prime ideals and maximal ideals

Irreducible

Let

F

be a field. A non-zero polynomial

p (x) \in F [x]

is irreducible over

F

deg (p) \geq 1

and there is no factorization

(p (x) = f (x) g (x))

where

f (x), g (x) \in 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) \in 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) \in F [x]

be irreducible. If

g (x) \in F [x]

is not constant, then either

gcd (p (x), g (x)) = 1

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 \in I

implies that

a \in I

b \in I

Irreducibility Criterion with Prime Ideals

f

is a field, then a non-zero polynomial

p (x) \in 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 \subset J \subset 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

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

F

is a field and

p (x) \in 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 $p (x) \in {(p (x))}_{⋄}$ .

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 $a \mapsto a + I$ is an isomorphism from $F$ to $F^{'} : = {a + I : a \in F} \subseteq E$ . We finally show that $E$ contains a root.

Consider $θ = x + I \in E$ , and write $p (x) = a_{0} + a_{1} x^{1} + \dots + a_{n} x^{n}$ , then $p (θ) & = (a_{0} + I) + (a_{1} + I) θ + \dots + (a_{n} + I) θ^{n} & = (a_{0} + I) + (a_{1} + I) (x + I) + \dots + (a_{n} + I) {(x + I)}^{n} & = (a_{0} + I) + (a_{1} x + I) + \dots + (a_{n} x^{n} + I) & = a_{0} + a_{1} x + \dots + a_{n} x^{n} + I & = p (x) + I & = 0 + I$ 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 \in F

, then we say that the polynomial

x - a \in F [x]

is a linear factor.

Splits Over

A polynomial

f (x) \in F [x]

splits over

F

if it is a product of linear factors in

F [x]

Splits iff Roots

f (x) \in 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 \in F

has a unique

p

-th root in

F

Let's first observe that that map $g : F \to F$ 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 $a \in F$ then there exists some $k \in F$ 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 $j \neq k$ 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.

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)