# Prime Ideal Integral domain

• May 18th 2009, 08:37 PM
curiousmuch
Prime Ideal Integral domain
Let R be a commutative ring with unity. Show that the ideal generated by <x>={xf(x)|f(x) is a polynomial in R[x]} is a prime ideal in R[x] if and only if R is an integral domain.

For the forward direction I tried to use fact that an ideal N of R is prime if and only if R/N is an integral domain. But got stuck after that.

And I am completely clueless about the backwards direction. Sorry. Any help would be appreciated.
• May 18th 2009, 08:43 PM
NonCommAlg
Quote:

Originally Posted by curiousmuch
Let R be a commutative ring with unity. Show that the ideal generated by <x>={xf(x)|f(x) is a polynomial in R[x]} is a prime ideal in R[x] if and only if R is an integral domain.

For the forward direction I tried to use fact that an ideal N of R is prime if and only if R/N is an integral domain. But got stuck after that.

And I am completely clueless about the backwards direction. Sorry. Any help would be appreciated.

define the map $\varphi: R[x] \longrightarrow R$ by $\varphi(f(x))=f(0).$ it's easy to see that $\varphi$ is a surjective ring homomorphism and $\ker \varphi = .$ thus $\frac{R[x]}{} \cong R$ and the result follows.
• May 18th 2009, 08:56 PM
curiousmuch
Oh wow this way both directions are taken care of at once.