Roots and Reducibility

In this section, we will use the function view point of a polynomial $f(x)\in F[x]$ introduced back in to define roots, and relate roots with divisibility.

Evaluation Map

Given a polynomial $f(x)=a_n{x}^n+\cdots +a_{1}x+a_0\in F[x]$, we can view $f(x)$ as a function $F\to F$ by evaluating $f(x)$ at $a\in F$: $f(a)=a_n{a}^n+\cdots +a_{1}a+a_0$. We make the following definition:

Definition:

Given a polynomial $f(x)\in F[x]$, the associated function $f:F\to F$ is the function $F\to F$ defined by $f(a)=a_n{a}^n+\cdots +a_{1}a+a_0$. By abuse of notation, we will denote the associated function of $f(x)$ as $f$.

Warning:

We have defined the Evaluation Map on Polynomial Ring, denoted as $e{v}_a$, as a morphism from $F[x]$ to $F$ that evaluates a polynomial at a value $a\in F$. This is different from the associated function $f(a)$, which is a function from $F$ to $F$.

Warning:

As mentioned in the definition, the notation $f:F\to F$ is an abuse of notation. It is possible that:

1. We start with a non-zero polynomial $f(x)\in F[x]$. The associated function $f$ is the zero function $F\to F$. For example, take $f(x)=x^2-x$ and $F={\mathbb{Z}}_2$.
2. We start with two different polynomials $f(x),g(x)\in F[x]$. The associated functions $f$ and $g$ are the same function $F\to F$.

The associated function of a polynomial is closely related to the divisibility of the polynomial by linear factors. We will prove the following theorem:

Theorem:

Given a polynomial $f(x)\in F[x]$ and $a\in F$, let $r\in F$ be the remainder of the division of $f(x)$ by $(x-a)$. Then:

$$r=f(a).$$

Proof of Theorem

We will prove the theorem by the division algorithm. Let $f(x)=q(x)(x-a)+r(x)$ be the division of $f(x)$ by $(x-a)$, where $q(x),r(x)\in F[x]$ and $\mathrm{deg}(r)<\mathrm{deg}(x-a)=1$. We can write $r(x)=r$ for some $r\in F$. We plug in $x=a$ to the equation:

$$f(a)=q(a)(a-a)+r=r.$$

This completes the proof.

Roots of a Polynomial

Definition:

Given a polynomial $f(x)\in F[x]$, we say $a\in F$ is a root of $f(x)$ if $f(a)=0$. We can also say $f(x)$ vanishes at $a$.

The root of a polynomial is again closely related with the divisibility of the polynomial of linear factors. We can restate the above theorem by:

Theorem:

Given a polynomial $f(x)\in F[x]$ and $a\in F$, $a$ is a root of $f(x)$ if and only if $f(x)$ is divisible by $(x-a)$.

The proof of the theorem is the same as the above theorem, we omit it.

One immediate corollary of the above theorem is:

Corollary:

Given a polynomial $f(x)\in F[x]$ with $\mathrm{deg}(f)\ge 1$, $f(x)$ has at most $\mathrm{deg}(f)$ roots in $F$.

Above theorem also gives us a way to check whether a polynomial is reducible:

Corollary:

If $f(x)\in F[x]$ has $\mathrm{deg}(f)\ge 2$ and $f(x)$ has a root in $F$, then $f(x)$ is reducible. Equivalently, if an $f(x)$ with $\mathrm{deg}(f)\ge 2$ is irreducible, then $f(x)$ has no roots in $F$.

Warning:

One may want to know if the converse of the corollary is true, i.e. if $f(x)$ is reducible, then $f(x)$ has a root in $F$. This is not true in general, as we will see in the following example:

$$f(x)=x^4+2x^2+1\in \mathbb{R}[x].$$

The polynomial $f(x)$ is reducible as $f(x)=(x^2+1{)}^2$, but $f(x)$ has no roots in $\mathbb{R}$.

The converse of the corollary is true in the following case due to low degree:

Theorem:

Given a polynomial $f(x)\in F[x]$ with $2\le \mathrm{deg}(f)\le 3$, if $f(x)$ has a root in $F$, then $f(x)$ is reducible.

Checking irreducibility of a polynomial is actually a hard problem in general. We will see a criterion for $\mathbb{Q}[x]$ in the next section.