Question

Question: Let R be an integral domain. Assume that the Division Algorithm always holds in $\mathit{R}\left[x\right]$. Prove that R is a field.

Short answer

It is proved thatR is a field.

Step-by-step solution

Integral domain

It is given that R is an integral domain, where an integral domain has an identity 1 in particular, where R is contained in $R\left[x\right]$.

Then, there exist polynomials with degree 0 in$R\left[x\right]$ for any $a\in R-\left\{0\right\}$, let $q\left(x\right)$and $p\left(x\right)$be quotient and remainder of the division of 1 by a.

Proof part

By using the definition of division algorithm, on dividing 1 by a, we get:

$1=q\left(x\right)a+r\left(x\right)$

When remainder is 0, we can say when $degr\left(x\right)>dega=0$, then,

$1=q\left(x\right)a$.

Assume that $q\left(x\right)=\sum _{i=0}^m{q}_i{x}^i$. Then, $1=q\left(x\right)a$becomes:

$1=\sum _{i=0}^{m}a{q}_i{x}^i$

By uniqueness of representation, we have $a{q}_0=1$and $a{q}_i=0$for all $i>0$.

As the denominator is not 0, that is $a\ne 0$and R is an integral domain, which implies $q_i=0$ for all $i>0$, then$q\left(x\right)=q_0$ which belongs to R, which is an inverse of a so, it is a unit.

Hence, R is a field.