# Quotient ring isomorphisms

Is it true that $k[x,y]/(xy−2)≃k[2/y,y]≃k[1/y,y]≃k$? If so, why? It seems one should be able to argue with exact sequences, but I can't find any appropriate homomorphisms. So I think I may be wrong in my claim...

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Can you give a brief example in a comment about reasoning with an exact sequence of ring morphisms? I'm not sure what you had in mind... – rschwieb Nov 15 '12 at 15:23

It looks like you are talking about isomorphisms of rings, but I can't be 100% sure until you say so.

$k[2/y,y]\cong k[1/y,y]$ is certainly true, but $k[1/y,y]\not\cong k$.

$k[1/y,y]$ is basically the domain $k[y]$ localized at $\{1,y,y^2\dots\}$, and this is clearly not a field (How could $1+y$ be a unit?), so it can't be isomorphic to $k$.

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I have to admit I don't know what to say about $k[x,y]/(xy-2)$. On one hand I'm tempted to believe it's isomorphic to $k[1/y,y]$, but on the other hand $(xy-2)$ looks maximal... – rschwieb Nov 14 '12 at 14:26
Well the title was supposed to indicate that these were ring isomorphisms. Can you explain the exact sequence leading to you to conclude that $k[2/y,y]\simeq k[1/y,y]$? This should be basic... – Wouter Zeldenthuis Nov 14 '12 at 14:35
@WouterZeldenthuis Sorry, you're right about the title. That's probably where I got the idea! :) I can honestly say I've never used an exact sequence of rings before... the fact here simply is that $k[2/y,y]=k[1/y,y]$ as sets as long as 2 is a unit in $k$. A field containing $k, 1/y$ and $y$ will always contain $2/y$, and a field containing $k, 2/y$ and $y$ will always contain $(2/y)\cdot(1/2)=1/y$ – rschwieb Nov 14 '12 at 15:09

Notice that $xy-2 = y(x-1)+(y-2)$, which means that $xy-2 \in (x-1,y-2)$ and therefore $(xy-2) \subset (x-1,y-2)$. The inclusion is strict, so $(xy-2)$ is not maximal and $k[x,y]/(xy-2)$ cannot be a field.

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Ah! Now it is so obvious... thanks for the clear explanation of why $(xy-2)$ isn't maximal! – rschwieb Nov 15 '12 at 15:21