Question

If $\mathbf{f}\left(\mathbf{x}\right)\in \mathbf{F}\left[\mathbf{x}\right]$has degree n, prove that there exists an extension field E of F such that$\mathbf{f}\left(x\right)\mathbf{=}{\mathbf{c}}_{\mathbf{0}}\left(x-c_1\right)\left(x-c_2\right)\mathbf{\cdots }\left(x-c_n\right)$ for some (not necessarily distinct) ${\mathbf{c}}_{\mathbf{i}}\in \mathbf{E}$. In other words, E contains all the roots of $\mathbf{f}\left(x\right)$.

Short answer

It is proved that there exists an extension field$E$ of$F$ such that$f\left(x\right)=c_0\left(x-c_1\right)\left(x-c_2\right)\cdots \left(x-c_n\right)$ for some $c_i\in E$.

Step-by-step solution

Statement of Corollary 5.12

Corollary 5.12states consider that$F$ as a fieldand$f\left(x\right)$ as a nonconstant polynomial in $F\left[x\right]$. Then therewillbe an extension field$K$ of $F$, which contains a root of $f\left(x\right)$.

Show that there exists an extension field E of F such that f(x)=c0(x-c1)(x-c2)⋯(x-cn)  for some ci∈E

Use induction on$n$ to demonstrate there exists an extension field E of F.

For $n=1$,the statement is trivially true.

Assume that the statement is true for polynomials of degrees less than $n$. According to corollary 5.12, there could be some extension $H$of $F$that has a root $c_n$of $f\left(x\right)$yield $f\left(x\right)=\left(x-c_n\right)g\left(x\right)$in $H\left[x\right]$. Through the inductive hypothesis, there will be an extension $K$of $H$in which$g\left(x\right)=c_0{\prod }_{i=1}^{n-1}\left(x-c_i\right)$ because $\mathrm{deg}g\left(x\right)=n-1<\mathrm{deg}f\left(x\right)$.

It follows that in $K\left[x\right]$, there is $f\left(x\right)=c_0{\prod }_{i=1}^n\left(x-c_i\right)$.

Hence, it is proved that there exists an extension field $E$of $F$such that$f\left(x\right)=c_0\left(x-c_1\right)\left(x-c_2\right)\cdots \left(x-c_n\right)$ for some $c_i\in E$.