Final review solutions
1. Suppose that R is a Euclidean ring and that a, b ∈ R. We say that m ∈ R is a least common
multiple ( lcm) of a and b if
• a m and b m
• If a m0 and b m0 , then m m0
Let (a) = aR be the principal ideal generated by a.
(a) Show that (a) ∩ (b) is an ideal and that if (m) = (a) ∩ (b), then m is a least common
multiple of a and b.
If r, s ∈ (a) ∩ (b), then r + s ∈ (a) (since (a) is closed under addition) and r + s ∈ (b), so
r + s ∈ (a) ∩ (b). Likewise, −r ∈ (a) ∩ (b), so (a) ∩ (b) is an additive subgroup of R.
If r ∈ (a) ∩ (b) and c ∈ R, then cr ∈ (a) and cr ∈ (b), so cr ∈ (a) ∩ (b), so (a) ∩ (b) is an
ideal.
The elements of (a) ∩ (b) are exactly those elements c ∈ R such that a c and b c —
the common multiples of a and b. Therefore, if (m) = (a) ∩ (b), then a m and b m.
0
0
Furthermore, if m
is0 a common multiple of a and b, then m ∈ (a) ∩ (b) = (m), so
0
m = mr, and m m .
(b) Show that if d is a greatest common divisor of a and b, then ab is divisible by d and ab/d
is a least common multiple of a and b.
Suppose that a = rd and that b = sd, so that m = ab/d = rsd. Then m = r(sd) = rb
and m = s(rd) = sa, so m is a common multiple of a and b.
0
ab Suppose that m0 is another common
0 multiple of a and b. We claim that d m . This is
equivalent to showing that ab m d. Let x, y ∈ R be such that d = ax + by. Then
m0 d = m0 ax + m0 by.
Then ab divides m0 ax (since b m0 ) and ab divides m0 by (since a m0 ), so ab divides
m0 d, as desired.
√
√
√
2. Show that R = Z[ −2] = {a + bi 2 | a, b ∈ Z} is a ring. Let d(a + bi 2) = a2 + 2b2 and
show that d(xy) = d(x)d(y). What are the units of R?
Show that R is a Euclidean ring with Euclidean function d. Find an example of a prime in R.
Show that 11 is not prime in R.
If we define a + bi = a − bi for all a, b ∈ R, then
zw = zw
and
√
√
√
d(a + bi 2) = (a + bi 2)(a + bi 2) = a2 + 2b2 .
Therefore,
d(xy) = xyxy = (xx)(yy) = d(x)d(y).
In particular, if x is a unit, then 1 = d(1) = d(xx−1 ) = d(x)d(x−1 ), so d(x−1 ) = d(x)−1 . Since
d(x) is always an integer, x can only be a unit if x = 1 or x = −1, and both of these are units.
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Now, we claim that R is a Euclidean ring. Let x, y ∈ R, x, y 6= 0. We claim that there are
q, r ∈ R such that x = qy + r and d(r) < d(y). Let z = x/y and let a and b be the real and
0
0
imaginary parts
√ of z, so z = a + 0bi. Let a be the closest
√integer to a and let b be the closest
0
multiple of 2 to b. Then |a − a | ≤ 1/2 and |b − b | ≤ 2/2. Let
q = a0 + b0 i
r = x − qy
Then x = qy + r, so we only need to bound d(r).
We can write
x
−q
y
= y (a − a0 ) − (b − b0 )i
r=y
and
d(r) = d(y)d((a − a0 ) − (b − b0 )i)


√ !2
2
2 
1
+
≤ d(y) 
2
2
3
≤ d(y) ≤ d(y).
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Finally, we consider primes in R. Just as with the Gaussian integers, if xy = z, then d(z) =
d(x)d(y). So if d(z) is prime and√xy = z, then either d(x) = 1 or d(y) = 1 — that is, either x
or y is a unit. So, for instance, i 2 is a prime in R.
On the other hand, 11 is not a prime in R because
√
√
(3 + i 2)(3 − i 2) = 9 + 2 · 12 = 11.
3. Suppose that R = { m
n | n is odd}. If m 6= 0 and n is odd, define
m
d
= the exponent of 2 in the prime factorization of m.
n
Show that
R is a ring. What elements of R are units? Show that if a, b ∈ R and d(a) ≤ d(b),
then a b. Show that R is a Euclidean ring.
If ab , dc ∈ R, then
a c
ad + bc
+ =
,
b d
bd
−a
and bd is odd, so this is an element of R. Similarly, ab · dc = ac
bd and b are elements of R.
