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Consider the following rings:

$A=\mathbb{C}\lbrace x,y,u \rbrace /(xy+x^3,y^2,xy^2+x^5) \ $ and

$B=\mathbb{C}\lbrace x,y,u \rbrace /(xy+x^3,y^2+ux^4,xy^2+x^5)$

There is an isomorphism of $\mathbb{C}$-algebras between $A$ and $B$?

Here $\mathbb{C}\lbrace x,y,u \rbrace$ denotes the formal series ring. I'm trying to prove that there is an isomorphism between $A$ and $B$, but I suspect that this isomorphism can not exist. Furthermore, I would like to see if there is at least one morphism between $A$ and $B$ or maybe between $B$ and $A$?

I would be happy with any suggestions.

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  • $\begingroup$ In $A$ $y$ is a zero-divisor but not in $B$? $\endgroup$ – Daniel Larsson Jun 8 '16 at 7:15
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    $\begingroup$ @Daniel Larsson -- it's not clear that the isomorphism is supposed to send $y$ to $y$. Otoniel Silva -- there are definitely morphisms in each direction because you can just send all the variables to zero. $\endgroup$ – znt Jun 8 '16 at 9:23
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    $\begingroup$ To try and prove they're not isomorphic I might grit my teeth (or pull out a computer algebra package) and compute the dimension of $A/m^n$ and $B/m^n$ for the first few values of $n$, where here $m$ is the maximal ideal $(x,y,u)$. $\endgroup$ – znt Jun 8 '16 at 9:25
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    $\begingroup$ @znt If I did this correctly (using Maple) then these dimensions are the same for $n\leq 12$. More precisely, if we use a pure lexicographic term order with $y<x<u$ then the Grobner bases for $A/\mathfrak{m}^n$ and $B/\mathfrak{m}^n$ have the same leading terms. $\endgroup$ – Neil Strickland Jun 8 '16 at 9:39
  • $\begingroup$ Does $B$ contain any non-zero nilpotent element? $\endgroup$ – Ilya Bogdanov Jun 8 '16 at 9:53
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Set $z=y+x^2$. Then $$ (xy+x^3,y^2,xy^2+x^5)=(xz,(z-x^2)^2,xz(y-x^2)+2x^5) =(xz,x^5,z^2+x^4) $$ and similarly $(xy+x^3,y^2+ux^4,xy^2+x^5)=(xz,x^5,z^2+(1+u)x^4)$. So, after setting $x'=x\root4\of{1+u}$ we get the required isomorphism.

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    $\begingroup$ You mean $z=y+x^2$? $\endgroup$ – Fedor Petrov Jun 8 '16 at 10:32
  • $\begingroup$ Does one just come up with this from playing around or is there some sort of Groebner Basis strategy for proving this? Well done, by the way. $\endgroup$ – znt Jun 8 '16 at 10:34
  • $\begingroup$ Fedor: Thanks, surely. znt: Something between; I did not try to find a Groebner basis, but followed some similar way, trying just to simplify the relations. $\endgroup$ – Ilya Bogdanov Jun 8 '16 at 10:38
  • $\begingroup$ Thank you for your answer, could you please write the isomorphism explicitly? The definition of $x^{'}$ result that $z$ also depend on $x^{'}$ , this may not affect the construction of the isomorphism? $\endgroup$ – Otoniel Silva Jun 9 '16 at 2:05
  • $\begingroup$ Well, it affects a bit: I've provided an automorphism of $\mathbb C[[x,z,u]]$ mapping the first ideal to the second. To get the isomorphism in the initial variables, you need to conjugate this automorphism by the change of variables. The result seems to be $x\mapsto x\root4\of{1+u}$, $u\mapsto u$, $y\mapsto y+x^2-x^2\sqrt{1+u}$. $\endgroup$ – Ilya Bogdanov Jun 9 '16 at 6:38
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This is getting too long for a comment, so I will make it an answer.

First note that in the third relation for $A$, the term $xy^2$ is zero by the second relation. So we have $$ A = \mathbb{C}[[x,y,u]]/(xy+x^3,y^2,x^5). $$ Next note that in $B$ we have $x^5=-xy^2=ux^5$ so $(1-u)x^5=0$ but $1-u$ is invertible so $x^5=0$. Using this we get $$ B = \mathbb{C}[[x,y,u]]/(xy+x^3,y^2+ux^4,x^5). $$ It follows that both $A$ and $B$ have plenty of nilpotents, and $A/\sqrt{0}=B/\sqrt{0}=\mathbb{C}[[u]]$.

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