Question

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.

Commutative Algebra
ANSWERED
asked 2021-03-07

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.

Answers (1)

2021-03-08

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.

0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours
...