Question

Let $\mathit{f}\left(x\right)\mathbf{,}\mathit{g}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$have degree $\mathbf{⩽}\mathit{n}$and let ${\mathit{c}}_{\mathbf{0}}\mathbf{,}{\mathit{c}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{c}}_{\mathbf{n}}$be distinct elements of F. If $\mathit{f}\left(c_i\right)\mathbf{=}\mathit{g}\left(c_i\right)\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{i}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{n}$, prove that $\mathit{f}\left(x\right)\mathbf{=}\mathit{g}\left(x\right)\mathbf{}\mathit{i}\mathit{n}\mathbf{}\mathit{F}\left[x\right]$.

Short answer

The corollary is proved that $f\left(x\right)=g\left(x\right)$.

Step-by-step solution

Shows that f(x)=g(x) in F[x]

Let’s consider a polynomial, $h\left(x\right)=f\left(x\right)-g\left(x\right)$and $\mathrm{deg}\left(h\left(x\right)\right)⩽n$.

For that, each distinct elements are $c_0,.....,c_{n}thath\left(c_i\right)=f\left(c_i\right)-g\left(c_i\right)=0$.

Therefore, h (x) has at least one root that is n + 1.

By using Corollary 4.17, then,

If h (x) = 0, it means that $f\left(x\right)-g\left(x\right)=0$and hence, $f\left(x\right)=g\left(x\right)$.

Therefore, it is proved.