Question

Let R be a field. Using sequence notation, prove that the polynomial ring R[x] is not a field.

Short answer

It is proved that $\mathrm{R}\left(\mathrm{x}\right)$is not a field.

Step-by-step solution

Definition of field

It is defined as any sets of elements such that they satisfies both addition and multiplication and is commutative.

Prove that polynomial ring R[x] is not a field

Consider $\mathrm{R}$be an integral domain and let $f\left(x\right)$and $g\left(x\right)$is any non- zero polynomials belongs to $R\left(x\right)$.

To show that $R\left(x\right)$is not a field, it is required to prove that not every polynomial in $R\left(x\right)$has a multiplicative inverse.

Consider a polynomial$f\left(x\right)=x$ and multiplicative inverse$g\left(x\right)$ then,

$xg\left(x\right)=I\left(x\right)$

It is given that$R$ is an Integral domain therefore,

$\begin{array}{c}\mathrm{deg}\left(xg\left(x\right)\right)=\mathrm{deg}\left(x\right)+\mathrm{deg}\left(g\left(x\right)\right)\\ =1+\mathrm{deg}\left(g\left(x\right)\right)\end{array}$

But,

$\mathrm{deg}\left(xg\left(x\right)\right)=\mathrm{deg}\left(I\left(x\right)\right)=0$

This is a contradiction this implies that for a polynomial$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}$ there may not exist a multiplicative inverse therefore$R\left(x\right)$ is not a field since not every polynomial in$R\left(x\right)$ has a multiplicative inverse.

Therefore, it is proved that$R\left(x\right)$ is not a field.