# 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.

Question
Commutative Algebra
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.

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.

### Relevant Questions

Suppose that R and S are commutative rings with unites, Let PSJphiZSK be a ring homomorphism from R onto S and let A be an ideal of S.
a. If A is prime in S, show that $$\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}={\left\lbrace{x}\in{R}{\mid}\phi{\left({x}\right)}\in{A}\right\rbrace}$$ is prime $$\displaystyle\in{R}$$.
b. If A is maximal in S, show that $$\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}$$ is maximal $$\displaystyle\in{R}$$.
Let R be a commutative ring. If I and P are idelas of R with P prime such that $$\displaystyle{I}!\subseteq{P}$$, prove that the ideal $$\displaystyle{P}:{I}={P}$$
Let R be a commutative ring with identity and let I be a proper ideal of R. proe that $$\displaystyle\frac{{R}}{{I}}$$ is a commutative ring with identity.
If R is a commutative ring with unity and A is a proper ideal of R, show that $$\displaystyle\frac{{R}}{{A}}$$ is a commutative ring with unity.
Let R be a commutative ring with unity and a in R. Then $$\displaystyle{\left\langle{a}\right\rangle}={\left\lbrace{r}{a}:{r}\in{R}\right\rbrace}={R}{a}={a}{R}$$
If A and B are ideals of a commutative ring R with unity and A+B=R show that $$\displaystyle{A}\cap{B}={A}{B}$$