Question

Let R be an integral domain. Then the Division Algorithm holds in R [x] whenever the divisor is monic by exercise 14 in 4.1. Use fact to show that the remainder and factor theorem holds in R [x].

Short answer

The factor theorem holds in R [x]is proved.

Step-by-step solution

Remainder holds the theorem

Let’s consider that, $f\left(x\right)\in R\left[x\right]$. The Divisor algorithm holds for monic linear polynomials x - a such that, there are $q\left(x\right),r\left(x\right)\in R\left[x\right]$so that, $f\left(x\right)=\left(x-a\right)q\left(x\right)+r\left(x\right)$.

Here, r(x) is either zero or degree less than the degree x - a, which means that, $r\left(x\right)=c\in R$is constant and it is computes as, $f\left(a\right)=\left(a-a\right)q\left(x\right)+r\left(a\right)=c$.

Hence, the remainder theorem holds.

Hold the factor theorem

Now, assume that a is a root of f (x) then, by the division algorithm, $f\left(x\right)=\left(x-a\right)q\left(x\right)+c$thus, $0=f\left(a\right)=\left(a-a\right)q\left(a\right)+c=c$and hence $f\left(x\right)=\left(x-a\right)q\left(x\right)$that means $\left.\left(x-a\right)\right|f\left(x\right)$.

Again, if $\left.\left(x-a\right)\right|f\left(x\right)$then $f\left(x\right)=\left(x-a\right)q\left(x\right)$and thus, $f\left(a\right)=\left(a-a\right)q\left(a\right)=0$therefore, a is root of f (x).

And hence, the factor theorem holds.