2p) is p itself, and so it must be that
x2 + y 2 = p.
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