Reducible and Irreducible Polynomials over a Field

# Reducible and Irreducible Polynomials over a Field

 Definition: Let $(F, +, \cdot)$ be a field and let $f \in F[x]$. Then $f$ is said to be Irreducible over $F$ if $f$ cannot be factored into a product of polynomials all of which having lower degree than $f$. If $f$ is not irreducible over $F$ then we say that $f$ is Reducible over $F$.

For example, over the field of real numbers the polynomial $x^2 + 1$ is irreducible.

However, over the the field of complex numbers the polynomial $x^2 + 1$ is reducible since:

(1)
\begin{align} \quad x^2 + 1 = (x + i)(x - i) \end{align}

And because $\deg (x + i) = 1$, $\deg (x - i) = 1$, both of which are less than $\deg (x^2 + 1) = 2$.

Thus when we talk about irreducibility of polynomials it is important to specify the field we are working over.

We will now look at a very nice theorem regarding irreducibility of polynomials and roots of that polynomial.

 Theorem 1: Let $(F, +, \cdot)$ be a field and let $f \in F[x]$ with $\deg (f) = 2$ or $\deg(f) = 3$. Then $f$ is irreducible over $F$ if and only if $f$ has no roots in $F$.
• Proof: We first consider the case when $\deg (f) = 2$.
• $\Rightarrow$ Suppose that $f$ is irreducible over $F$. Then $f$ cannot be written as a product of polynomials all having degree less than $f$. So $f$ cannot be written as a product of two linear factors. But a polynomial has a root $c \in F$ if and only if $(x - c)$ is a factor of $f$. So $(x - c)$ is not a factor of $f$ for all $c \in F$. So $f$ has no roots in $F$.
• $\Leftrightarrow$ Suppose that $f$ has no roots in $F$. Then $(x - c)$ is not a factor of $f$ for all $c \in F$. So $f(x) \neq (x - c_1)(x - c_2)$ for any $c_1, c_2 \in F$ so $f$ is irreducible over $F$.
• We now consider the case when $\deg (f) = 3$.
• Note that any factorization of $f$ into polynomials all of degree less than $f$ must contain a polynomial of degree 1 (a linear factor). So $f$ is irreducible if and only if $f$ contains no roots in $F[x]$. $\blacksquare$