Question

Let $\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$ and assume that $\mathit{f}\left(x\right)\mathbf{|}\mathit{g}\left(x\right)$ for every non-constant$\mathbf{g}\left(x\right)\mathbf{\in }\mathbf{F}\left[x\right]$. Show that $\mathit{f}\left(x\right)$ is a constant polynomial. [Hint: localid="1654517113267" $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ must divide both localid="1654517101344" $\mathbf{x}\mathbf{+}\mathbf{1}$and x.]

Short answer

It is proved that $f\left(x\right)$ is a constant polynomial.

Step-by-step solution

Statement of Theorem 4.7

Theorem 4.7 states that consider that $F$ as a field and $a\left(x\right),b\left(x\right)\in F\left(x\right)$ with $b\left(x\right)$non-zero.

1. When $b\left(x\right)$divides $a\left(x\right)$, then $cb\left(x\right)$ divide for some non-zero $c\in F$.
2. Any divisor of $a\left(x\right)$contains a degree that is less than or equal to $a\left(x\right)$.

Show that f(x) is a constant polynomial

When $f\left(x\right)$ divides any non-contact polynomial in particular, it divides $x+1$and $x$.

According to theorem 4.7 (2), this implies that $degf\left(x\right)\le 1$ and $f\left(x\right)=a_{1}x+a_0$ with some $a_1,a_0\in F$.

Moreover, since the degree of products of polynomials over fields would be the sum of the degrees of the factors when $deg\left(f\left(x\right)\right)=1$, or correspondingly , $a_1\ne 0$there exists $r,s\in F$ such that,

role="math" localid="1648088224049" $$\begin{gathered} x+1=r{a}_{1}a+a_0 \\ x=s{a}_{1}x+s{a}_0 \end{gathered}$$

Furthermore, $s{a}_1=1$ entails $s\ne 0$ and because $s{a}_0=0$ it could deduce that$a_0=0$ .

However, $1=r{a}_0$ and it deduces that role="math" localid="1648088554769" $1=0$, which is a contradiction.

As a result, $degf\left(x\right)=0$ and $f\left(x\right)$ would be a constant polynomial.

Hence, it is proved that $f\left(x\right)$is a constant polynomial.