Basic Theorems Regarding Polynomials over a Field

# Basic Theorems Regarding Polynomials over a Field

Recall from the Polynomials over a Field page that if $(F, +, \cdot)$ is a field then a polynomial over $F$ is an equation of the form:

(1)\begin{align} \quad f(x) = a_0 + a_1x + ... + a_nx^n \end{align}

The coefficients $a_0, a_1, ..., a_n \in F$. We also said that if $a_n \neq 0$ then the degree of $f$ is $n$ and is denoted $\deg (f) = n$. Furthermore, if $a_n = 1$ then $f$ is said to be monic.

We will now look at some basic theorems regarding polynomials over a field.

Theorem 1: Let $(F, +, \cdot)$ be a field and let $f, g \in F[x]$. If $f(x) \neq 0$ and $g(x) \neq 0$ then $f(x)g(x) \neq 0$. |

**Proof:**Let $f, g \in F[x]$ with $f(x) \neq 0$ and $g(x) \neq 0$. Then $f(x)$ and $g(x)$ are of the following forms (with $a_n \neq 0$ and $b_m \neq 0$):

\begin{align} \quad f(x) &= a_0 + a_1x + ... + a_nx^n \\ \quad g(x) &= b_0 + b_1x + ... + b_mx^m \end{align}

- Taking their product yields a polynomial of the form:

\begin{align} \quad f(x)g(x) &= (a_0 + a_1x + ... + a_nx^n)(b_0 + b_1x + ... + b_mx^m) \\ &= a_0b_0 + ... + a_nb_mx^{m+n} \end{align}

- Since $a_n, b_m \neq 0$, we have that $a_nb_m \neq 0$ so $f(x)g(x) \neq 0$. $\blacksquare$

Theorem 2: Let $(F, +, \cdot)$ be a field and let $f, g, h \in F[x]$. If $f(x)h(x) = g(x)h(x)$ and $h(x) \neq 0$ then $f(x) = g(x)$. |

**Proof:**Let $f, g, h \in F[x]$ with $f(x)h(x) = g(x)h(x)$ and $h(x) \neq 0$. Then:

\begin{align} \quad f(x)h(x) - g(x)h(x) &= 0 \\ \quad h(x)[f(x) - g(x)] &= 0 \end{align}

- Since $h(x) \neq 0$, the previous theorem tells us that $f(x) - g(x) = 0$. So $f(x) = g(x)$. $\blacksquare$