Then
p
(
λ
) = 0, so there is a monic polynomial in
K
[
z
] with
λ
as a root.
Pick a monic polynomial
q
∈
K
[
z
] of minimal degree such that
q
(
λ
) = 0. Then obviously
q
is irreducible.
Uniqueness:
Suppose
p, q
∈
K
[
z
] are both monic polynomials of minimal degree such that
p
(
λ
) =
q
(
λ
) = 0. Then deg(
p
) = deg(
q
), so by the Euclidean algorithm, there are
k, r
∈
K
[
z
]
with deg(
r
)
<
deg(
q
) such that
p
=
kq
+
r
. Now
p
(
λ
) =
k
(
λ
)
q
(
λ
) +
r
(
λ
), so
r
(
λ
) = 0. This
is only possible if
r
= 0 as deg(
r
)
<
deg(
q
). Hence
p
=
kq
with deg(
p
) = deg(
q
), so
k
is a
constant. As
p, q
are both monic,
k
= 1, and
p
=
q
.
Corollary 4.3.3.
Suppose
λ
∈
F
. Then
λ
∈
K
if and only if
deg(Irr
K
,λ
) = 1
.
Corollary 4.3.4.
Suppose
λ
∈
F
and
dim
K
(
F
) =
n
. Then
deg(Irr
K
,λ
)
≤
n
.
Lemma 4.3.5.
Suppose
λ
∈
F
is a root of
p
∈
K
[
x
]
. Then
Irr
K
,λ

p
.
Proof.
The proof is similar to 4.3.2. Without loss of generality, we may assume
p
is monic (if
not, we replace
p
with
p/
LT(
p
). Now since deg(Irr
K
,λ
)
≤
deg(
p
), by the Euclidean algorithm,
there are
k, r
∈
K
[
z
] with deg(
r
)
<
deg(Irr
K
,λ
) such that
p
=
k
Irr
K
,λ
+
r
. We immediately
see
r
(
λ
) = 0, so
r
= 0.
Example 4.3.6.
We know that
C
is a two dimensional real vector space. If
λ
∈
C
is a root
of
p
∈
R
[
z
], then
λ
is also a root of
p
. In fact, since
p
(
λ
) = 0, we have that
p
(
λ
) =
p
(
λ
) = 0
since the coefficients of
p
are all real numbers. Hence, the irreducible polynomial in
R
[
z
] for
λ
∈
C
is given by
Irr
R
,λ
(
z
) =
(
z

λ
if
λ
∈
R
(
z

λ
)(
z

λ
) =
z
2

2 Re(
λ
)
z
+

λ

2
if
λ /
∈
R
.
Definition 4.3.7.
The polynomial
p
∈
F
[
z
] splits (into linear factors) if there are
λ
1
, . . . , λ
n
∈
F
such that
p
(
z
) =
n
Y
i
=1
(
z

λ
i
)
.
Examples 4.3.8.
(1) Every constant polynomial trivially splits in
F
[
z
].
(2) The polynomial
z
2
+ 1 splits in
C
[
z
] but not in
R
[
z
].
Definition 4.3.9.
The field
F
is called algebraically closed if every
p
∈
F
[
z
] splits.
51
4.4
The Fundamental Theorem of Algebra
Definition 4.4.1.
A function
f
:
C
→
C
is entire if there is a power series representation
f
(
z
) =
∞
X
n
=0
a
n
z
n
where
a
n
∈
C
for all
n
∈
Z
≥
0
that converges for all
z
∈
C
.
Examples 4.4.2.
(1) Every polynomial in
C
[
z
] is entire as its power series representation is itself.
(2)
f
(
z
) =
e
z
is entire as
e
z
=
∞
X
j
=0
z
n
n
!
(3)
f
(
z
) = cos(
z
) is entire as
cos(
z
) =
∞
X
j
=0
(

1)
n
z
2
n
(2
n
)!
(4)
f
(
z
) = sin(
z
) is entire as
sin(
z
) =
∞
X
j
=0
(

1)
n
z
2
n
+1
(2
n
+ 1)!
In order to prove the Fundamental Theorem of Algebra, we will need a theorem from
complex analysis. We state it without proof.
Theorem 4.4.3
(Liouville)
.
Every bounded, entire function
f
:
C
→
C
is constant.
Theorem 4.4.4
(Fundamental Theorem of Algebra)
.
Every nonconstant polynomial in
C
[
x
]
has a root.
Proof.
Let
p
be a polynomial in
C
[
z
]. If
p
has no roots, then 1
/p
is entire and bounded. By
Liouville’s Theorem, 1
/p
is constant, so
p
is constant.
Remark
4.4.5
.
Let
p
∈
C
[
z
] be a nonconstant polynomial, and let
n
= deg(
p
).
Then by
4.4.4,
p
has a root, say
μ
1
, so there is a
p
1
∈
C
[
z
] such that
p
(
z
) = (
z

μ
1
)
p
1
(
z
). If
p
1
is
nonconstant, then apply 4.4.4 again to see
p
1
has a root
μ
2
, so there is a
p
2
∈
C
[
z
] with
deg(
p
2
) =
n

2 such that
p
(
z
) = (
z

μ
1
)(
z

μ
2
)
p
2
(
z
). Repeating this process, we see that
p
n
must be constant as a polynomial has at most
n
roots. Hence there are
μ
1
, . . . , μ
n
∈
C
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 Fall '08
 GUREVITCH
 Linear Algebra, Algebra, Matrices, Vector Space, Sets, WI