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.