Question

If $\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$, show that every non-zero constant polynomial divides role="math" localid="1648073927112" $\mathit{f}\left(x\right)$.

Short answer

It is proved that every non-zero constant polynomial divides $f\left(x\right)$.

Step-by-step solution

Definition

Let $a\left(x\right),b\left(x\right)\in F\left[x\right]withb\left(x\right)\ne 0$ then$b\left(x\right)$ divides$a\left(x\right)$ , ifF is a field.

Proof part

It is given that F is a field. Let $a_0\in F\left[x\right]$, non-zero constant polynomial be a unit having multiplicative inverse${a_0}^{-1}$ . Then, for any $\sum _{i=0}^m{b}_i{x}^i\in F\left[x\right]$.

We have,

role="math" localid="1648074743867" $\sum _{i=0}^m{b}_i{x}^i=a_0\left({a_0}^{-1}{b}_i{x}^i\right)$

As $a_0$is a factor of role="math" localid="1648075227355" $\sum _{i=0}^m{a_0}^{-1}{b}^i{x}^i$which implies that $a_0$will divide $\sum _{i=0}^m{a_0}^{-1}{b}_i{x}^i$, so, every non-zero constant polynomial divides $f\left(x\right)$.