Question

Let R be an integral domain. Assume that the Division Algorithm always holds in R[x]. 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[x].

Then, there exist polynomials with degree 0 in R[x] for any $a\in R-\left\{0\right\}$, let q(x) and r(x) 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(x)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$a{q}_1=0$ 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.