# Prove that the principal ideal generated by $x$ in the polynomial ring $R[x]$ is a prime ideal iff $R$ is an integral domain.

I am going through Dummit and Foote's Abstract Algebra textbook. In the Properties of Ideals section (7.4) the complete question asks:

Let $$R$$ be a commutative ring with $$1$$. Prove that the principal ideal generated by $$x$$ in the polynomial ring $$R[x]$$ is a prime ideal iff $$R$$ is an integral domain. Prove that $$(x)$$ is a maximal ideal iff $$R$$ is a field.

I assume that P is a prime and principal ideal which is generated by $$x$$ in $$R[x]$$. By Prop 13 in D&F:

P is prime iff $$R/P$$ is an integral domain.

Then use the first isomorphism thm. Take a ring homomorphism $$\phi:R[x] \to R$$ where the kernel is an ideal of R[x]. $$ker(\phi) = P$$ and there exists an isomorphism $$R[X]/P\cong\phi(R[x])\cong R$$

But I'm struggling to see how this implies that R is also an integral domain.

In the other direction, R is an integral domain. Does this imply $$R[x]/P$$ is an integral domain which gives that P must be prime?

Thanks for the help.

Yea, if $$R$$ is an integral domain and $$\varphi : R \cong R'$$ then $$R'$$ is an integral domain.
For any two $$r,s \in R$$ such that $$r\neq 0$$ and $$s \neq 0$$ we have that $$\varphi$$ is an $$\cong$$ implies that $$\varphi(r)\neq 0$$ and $$\varphi(s) \neq 0$$; furthermore since $$R$$ is an integral domain we have that $$rs \neq 0$$ and therefore $$\varphi(rs) \neq 0$$ but then $$\varphi(r) \varphi(s) \neq 0$$ as well (since $$\varphi$$ is a morphism).
Since $$\varphi$$ is unto then this must hold in the opposite direction so that for any $$r',s' \in R'$$ such that $$r'\neq 0$$ and $$s ' \neq 0$$ there is two $$\varphi^{-1}(r'),\varphi^{-1}(s') \in R$$ such that $$\varphi^{-1}(r')\neq 0$$ and $$\varphi^{-1}(s') \neq 0$$ and therefore we have that $$r' s' \neq 0$$ as was needed.
Setting $$R' = R[x]/(x)$$ you get the last piece of your proof.