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.


1 Answer 1


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.


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