Question

Use mathematical induction to prove Corollary 4.17.

Short answer

It is proved that $f\left(x\right)$ contains most $n$ roots in$F$ .

Step-by-step solution

Statement of Corollary 4.17

Corollary4.17considers $F$ as a field, and $f\left(x\right)$ a non-zero polynomial of degree $n$ in $F\left[x\right]$. Then, $f\left(x\right)$ contain at most $n$ roots in $F$.

Use mathematical induction to prove corollary 4.17

Consider $f\left(x\right)\in F\left[x\right]$ as the non-zero polynomial of degree $n$. Use induction on $n$ to demonstrate that $f\left(x\right)$ has at most $n$ roots.

For any $r\in F$, there is $f\left(r\right)=c\ne 0$ since $f\left(x\right)$ is a non-zero constant polynomial $c$ for $n=0$. As a result, $f\left(x\right)$ contain 0 roots.

Now, assume that the conclusion holds for a polynomial of degree $k<n$ and consider that $f\left(x\right)$ contain degree $n>0$.

Assume that $f\left(x\right)$ contain a root $a\in F$(When it does not contain 0 roots, we were done because $0<n$). According to the factor theorem, $\left.\left(x-a\right)\right|f\left(x\right)$ and therefore, there is an $h\left(x\right)\in F\left[x\right]$ in which $f\left(x\right)=\left(x-a\right)h\left(x\right)$.

It is observed that in particular, an integral domain because $F$ is a field. When $f\left(x\right)$ contains a root $b$ which is not $a$ then, $b-a\ne 0$. $b$is a root of $h\left(x\right)$.

Moreover, $\mathrm{deg}\left(f\left(x\right)\right)=1+\mathrm{deg}\left(h\left(x\right)\right)$ because $F$ is a field. As a result, $\mathrm{deg}\left(h\left(x\right)\right)<\mathrm{deg}\left(f\left(x\right)\right)=n$ and therefore, the inductive hypothesis $h\left(x\right)$ contains at most $n-1$ roots. It concludes that $f\left(x\right)$ contains at most $n$ roots because roots of $f\left(x\right)$ are either a or a root of $h\left(x\right)$.

Hence, it is proved that $f\left(x\right)$ contains at most $n$ roots in $F$.