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I don't think so but could not prove or disprove.

Since I dont believe it.


I tried to use

  1. We can understand that $k(x)\not\cong k[x,x^{-1}]$ one of them is field but other is not.

However $S[x]\cong T[x] \not\Rightarrow S\cong T$ since, in general see: Does $R[x] \cong S[x]$ imply $R \cong S$?


  1. Like $k[x][y]=k[y][x]$ I try to change $k(x)[y]=(k[y])(x)$ but again one of them is field but other is not.

  1. I try to think about $R[x]$ and formal polynomial series $R[[x]]$ and thought that if a polynomail in $R[[x]]$ is not invertible then it cannot be invertible in $R[x]$ but every invertible element in formal series are with nonzero constant if we have purely x polynomial it can be inverted in in subring $R[x]$ considering canonical injection into $R[[x]]$

  1. Try to find some examples as: $x+1$ is invertible in $k(x)[y]$ since $1/(x+1)$ in here but $x+1$ is not invertible in $k[x,x^{-1},y]$ since if it was invertible then $$1=(x+1)\left(\sum_{n\in \mathbb Z}a_i(y)x^n\right)$$ where $a_i(y)\in k[y]$ and finitely many $a_i(y)$ are nonzero. We can see that $x+1$ is non invertible in $k[x,x^{-1},y]$ but I cannot be sure about $x+1$ should be mapped to $x+1$.

I thought that if there was a isomorphism $\phi :k(x)[y]\to k[x,x^{-1},y]$ then $\phi(x+1)$ should be invertible in $k[x,x^{-1},y]$ but we dont know where this $\phi$ map our $x+1$ to?

Edit:

Alex Wertheim's comment is vanished with some deleted answer so I am adding his good coment here (If he likes, please edit and delete your comment here)

Alex Wertheim : "It would probably be easier to note that $R/yR≅k[x,x^{-1}]$, and so $R$ contains a nonzero prime ideal which is not maximal, and hence cannot be a PID."

Jale'de jaled
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2 Answers2

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As @Douglas Molin suggested in the comments, because $k(x)$ is a field (even if $k$ is not algebraically closed), there is a well-known theorem that says $R[x]$ is a PID iff $R$ is a field, which means $k(x)[y]$ is a PID.

But $k[x,x^{-1},y]$ is not a PID since $k[x,x^{-1}]$ is not a field (look at $x+1$).

Or Shahar
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    Your justification for why $R := k[x, x^{-1}, y]$ is not a PID is not correct. The ideal generated by $x, y$ is all of $R$, since $x$ is a unit. It would probably be easier to note that $R/yR \cong k[x, x^{-1}]$, and so $R$ contains a nonzero prime ideal which is not maximal, and hence cannot be a PID. – Alex Wertheim Nov 29 '21 at 20:15
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Hint how you can complete your argument: $\phi(x)$ has to be an element which has no square root. In $k[x,x^{-1},y]$, if an element $\alpha$ has no square root, then it is not in $k$, so either $\alpha$ or $\alpha+1$ is not invertible (to see this, prove that $\alpha$ is invertible iff $x^k\cdot \alpha$ is invertible in $k[x,y]$ for some $k\in \mathbf Z$).

(Note that the other argument gives you a stronger theorem, as you don't need $k$ to be algebraically closed.)

tomasz
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