Irreducible
Let be a field. A non-zero polynomial
p
(
x
)
∈
F
[
x
]
is irreducible over
F
if
deg
(
p
)
≥
1
and there is no factorization
(
p
(
x
)
=
f
(
x
)
g
(
x
)
)
where
f
(
x
)
,
g
(
x
)
∈
F
[
x
]
with
deg
(
f
)
,
deg
(
g
)
<
deg
(
p
)
Note that the field plays into whether or not a polynomial is irreducible, for example
x
2
+
1
is irreducible over
R
, but factors in
C
Degree One Polynomials are Irreducible
As per title.
Observer that the degree of any polynomial is bounded below by zero. Now let
p
(
x
)
be a degree one polynomial, and suppose there were two polynomials
a
(
x
)
,
b
(
x
)
such that
p
(
x
)
=
a
(
x
)
b
(
x
)
where
deg
(
a
)
,
deg
(
b
)
<
deg
(
p
)
=
1
this means that both
a
(
x
)
,
b
(
x
)
have degree zero, so then they are constant, so their product is constant, which is a contradiction, therefore
p
(
x
)
is irreducible.
Polynomial of Degree 2 or 3 is Irreducible Iff It has no Roots
A polynomial
p
(
x
)
∈
F
[
x
]
of degree 2 or 3 is irreducible over
F
if and only if
F
contains no root of
p
(
x
)
Note that the above doesn't hold for degree 4 because the polynomial
(
x
2
+
1
)
2
factors as
(
x
2
+
1
)
(
x
2
+
1
)
in
R
[
x
]
(and is thus not irreducible) but has no real roots.
Irreducible Either has GCD 1 or Divides any other Non-Constant Polynomial
Let
p
(
x
)
∈
F
[
x
]
be irreducible. If
g
(
x
)
∈
F
[
x
]
is not constant, then either
gcd
(
p
(
x
)
,
g
(
x
)
)
=
1
or
p
(
x
)
∣
g
(
x
)
Prime Ideal
An ideal
I
in a ring
R
is called a prime ideal if it is a proper ideal and
a
b
∈
I
implies that
a
∈
I
or
b
∈
I
Irreducibility Criterion with Prime Ideals
If
f
is a field, then a non-zero polynomial
p
(
x
)
∈
F
[
x
]
is irreducible iff
(
p
(
x
)
)
⋄
is a prime ideal
A Proper Ideal is Prime iff the Quotient Ring is a Domain
Suppose that
I
is a proper ideal of
R
then
I
is a prime ideal iff
R
/
I
is a domain
Maximal Ideal
An ideal
I
in a crone
R
is a maximal ideal if it is a proper ideal and there is no ideal
J
such that
I
⊂
J
⊂
R
Maximal iff Quotient Ring is a Field
A proper ideal
I
in a crone
R
is a maximal ideal if and only if
R
/
I
is a field.
Maximal Implies Prime
Every maximal ideal
I
in a crone
R
is a prime ideal
In a PID every Non-Zero Prime Ideal is Maximal
If
R
is a principal ideal domain then every non-zero prime ideal
I
is a maximal ideal
Field Polynomials Mod an Irreducible Yields a Root
If
F
is a field and
p
(
x
)
∈
F
[
x
]
is irreducible, then the quotient ring
F
[
x
]
/
(
p
(
x
)
)
⋄
is a field containing (an isomorphic copy of)
F
and a root of
p
(
x
)
Since
p
(
x
)
is irreducible, then it has degree at least one, and therefore is non-zero thus we know that
(
p
(
x
)
)
⋄
is a prime ideal, it is non-zero since
p
(
x
)
∈
(
p
(
x
)
)
⋄
.
Since
F
is a field, then we know that
F
[
x
]
is a PID, therefore since we've just seen that
I
is a non-zero prime ideal then we know that
I
is a maximal ideal.
Since
I
is a maximal ideal, then we know that
E
:=
F
[
x
]
/
I
is a field. Observe that the map
a
↦
a
+
I
is an isomorphism from
F
to
F
′
:=
{
a
+
I
:
a
∈
F
}
⊆
E
. We finally show that
E
contains a root.
Consider
θ
=
x
+
I
∈
E
, and write
p
(
x
)
=
a
0
+
a
1
x
1
+
⋯
+
a
n
x
n
, then
p
(
θ
)
=
(
a
0
+
I
)
+
(
a
1
+
I
)
θ
+
⋯
+
(
a
n
+
I
)
θ
n
=
(
a
0
+
I
)
+
(
a
1
+
I
)
(
x
+
I
)
+
⋯
+
(
a
n
+
I
)
(
x
+
I
)
n
=
(
a
0
+
I
)
+
(
a
1
x
+
I
)
+
⋯
+
(
a
n
x
n
+
I
)
=
a
0
+
a
1
x
+
⋯
+
a
n
x
n
+
I
=
p
(
x
)
+
I
=
0
+
I
Since
0
+
I
is the zero element of
F
[
x
]
/
I
then
θ
is a root of
p
(
x
)
Linear Factor
Suppose that
F
is a field and
a
∈
F
, then we say that the polynomial
x
−
a
∈
F
[
x
]
is a linear factor.
Splits Over
A polynomial
f
(
x
)
∈
F
[
x
]
splits over
F
if it is a product of linear factors in
F
[
x
]
Splits iff Roots
f
(
x
)
∈
F
[
x
]
splits over
F
[
x
]
if and only if
F
contains all the roots of
f
(
x
)
.
Prime Field
The prime field of a field
F
is the intersection of all the subfields of
F
Characteristic
Suppose that
p
is prime, and that
F
is a field, then we say that
F
has
characteristic
p
if its
prime field is isomorphic to
Z
p
%
Unique Roots in a Field of Characteristic p
Let
F
be a finite field of characteristic
p
, show that each element
a
∈
F
has a unique
p
-th root in
F
Let's first observe that that map
g
:
F
→
F
defined by
f
(
x
)
=
x
p
is an injective mapping, suppose that
x
p
=
y
p
⟺
x
p
−
y
p
=
0
then by the properties of a field of characteristic
p
we have
(
x
−
y
)
p
=
0
⟺
x
−
y
=
0
⟺
x
=
y
, so that
g
is injective.
Recall that an injection over a finite set becomes a surjection, since
a
∈
F
then there exists some
k
∈
F
such that
g
(
k
)
=
a
which is to say
k
p
=
a
thus we know that there exists a root of
a
. Let's prove that it's unique
Suppose that
j
≠
k
is another
p
-th root of
a
therefore we have that
j
p
=
a
=
k
p
but then we have
j
p
−
k
p
=
0
so we have that
(
j
−
k
)
p
=
0
⟺
j
=
k
which is a contradiction, so therefore the root is unique.