Let R be commutative ring with unity such that R[x] is a PID then R is a field.
This problem is already there in stackexchange. But i am not able to get notation. I am a beginner. My book notation is different. Pls clarify me the following
My attempt:-
We have $\frac{R[x]}{\langle x\rangle} \simeq R$
My idea is to if $\langle x\rangle$ is maximal we are done. (We know that if M is maximal ideal then $R/M$ is field)
Let $I=\langle f(x)\rangle $
$x \in I \implies x=f(x)g(x) $ then
case 1:- $f(x)=1, g(x)=x \implies I=R$
case 2:- $f(x) = \alpha x, g(x)=\alpha^{-1} \implies I=\langle x\rangle$
This case exists when $\alpha^{-1}$ exists
case 3:- $f(x)=\alpha, g(x) = \alpha^{-1}x \implies I=\langle\alpha\rangle$
**Here $I=\langle\alpha\rangle$ becomes ideal between $\langle x\rangle$ and R if $\alpha^{-1}$ does not exist. **
In all other case1, 2 $\langle x\rangle$ becomes maximal ideal. So R becomes field. But how to do case 3?
Best Answer
Here's a simpler approach:
If $R[x]$ is a PID then $R$ is an integral domain because it is a subring of $R[x]$. Now since $R\cong R[x]/(x)$, this implies $(x)$ is a prime ideal. Can you finish the proof from here?