Let R be commutative ring with unity such that R[x] is a PID then R is a field.

Question

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?

Answer 1

If I imitate your proof, then I would assume that $I=\big\langle f(x)\big\rangle$ is a maximal ideal containing the proper ideal $\langle x\rangle$ of $R$ ($I$ exists by Zorn's Lemma and due to the assumption that $R$ is unital, and $I$ is principal due to the hypothesis that $R[x]$ is a principal ideal domain). Thus, $x=g(x)\,f(x)$ for some $g(x)\in R[x]$. So, is this what you are trying to do? I don't understand your work well, but here is my guess (with supplementary arguments).

If $f$ and $g$ are both nonconstant, then the degree of $g(x)\,f(x)$ must be at least $2$, since $R$ is an integral domain, and we end up with a contradiction. Thus, $f(x)$ or $g(x)$ is constant. Ergo, one of $f(x)$ and $g(x)$ must be of the form $\alpha x+\beta$ and the other is a constant $\gamma$. Clearly, $\alpha$ and $\gamma$ are nonzero, and $\alpha\gamma=1$. Also, $\beta=0$. Thus, either $f(x)=\alpha x$ or $f(x)=\gamma$.

Now, $f(x)\neq \gamma$ because $\alpha$ is the inverse of $\gamma$, so $I=\langle \gamma \rangle = \langle 1\rangle =R$, contradicting the assumption that $I$ is a maximal (whence proper) ideal. Thus, $f(x)=\alpha x$, but as $\alpha\gamma=1$, we get $I=\langle \alpha x\rangle=\langle x\rangle$, which means $\langle x\rangle$ is a maximal ideal (as we have assumed that $I$ is a maximal ideal). However, Alex Mathers's suggestion will make things a lot simpler.

Answer 2

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?

Answer 3

Here's a hint for another approach: note that $R$ is a domain; take any proper ideal $I\unlhd R$ and consider the generator $\pi$ of the ideal $(I,x)\unlhd R[x]$. Since $R$ is a domain, the highest order coefficient of $\pi$ is not a zero divisor. Conclude that $I=0$.