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.

I am having a hard time writing a bijective map between the two rings: $$ R = \dfrac{k[x,y,z,u,v]}{\left<(x-y)z+uv\right>} \cong \dfrac{k[x,y,z,u,v]}{\left<(x-y)z\right>} = S. $$

I know that we need to explicitly write down where the generators go from left to right and then from right to left.

Any help is greatly appreciated.

$$ $$

Note: in one direction, we could map $k[x,y,z,u,v]\rightarrow k[x,y,z,u,v]/\left< (x-y)z\right>$, with $$ x \mapsto x, y \mapsto y, z\mapsto z, $$ but where would $u$ and $v$ go? Would $u\mapsto x-y$ and $v\mapsto -z$?

As for the other direction, $k[x,y,z,u,v]\rightarrow k[x,y,z,u,v]/\left< (x-y)z+uv\right>$, do we have $$ x\mapsto x, y\mapsto y, z\mapsto z, u\mapsto \not 1, v\mapsto \not 1? $$

share|improve this question
    
If indeed your isomorphism holds, I would suggest proving it by using the first isomorphism theorem. –  user38268 Jun 25 '12 at 0:41
    
I do not think my second map is correct. Is it? That is why I "crossed out" the $1$'s because I wasn't sure. –  math-visitor Jun 25 '12 at 0:43
2  
Are you sure they're isomorphic? Polynomial rings over $k$ are unique factorization domains. The modulus defining $S$ has a nontrivial factorization, and so $S$ has zero divisors. The modulus defining $R$, however, is irreducible, and so $R$ is a domain. –  Hurkyl Jun 25 '12 at 1:19
    
Thank you Hurkyl for that information. I could be wrong. I will check again. –  math-visitor Jun 25 '12 at 1:21
2  
The isomorphism you are proposing cannot work: if you map $x$, $y$, and $z$ to themselves, then $(x-y)z$ will be in the kernel. But $(x-y)z\notin\langle (x-y)z + uv\rangle$. –  Arturo Magidin Jun 25 '12 at 3:19

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.