Therefore R is an abelian group under addition and is closed under multiplication. Furthermore, multiplication satisfies the associative and distributive laws because the real numbers
satisfy the associative and distributive laws. Therefore R is a ring.
m
If m
n ∈ R, then n ∈ R if and only if m is odd, so the units are the fractions with odd
numerators and denominators.
p
k 0
l 0
Let a = m
n and b = q ∈ R, so that n and q are odd. Suppose that m = 2 m , p = 2 p , where
0
0
m and p are odd. If d(a) ≤ d(b), then k ≤ l, and
b
pn
2l−k p0 n
=
=
.
a
qm
qm0
2
Since the denominator qm0 is odd and the numerator is an integer, this is an element of R.
To show that this is a Euclidean ring, we need to show two things: that d(a) ≤ d(ab) for every
a, b 6= 0 and that for every a, b 6= 0, there are q, r ∈ R such that
a = bq + r
(1)
and either r = 0 or d(r) < d(b). First, note that if m, n, p, and q are odd, then
j 2 m
d
=j
n
k 2 p
=k
d
q
p k 2 2 p
d
= j + k,
·
q
q
so d(ab) = d(a) + d(b) for all a, b 6= 0. In particular, d(ab) ≥ d(a).
Finally, if a, b ∈ R and a, b 6= 0, then there are two possibilities. If d(a) < d(b), then
substituting q = 0 and r = a into (1) gives
a=b·0+a
and d(r) = d(a) < d(b). On the other hand, if d(a) ≥ d(b), then a is a multiple of b, so we
can let q = ab and r = 0. Then (1) is satisfied and r = 0.
4. Factor 30 into a product of Gaussian primes.
We start by factoring 30 = 2 · 3 · 5. Then 3 is already a Gaussian prime. The other two
factors can be written as products: 2 = (1 + i)(1 − i) and 5 = (2 + i)(2 − i), so 30 =
(1 + i)(1 − i) · 3 · (2 + i)(2 − i). All of these are Gaussian primes – d(1 + i) = d(1 − i) = 2 is
prime, d(2 + i) = d(2 − i) = 5 is prime, and 3 is a prime congruent to 3 mod 4.
5. Suppose that R is a unital ring and that x, y ∈ R generate the same principal ideal. Prove
that x and y are associates
If (x) = (y), then there are a, b ∈ R such that x = ay and y = bx. That is, x y and y x.
This implies that x and y are associates.
6. Describe all of the ring homomorphisms from Z to Z20 . What are their kernels and images?
Any ring homomorphism from Z to Z20 is a homomorphism of additive groups, so it is of the
form
f (n) ≡ an (mod 20)
for some n. This automatically satisfies f (x + y) = f (x) + f (y). It satisfies f (xy) = f (x)f (y)
if axy = (ax)(ay) (mod 20). In particular, we need
a ≡ a2
(mod 20)
2
a −a≡0
a(a − 1) ≡ 0.
We can find the homomorphisms by checking all 20 possible values of a. Alternatively, we can
narrow things down by noticing that if a(a − 1) is a multiple of 20, then one of a and a − 1
3
must be divisible by 5, so we only need to check a = 0, 1, 5, 6, 10, 11, 15, 16. The only values
that satisfy the equation are a = 0, 1, 5, 16, so there are four homomorphisms from Z to Z20 ,
corresponding to multiplication by 0, 1, 5, and 16.
7. Let R = Z[t] and let U = (t2 + 1) be the principal ideal of R generated by t2 + 1. Show that
every coset of U contains exactly one polynomial of degree < 2 and show that
(U + at + b)(U + ct + d) = U + (ad + bc)t + (bd − ac).
Conclude that the map f : R/U → Z[i]
f (U + at + b) = b + ai
is an isomorphism.
Any coset of U can be written U + f , where f ∈ Z[t]. We can use polynomial long division to
write
f = q · (t2 + 1) + r,
where deg(r) < deg(t2 + 1) = 2. Then f − r = q · (t2 + 1) ∈ U , so r ∈ U + f .
On the other hand, if r and r0 are polynomials of degree less than 2 and r and r0 are both in
U + f , then r − r0 ∈ U and deg(r − r0 ) < 2. But any element of U is of the form (t2 + 1)g,
so the only element of degree less than 2 is 0. Therefore, r = r0 , and there is exactly one
polynomial of degree < 2 in U + f .
By the definition of multiplication in R/U , we have
(U + at + b)(U + ct + d) = U + (at + b)(ct + d)
= U + act2 + (ad + bc)t + bd
= U + ac(t2 + 1) − ac + (ad + bc)t + bd
= U + (ad + bc)t + (bd − ac).
In particular, if f is the map given, then f is well-defined, because each coset of U can be
written uniquely as U + at + b, it is a homomorphism because
f ((U + at + b)(U + ct + d)) = f (U + at + b)f (U + ct + d)
and
f ((U + at + b) + (U + ct + d)) = f (U + at + b) + f (U + ct + d),
and it is an isomorphism because its inverse is the map
f −1 (a + bi) = U + b + ai.
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