Irreducible

Let \( F \) be a field. A non-zero polynomial \( p \left( x \right) \in F \left[ x \right] \) is irreducible over \( F \) if \( \operatorname{ deg } \left( p \right) \ge 1 \) and there is no factorization \( \left( p \left( x \right) = f \left( x \right) g \left( x \right) \right) \) where \( f \left( x \right) , g \left( x \right) \in F \left[ x \right] \) with \( \operatorname{ deg } \left( f \right) , \operatorname{ deg } \left( g \right) \lt \operatorname{ deg } \left( p \right) \)

Note that the field plays into whether or not a polynomial is irreducible, for example \( x ^ 2 + 1 \) is irreducible over \( \mathbb{ R } \), but factors in \( \mathbb{ C } \)

Degree One Polynomials are Irreducible

As per title.

Observer that the degree of any polynomial is bounded below by zero. Now let \( p \left( x \right) \) be a degree one polynomial, and suppose there were two polynomials \( a \left( x \right) , b \left( x \right) \) such that \( p \left( x \right) = a \left( x \right) b \left( x \right) \) where \( \operatorname{ deg } \left( a \right) , \operatorname{ deg } \left( b \right) \lt \operatorname{ deg } \left( p \right) = 1 \) this means that both \( a \left( x \right) , b \left( x \right) \) have degree zero, so then they are constant, so their product is constant, which is a contradiction, therefore \( p \left( x \right) \) is irreducible.

Polynomial of Degree 2 or 3 is Irreducible Iff It has no Roots

A polynomial \( p \left( x \right) \in F \left[ x \right] \) of degree 2 or 3 is irreducible over \( F \) if and only if \( F \) contains no root of \( p \left( x \right) \)

Note that the above doesn't hold for degree 4 because the polynomial \( \left( x ^ 2 + 1 \right) ^ 2 \) factors as \( \left( x ^ 2 + 1 \right) \left( x ^ 2 + 1 \right) \) in \( \mathbb{ R } \left[ x \right] \) (and is thus not irreducible) but has no real roots.

Irreducible Either has GCD 1 or Divides any other Non-Constant Polynomial

Let \( p \left( x \right) \in F \left[ x \right] \) be irreducible. If \( g \left( x \right) \in F \left[ x \right] \) is not constant, then either \( \gcd \left( p \left( x \right) , g \left( x \right) \right) = 1 \) or \( p \left( x \right) \mid g \left( x \right) \)

Prime Ideal

An ideal \( I \) in a ring \( R \) is called a **prime ideal** if it is a proper ideal and \( ab \in I \) implies that \( a \in I \) or \( b \in I \)

Irreducibility Criterion with Prime Ideals

If \( f \) is a field, then a non-zero polynomial \( p \left( x \right) \in F \left[ x \right] \) is irreducible iff \( \left( p \left( x \right) \right) _ \diamond \) 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

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 \( p \left( x \right) \in F \left[ x \right] \) is irreducible, then the quotient ring \( F \left[ x \right] / \left( p \left( x \right) \right) _ \diamond \) is a field containing (an isomorphic copy of) \( F \) and a root of \( p \left( x \right) \)

Since \( p \left( x \right) \) is irreducible, then it has degree at least one, and therefore is non-zero thus we know that \( \left( p \left( x \right) \right) _ \diamond \) is a prime ideal, it is non-zero since \( p \left( x \right) \in \left( p \left( x \right) \right) _ \diamond \).

Since \( F \) is a field, then we know that \( F \left[ x \right] \) 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 \( E := F \left[ x \right] / I \) is a field. Observe that the map \( a \mapsto a \boldsymbol{+} I \) is an isomorphism from \( F \) to \( F ^ \prime := \left\{ a \boldsymbol{+} I : a \in F \right\} \subseteq E \). We finally show that \( E \) contains a root.

Consider \( \theta = x \boldsymbol{+} I \in E \), and write \( p \left( x \right) = a_0 + a_1 x ^ 1 + \cdots + a_n x ^ n \), then \[ \begin{align} p \left( \theta \right) &= \left( a _ 0 \boldsymbol{+} I \right) + \left( a _ 1 \boldsymbol{+} I \right) \theta + \cdots + \left( a _ n \boldsymbol{+} I \right) \theta ^ n \\ &= \left( a _ 0 \boldsymbol{+} I \right) + \left( a _ 1 \boldsymbol{+} I \right) \left( x \boldsymbol{+} I \right) + \cdots + \left( a _ n \boldsymbol{+} I \right) \left( x \boldsymbol{+} I \right) ^ n \\ &= \left( a _ 0 \boldsymbol{+} I \right) + \left( a _ 1 x \boldsymbol{+} I \right) + \cdots + \left( a _ n x ^ n \boldsymbol{+} I \right) \\ &= a_0 + a_1 x + \cdots + a_n x ^ n \boldsymbol{+} I \\ &= p \left( x \right) \boldsymbol{+} I \\ &= 0 \boldsymbol{+} I \\ \end{align} \] Since \( 0 \boldsymbol{+} I \) is the zero element of \( F \left[ x \right] / I \) then \( \theta \) is a root of \( p \left( x \right) \)

Linear Factor

Suppose that \( F \) is a field and \( a \in F \), then we say that the polynomial \( x - a \in F \left[ x \right] \) is a linear factor.

Splits Over

A polynomial \( f \left( x \right) \in F \left[ x \right] \) splits over \( F \) if it is a product of linear factors in \( F \left[ x \right] \)

Splits iff Roots

\( f \left( x \right) \in F \left[ x \right] \) splits over \( F \left[ x \right] \) if and only if \( F \) contains all the roots of \( f \left( x \right) \).

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 \( \mathbb{ Z } _ p ^ \% \)

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 \( f \left( x \right) = x ^ p \) is an injective mapping, suppose that \( x ^ p = y ^ p \iff x ^ p - y ^ p = 0 \) then by the properties of a field of characteristic \( p \) we have \( \left( x - y \right) ^ p = 0 \iff x - y = 0 \iff x = y\), 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 \( g \left( k \right) = a \) 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 \( \left( j - k \right) ^ p = 0 \iff j = k \) which is a contradiction, so therefore the root is unique.