Question

Let R be an integral domain. Using sequence notation, prove that the polynomial ring R[x] is also an integral domain.

Short answer

It is proved that$R\left[x\right]$ is an integral domain.

Step-by-step solution

Definition of integral domain

The commutative ring with identity with no zero divisors is known as integral domain.

Prove that the polynomial ring R[x] is also an integral domain 

Consider the polynomial $\mathrm{f}\left[\mathrm{x}\right]\ne 0$as$\mathrm{f}\left[\mathrm{x}\right]={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+...+{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$ , where${\mathrm{a}}^{\mathrm{n}}\ne 0$ in$\mathrm{R}\left[\mathrm{x}\right]$ .

The degree of$\mathrm{f}\left[\mathrm{x}\right]$ is equal to the index of highest non-zero coefficient of $\mathrm{f}\left[\mathrm{x}\right]$.

If $R$is an integral domain then,

$\mathrm{deg}(\mathrm{f}\left(\mathrm{x}\right)\cdot \mathrm{g}\left(\mathrm{x}\right))=\mathrm{degf}\left(\mathrm{x}\right)+\mathrm{degg}\left(\mathrm{x}\right)$

But$\mathrm{f}\left[\mathrm{x}\right]\ne 0$ and $\mathrm{g}\left[\mathrm{x}\right]\ne 0$, then it is impossible to get $f\left[x\right]=0=g\left[x\right]$.

Therefore, it is proved that $\mathrm{R}\left[\mathrm{x}\right]$is an integral domain.