 Let R be a commutative ring with unit element .if f(x) is a prime ideal of R[x] then show that R is an integral domain. Amari Flowers 2021-03-07 Answered

Let R be a commutative ring with unit element .if $$f(x)$$ is a prime ideal of $$R[x]$$ then show that R is an integral domain.

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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.