THE UNIVERSITY OF SYDNEY
Math2968 Algebra (Advanced)
Semester 2
Tutorial Solutions Week 8
2008
1.
(for general discussion) Let
R
[
x
] denote the set of all polynomials in an indeterminate
x
with real
coefficients, which has the usual addition and multiplication of polynomials.
(a) Let
φ
:
R
[
x
]
→
C
be the ‘evaluation map’ at
i
=
√
−
1:
φ
:
p
(
x
)
mapsto→
p
(
i
)
for all
p
(
x
)
∈
R
[
x
]
.
Clearly
φ
preserves addition and multiplication. In particular,
φ
is a homomorphism between addi
tive abelian groups. Describe the kernel and image of
φ
. Interpret the Fundamental Homomorphism
Theorem in this case.
(b) Let
ψ
:
R
[
x
]
→
Mat
2
(
R
) be the ‘evaluation map’ at
M
=
bracketleftbigg
0
1
−
1
0
bracketrightbigg
:
ψ
:
p
(
x
)
mapsto→
p
(
M
)
for all
p
(
x
)
∈
R
[
x
]
.
Again
ψ
is a homomorphism between additive abelian groups. Describe the kernel and image of
ψ
.
Interpret the Fundamental Homomorphism Theorem in this case. Glue this and the previous case
together to find a representation of complex numbers by 2
×
2 real matrices.
Solution
(a) If
p
(
x
)
∈
(
x
2
+ 1)
R
[
x
] then
p
(
x
) = (
x
2
+ 1)
q
(
x
) for some
q
(
x
)
∈
R
[
x
], so that
p
(
i
) = (
i
2
+ 1)
q
(
i
) = 0
.
Thus (
x
2
+ 1)
R
[
x
]
⊆
ker
φ
. Suppose now
p
(
x
)
∈
ker
φ
, so that
p
(
i
) = 0. Hence
x
−
i
is a factor of
p
(
x
). But the coefficients of
p
(
x
) are real, so its roots come in complex conjugate pairs. Thus
x
+
i
is also a factor, so that
x
2
+ 1 = (
x
−
i
)(
x
+
i
)
is a factor of
p
(
x
). Hence
p
(
x
)
∈
(
x
2
+ 1)
R
[
x
]. This proves ker
φ
= (
x
2
+ 1)
R
[
x
]. Clearly, if
a,b
∈
R
then
a
+
bi
=
p
(
i
) where
p
(
x
) =
bx
+
a
, so that
φ
is surjective. Hence im
φ
=
C
. By the Fundamental
Homomorphism Theorem applied to these additive groups,
C
= im
φ
∼
=
R
[
x
]
/
ker
φ
=
R
[
x
]
/
(
x
2
+ 1)
R
[
x
]
,
which is just addition mod
x
2
+ 1 in disguise.
(b) If
p
(
x
)
∈
(
x
2
+ 1)
R
[
x
] then
p
(
x
) = (
x
2
+ 1)
q
(
x
) for some
q
(
x
)
∈
R
[
x
], so that
p
(
M
) = (
M
2
+
I
)
q
(
M
) = (
−
I
+
I
)
q
(
M
) = 0
.
Thus (
x
2
+ 1)
R
[
x
]
⊆
ker
ψ
.
Suppose now
p
(
x
)
∈
ker
ψ
, so that
p
(
M
) = 0.
Express
p
(
x
) =
(
x
2
+ 1)
q
(
x
) +
r
(
x
) for some quotient
q
(
x
) and remainder
r
(
x
) =
bx
+
a
. Then
0 =
p
(
M
) = (
M
2
+
I
)
q
(
M
) +
r
(
M
) = 0
q
(
M
) +
bM
+
aI
=
bracketleftbigg
a b
−
b a
bracketrightbigg
,
so that
a
=
b
= 0. Hence
p
(
x
)
∈
(
x
2
+ 1)
R
[
x
]. This proves ker
ψ
= (
x
2
+ 1)
R
[
x
]. To find the image,
again use quotients and remainders to see that
im
ψ
=
{
p
(
M
)

p
(
x
)
∈
R
[
x
]
}
=
{
(
M
2
+
I
)
q
(
M
) +
r
(
M
)

q
(
x
)
∈
R
[
x
]
, r
(
x
) a linear polynomial
}
=
{
(
−
I
+
I
)
q
(
M
) +
r
(
M
)

q
(
x
)
∈
R
[
x
]
, r
(
x
) =
bx
+
a, a,b
∈
R
}
=
{
bM
+
aI

a,b
∈
R
}
=
braceleftBig
bracketleftbigg
a b
−
b a
bracketrightbigg
vextendsingle
vextendsingle
vextendsingle
a,b
∈
R
bracerightBig
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 One '09
 easdown
 Algebra, Polynomials, Addition, Normal subgroup, Subgroup, Cyclic group, Coset, Fundamental Homomorphism Theorem

Click to edit the document details