Given R be a commutative ring with unit element.

If f(x) is a prime ideal of R[x] then we have to show that R is an integral domain.

That is we have to prove \(\displaystyle{R}=\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) is an integral domain.

Now, take non zero element in \(\displaystyle\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) say \(\displaystyle{f{{\left({x}\right)}}}+{\left({x}\right)}{\quad\text{and}\quad}{g{{\left({x}\right)}}}+{\left({x}\right)}\)

Now,

\(\displaystyle{\left({f{{\left({x}\right)}}}+{\left({x}\right)}\right)}{\left({g{{\left({x}\right)}}}+{\left({x}\right)}\right)}={\left({x}\right)}\)

\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\)

\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\in{\left({x}\right)}\)

Since, (x) is assumed to be a prime ideal

Hence, we have \(\displaystyle{f{{\left({x}\right)}}}\in{\left({x}\right)}\) or \(\displaystyle{g{{\left({x}\right)}}}\in{\left({x}\right)}\)

This implies that either \(\displaystyle{f{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\) or \(\displaystyle{g{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\)

Which roves that \(\displaystyle{R}=\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) is an integral domain.

If f(x) is a prime ideal of R[x] then we have to show that R is an integral domain.

That is we have to prove \(\displaystyle{R}=\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) is an integral domain.

Now, take non zero element in \(\displaystyle\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) say \(\displaystyle{f{{\left({x}\right)}}}+{\left({x}\right)}{\quad\text{and}\quad}{g{{\left({x}\right)}}}+{\left({x}\right)}\)

Now,

\(\displaystyle{\left({f{{\left({x}\right)}}}+{\left({x}\right)}\right)}{\left({g{{\left({x}\right)}}}+{\left({x}\right)}\right)}={\left({x}\right)}\)

\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\)

\(\displaystyle\Rightarrow{f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\in{\left({x}\right)}\)

Since, (x) is assumed to be a prime ideal

Hence, we have \(\displaystyle{f{{\left({x}\right)}}}\in{\left({x}\right)}\) or \(\displaystyle{g{{\left({x}\right)}}}\in{\left({x}\right)}\)

This implies that either \(\displaystyle{f{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\) or \(\displaystyle{g{{\left({x}\right)}}}+{\left({x}\right)}={\left({x}\right)}\)

Which roves that \(\displaystyle{R}=\frac{{{R}{\left[{x}\right]}}}{{{x}}}\) is an integral domain.