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Suppose $R$ is an integral domain, $R[x,y,z]$ is polynomial ring over $R$, and $$Q=R[x,y,z]/(xyz-1)$$ is a quotient ring. How to prove, that $Q$ is not principal ideal ring? I was trying to compose an ideal that cannot be generated by one element, but fruitless yet.

Edit: I guess this fact could be proven without constructing non-principal ideal, but I'm mostly interested in a proof in which such construction is made.

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    $\begingroup$ I suggest to try first $R$ a field and $I=(x-1,y-1)$. $\endgroup$ – user26857 Jun 1 '16 at 19:54
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Hint: $Q \simeq R[x,y]_{xy}$. Now try finding a non-principal ideal in $Q$, while recalling how ideals arise in localisations.

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  • $\begingroup$ Thank you for your answer! What does the lower index $xy$ mean in $R[x,y]_{xy}$? $\endgroup$ – Glinka Jun 1 '16 at 19:23
  • $\begingroup$ Um, that means the localisation of $R[x,y]$ at $xy$. I take it that you've not seen localisations yet ? $\endgroup$ – Hmm. Jun 1 '16 at 19:25
  • $\begingroup$ I didn't see this notatioon before, it was like $S^{-1}R$ $\endgroup$ – Glinka Jun 1 '16 at 19:28
  • $\begingroup$ Oh, in this case $S=(1,xy,x^2y^2...)$.In any case, think about an ideal in $R[x,y,z]$, a simple looking one, such that any element belonging to that ideal vanishes at (1,1,1). Can you find one ? $\endgroup$ – Hmm. Jun 1 '16 at 19:28
  • $\begingroup$ A simple looking ideal in $R[x,y,z]$, you mean $(xyz-1)$? $\endgroup$ – Glinka Jun 1 '16 at 19:36

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