Laurent polynomial ring which is a PID

Let $$R$$ be a commutative ring with identity. We know that a polynomial ring $$R[x]$$ is a PID if and only if $$R$$ is a field. (You may check the proof here.) The ring of formal series $$R[[x]]$$ satisfies a similar property with a similar proof. Meanwhile, when it comes to the Laurent polynomial ring, it is true that if $$R$$ is a field, then $$R[x,x^{-1}]$$ is a PID. (The proof can be found here.)

Is it true that $$R[x,x^{-1}]$$ is a PID if and only if $$R$$ is a field? (In other words, does the converse hold?)

Yes, unfortunately we can't use the factor ring $$R[x,x^{-1}]/\langle x\rangle$$ since $$x$$ is a unit in the Laurent polynomial ring. Does the converse still hold in this case? Thank you very much.

• What are you assuming on the ring $R$? Commented Mar 24, 2021 at 9:55
• @DietrichBurde the original assumption I used is $R$ as a commutative ring with identity. (But I think it might be specified into integral domains.) Commented Mar 24, 2021 at 9:57

Assuming that $$R$$ is Noetherian, the Krull dimension is given by $$\dim (R[x,x^{-1}])=1+\dim (R).$$ Assume that $$R[x,x^{-1}]$$ is a PID. Then its Krull dimension is $$1$$, so that it follows $$\dim(R)=0$$. If $$R$$ is a commutative ring with $$1$$, then $$R$$ is a field.

Reference:

Dimension of a quotient ring

Edit: If $$R$$ is not Noetherian, then the Krull dimension can be bigger and we have (also for $$R[x,x^{-1}]$$ instead of $$R[x]$$) $$\dim R+1\le \dim R[x] \le 2\dim R+1.$$ See the papers by Seidenberg. If $$1+\dim (R)\le \dim (R[x,x^{-1}])=1$$, then again $$\dim (R)=0$$.

• Regarding to the papers by Seidenberg, I found some papers related to dimension theory: On the dimension theory of rings II ( msp.org/pjm/1954/4-4/pjm-v4-n4-p09-p.pdf ), and A note on the dimension theory of rings ( projecteuclid.org/journals/pacific-journal-of-mathematics/… ) Are those the paper you recommended to look for examples? Thank you in advance. Commented Mar 24, 2021 at 10:11
• Yes, we have the estimate $\dim(R)+1\le \dim (R[x])\le 2\dim (R)+1$ for $R$ a commutative ring, and all dimensions are possible in this interval, as Seidenberg shows. So we could do the same with $R[x,x^{-1}]$. Commented Mar 24, 2021 at 10:38
• Do you use that $R$ is an integral domain? Otherwise we can only conclude that $R$ is Artinian. Or is that implied by $R[x, x^{-1}]$ being a PID? Commented Mar 25, 2021 at 13:32
• @DietrichBurde It is very easy to show that $\dim R+1\le\dim R[X,X^{-1}]$ for every commutative ring $R$. If $p_0\subset p_1\subset\cdots\subset p_n$ is a chain of prime ideals in $R$, then this extends to a chain of prime ideals in $R[X,X^{-1}]$ (we first extend every prime $p_i$ to $R[X]$ and notice that $X\notin p_i[X]$), and $p_nR[X,X^{-1}]$ is not maximal since $R[X,X^{-1}]/pR[X,X^{-1}]\simeq (R/p)[X,X^{-1}]$ which is not a field. Commented Mar 25, 2021 at 18:36
• @user26857 Thank you, this is very useful here. Commented Mar 26, 2021 at 13:17