Question

If R is an integral domain and $f\left(x\right)$is a nonzero polynomial of degree n in$R\left[x\right]$

, prove that$f\left(x\right)$ has at most n roots in $R$. [Hint: Exercise 20.]

Short answer

It is proved that, $f\left(x\right)$has roots in $R$.

Step-by-step solution

Determine that fx has roots

Consider that, $f\left(x\right)\in R\left[x\right]$is a non-zero polynomial of degreerole="math" localid="1648806296034" $\mathrm{n}$ .

For $n=0$, contains that $f\left(x\right)$is a non-zero constant polynomial thus, for all $r\in R$.

Given that, $f\left(r\right)=c\ne 0$, hence $f\left(x\right)$has 0 roots.

Let’s assume that, the conclusion holds for polynomial of degree $k<n$and $f\left(x\right)$has degree.

Determine that fx has at most n roots in R

Now let’s assume that, $f\left(x\right)$has a root$a\in R$. Using factor theorem,$\left.\left(x-a\right)\right|f\left(x\right)$and here, $h\left(x\right)\in R\left[x\right]$such that $f\left(x\right)=\left(x-a\right)h\left(x\right)$.

$R$is an integral domain. If $f\left(x\right)$has a root not then, as$b-a\ne 0$and $b$is a

root of $h\left(x\right)$.

Similarly, $R$is an integral domain then $\mathrm{deg}\left(f\left(x\right)\right)-1+\mathrm{deg}\left(h\left(x\right)\right)$, the$\mathrm{deg}\left(h\left(x\right)\right)<\mathrm{deg}\left(f\left(x\right)\right)-n$.

Thus, by inductive hypothesis $h\left(x\right)$has role="math" localid="1648807202772" $n-1$roots. As root of $f\left(x\right)$are either or root of $h\left(x\right)$.

Therefore, $f\left(x\right)$has$n$roots in $R$.