Solutions to Exercise Sheet 3
for 6CCM350A/7CCM350B
S. Vigni, 26/10/2011
Solution to Exercise 1
Define the map
ϕ : R −→ C,
� a −b �
b a
�−→ a + bi
where i ∈ C is such that i2 = −1. A straightforward computation shows that ϕ is a
ring homomorphism. On the other hand, ϕ is visibly bijective and so gives the desired
isomorphism.
Solution to Exercise 2
For notational convenience, set f := X 2 + X + 1̄. Since R is a field, the polynomial ring
R[X] is a PID, hence the ideal I = (f ) is maximal (that is, R[X]/I is a field) if and only if
f is an irreducible element of R[X]. But one checks that
¯ + 4̄ + 1̄ = 21
¯ = 0̄
f (4̄) = 16
in R = Z/(7), hence f has a linear factor in R[X] and so is not irreducible. Actually, one
can write f = (X + 3̄) · (X + 5̄). Therefore I is not a maximal ideal of R[X], which means
that R[X]/I is not a field.
Solution to Exercise 3
Parts (a) and (b) are completely elementary
� 2 0 � and we do not give details about them. To
answer (c) observe that, for example, 0 5 ∈ I but we can write
�2 0� �2 0� �1 0�
0 5 = 0 1 · 0 5
� �
� �
with 20 01 �∈ I and 10 05 �∈ I. Therefore I is not a prime ideal of R.
Solution to Exercise 4
Let us consider the ring homomorphism
ϕ : R[X] −→ C,
f �−→ f (i)
where, as usual, i ∈ C is such that i2 = −1. Of course, the map ϕ is surjective (in fact, it is
surjective even when restricted to the R-vector subspace of R[X] consisting of all polynomials
of degree ≤ 1). Let us study its kernel. First of all, note that R[X] is a PID because R is a
field, hence ker(ϕ) can be generated by a single element. By definition of i, the polynomial
X 2 + 1 belongs to ker(ϕ). On the other hand, since i �∈ R, no polynomial of degree ≤ 1
belongs to ker(ϕ) (check it!). It is now immediate to conclude that ker(ϕ) = (X 2 + 1), so ϕ
induces the desired ring isomorphism.
Solution to Exercise 5
(a) A routine verification.
(b) Let us consider the map
√
N : Z[ 2] �−→ Z,
√
a + b 2 �−→ a2 − 2b2 .
Then
√
√ �
√ �
�
�
N (a + b 2) · (c + d 2) = N ac + 2bd + (ad + bc) 2
= (ac + 2bd)2 − 2(ad + bc)2
√ �
√ �
�
�
= (a2 − 2b2 ) · (c2 − 2d2 ) = N a + b 2 · N c + d 2 .
(1)
�
√
√
Suppose that x = a + b 2 ∈ Z[ 2]× . Since N (1) = 1, it follows from (1) that N (x)�1 in Z,
hence N (x) = a2 − 2b2 ∈ {±1}.
√
2
2
Now assume
that
a
−
2b
∈
{±1}
with
a,
b
∈
Z
and
set
x
:=
a
+
b
2: we want to show
√
that x ∈ Z[ 2]× . If a2 − 2b2 = 1 then
√ � �
√ �
�
a + b 2 · a − b 2 = a2 − 2b2 = 1,
√
√
√
hence x ∈ Z[ 2]× since a − b 2 ∈ Z[ 2]. On the other hand, if a2 − 2b2 = −1 then
√ � � √
�
�
a + b 2 · b 2 − a = 2b2 − a2 = 1,
√
√
√
which shows that x ∈ Z[ 2]× because b 2 − a ∈ Z[ 2].
√
√ ×
(c) Write 3 = x · y with x, y ∈ Z[ 2]. We
want
to
see
if
we
can
choose
x,
y
∈
�
Z[
2] ,
√
in which case 3 would be reducible in Z[ 2]. The
multiplicative
nature
of
the
function
�
N that was established in (1) implies that N (x)�N (3) = 9, so N (x) ∈ {±1, ±3, ±9}. If
√
N (x) ∈√{±1} then x ∈ Z[ 2]× by part
√ (b), while if N (x) ∈ {±9} then N (y) ∈ {±1} and
×
y ∈ Z[ 2] . Thus, writing x = a + b 2, we are wondering whether at least one of the two
equations
a2 − 2b2 = 3,
a2 − 2b2 = −3
(2)
admits solutions a, b ∈ Z. A convenient way to study solutions to the equations in (2) is to
“go modulo 3”, that is, to search for solutions in the finite ring Z/(3). In other words, we
are looking for solutions to the equation
ā2 − 2̄b̄2 = 0̄
(3)
in Z/(3). As a preliminary remark, observe that the existence of solutions in Z/(3) to
equation (3) is a necessary (but not sufficient!) condition for the existence of solutions in
Z to the equations in (2) (in other words, every solution in Z to one of the equations in
(2) gives “by passing to the quotient” a solution to (3), but it may well be the case that
no solution in Z/(3) to (3) “lifts” to a solution in Z to one of the equations in (2)). The
squares in Z/(3) = {0̄, 1̄, 2̄} are 0̄ and 1̄, and an immediate inspection reveals that equation
(3) has the unique solution in Z/(3) given by ā = 0̄, b̄ = 0̄. This means that both a and b
are divisible by 3 in Z, hence a2 − 2b2 is divisible by 9 and so cannot be equal to ±3.
This shows that the two equations in (2) √
have no solutions a, b ∈ Z, therefore we can
conclude that 3 is an irreducible element of Z[ 2].
Solution to Exercise 6
We already know that if R is a field then R[X] is a euclidean domain, hence a PID. Now let
us assume that R[X] is a PID and prove the converse.
To begin with, let us show that X is an irreducible element of R[X]. Write X = f · g with
f, g ∈ R[X]; since the degree of the product of two polynomials is the sum of the degrees
(R is an integral domain!), without loss of generality we can assume that deg(f ) = 0 and
deg(g) = 1, so that f ∈ R. Let g = aX + b with a, b ∈ R and a �= 0. It follows that, in
particular, a·f = 1, hence f ∈ R× ⊂ R[X]× , which shows that X is irreducible in R[X]. But
R[X] is a PID by assumption, therefore (X) is a maximal ideal of R[X]. Finally, consider
the ring homomorphism
ϕ : R[X] −→ R,
h �−→ h(0).
It is immediate to check that ϕ is surjective and ker(ϕ) = (X), hence R ∼
= R[X]/(X) is a
field.