Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
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
add comment

2 Answers

up vote 1 down vote accepted

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 $k$.

share|improve this answer
    
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
1  
@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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.