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I need to find the condition on $m,n\in\mathbb{Z}^+$ under which the following ring isomorphism holds: $$ \mathbb{Z}[x,y]/(x^2-y^n)\cong\mathbb{Z}[x,y]/(x^2-y^m). $$

My strategy is to first find a homomorphism $$ h:\mathbb{Z}[x,y]\rightarrow\mathbb{Z}[x,y]/(x^2-y^m) $$ and then calculate the kernel of $h$.

To achieve this, I furthermore try to identify the isomorphism between $\mathbb{Z}[x,y]$ and itself, which I guess is $$ f:p(x,y)\mapsto p(ax+by,cx+dy) $$

where $ad-bc=\pm 1,a,b,c,d\in\mathbb{Z}$.

Then $f$ induces a homomorphism $h$. But from here I failed to move on.

I believe there is some better idea, can anyone help?

Updated:

It should be isomorphism between quotient rings, not groups. Very sorry for such mistake.

hxhxhx88
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  • I don't think such an isomorphism holds whenever $n\neq m$. – Potato Jun 24 '13 at 07:50
  • @Potato, it is a problem from a formal examination, I don't think it could be wrong, at least I want to try.. – hxhxhx88 Jun 24 '13 at 08:00
  • Yes, and I think the point is to show that no isomorphisms can exist for distinct $n$ and $m$. – Potato Jun 24 '13 at 08:01
  • Reading the OP's last comment I think oxeymon is right: if they means as $,groups then any pair of natural numbers will do. – DonAntonio Jun 24 '13 at 08:56
  • @YACP, yes, it should be ring isomorphism, my fault. – hxhxhx88 Jun 24 '13 at 14:08
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    When $n$ is odd, and $m$ is even(or vice-versa), then clearly the two rings are not isomorphic, because then the ring on the left will be a domain and the ring on the right will not be a domain(or vice-versa). – messi Jun 26 '13 at 14:35

2 Answers2

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Well, the additive group of $\mathbb{Z}[x,y]/(x^2 - y^n)$ is just a free abelian group on the generators $1,x, y, xy, xy^2, xy^3,\ldots$ - ie its the free abelian group on countably many generators, which is independent of $n$.

Hence, $m,n$ can be anything.

Basically, the $\mathbb{Z}$-module $\mathbb{Z}[x,y]$ is generated by all monomials with coefficient 1, ie, $1, x, y, x^2, y^2, xy, x^3, y^3, x^2y, xy^2,\ldots$. The relation $x^2 - y^n$ just allows you to replace any $x^k$ you see with $x^{k-2}y^n$ (for $k\ge 2$). However, in both cases you still end up having countably infinitely many generators (and hence countably infinite rank), and since a free abelian group is determined up to isomorphism by its rank, they're isomorphic.

If you're talking about ring isomorphisms, then Potato is right - if $n\ne m$, then the two rings are not isomorphic.

Let $R_n = \mathbb{Z}[x,y]/(x^2 - y^n)$.

As to why they're not isomorphic as rings, this seems to me to be a rather deep question, and I feel like the best explanation is through algebraic geometry. Essentially, the polynomial $x^2 - y^n$ defines a curve $C_n$ in the plane (namely the set of points (a,b) where $a^2 - b^n = 0$). These curves $C_n$ are birationally defined by their function fields, which in this case is just the quotient field of your ring $R_n$. If the rings $R_n,R_m$ are isomorphic, then their quotient fields must be isomorphic as well, and so the curves $C_n,C_m$ they define must be birationally equivalent. However, it can be computed via the Riemann-Hurwitz formula on the coordinate function $y$, viewed as a function from your curve to $\mathbb{P}^1$ that the curve associated to $R_n$ has geometric genus $(n-1)(n-2)/2$ (as long as $n\ge 1$, see exercise 2.7 in Silverman's book "The Arithmetic of Elliptic Curves"), which being a birational invariant, tells you that the function fields for your curves $C_n,C_m$ are not isomorphic for $n\ne m$, and hence $R_n, R_m$ could not be isomorphic either.

Finally it's easy to see that $R_0$ is not isomorphic to $R_n$ for any $n\ge 1$ since $R_0$ has nilpotent elements, and $R_n$ for $n\ge 1$ does not.

I can think of some other proof ideas, but essentially they all rely on some form of algebraic geometry. Many of these ideas I could phrase purely ring-theoretically, but it would seem complicated and completely unmotivated without explaining the connection to geometry.

oxeimon
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  • I think this is nice, yet I don't think the free abelian group's generators we get is independient of $,n,$: for example, $,y^n=x^2,$, which we need to generate all the elements in the group with $,x-$degree greater than one... – DonAntonio Jun 24 '13 at 08:12
  • No the particular generators aren't the same, per se, but both cases are free abelian groups of countably infintie rank, and since a free abelian group is determined up to isomorphism by the cardinality of a basis (in this case both are countably infinite), the two groups are isomorphic. – oxeimon Jun 24 '13 at 08:50
  • Oh, I see...very nice indeed. +1 Now we only have to make sure whether the OP meant group or ring isormophism – DonAntonio Jun 24 '13 at 08:52
  • I think he meant group isomorphism, since the title of the post is "isomorphisms between quotient group of Z[x,y]" – oxeimon Jun 24 '13 at 08:54
  • @DonAntonio, It should be ring isomorphism, so sorry – hxhxhx88 Jun 24 '13 at 14:06
  • @oxeimon, it should be ring isomorphism, sorry. – hxhxhx88 Jun 24 '13 at 14:07
  • Well @hxhxhx88, then perhaps my answer fits better, though for the group thing Oxeymon's is simple and nice. – DonAntonio Jun 24 '13 at 14:23
  • Your proof on group is beautiful, thanks! – hxhxhx88 Jun 24 '13 at 15:22
  • @oxeimon Could you add a proof that they aren't isomorphic as rings? – Potato Jun 24 '13 at 17:54
  • I added a proof that they're not isomorphic as rings. – oxeimon Jun 26 '13 at 06:40
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    Your proof incorrectly assumes that the corresponding curve is nonsingular, which is never the case when $n > 1$ (it will always have a singularity at $[1:0:0]$). For example, $x^2 - y^3$ defines a curve of genus $0$, not $1$. – bzc Jul 10 '13 at 16:37
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    Correction: The singularity will be at $[0:0:1]$ for $n > 1$ (although there will also be one at $[1:0:0]$ for $n > 2$). – bzc Jul 10 '13 at 16:51
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    I agree with Brandon's criticism. Also, if $n$ or $m$ is even these rings are not integral domains, so talking about their fields of fractions seems a bit strange :-). I did upvote this for the group part. – Jyrki Lahtonen Jul 14 '13 at 06:23
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    I have retracted my upvote, as the OP did not react to the criticism. When $n$ is even, the algebraic set defined by $x^2-y^n=0$ is reducible (two component curves with $x=\pm y^{n/2}$. When $n$ is odd, this curve is birationally equivalent to a curve of genus zero irrespective of the value of $n$. – Jyrki Lahtonen Jul 22 '13 at 09:22
  • Ah damn, I just noticed these comments, and that's a good point, though I'm not sure how to fix it... – oxeimon Sep 14 '13 at 22:25
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We can write $\;x^2=y^k\;$ in the ring $\,R_k:=\Bbb Z[x,y]/(x^2-y^k)\;$ , so that any polynomial in $\;\Bbb Z[x,y]\;$ is mapped to a polynomial with $\,y-$ degree at most $\,k\,$ , for example in $\,R_3\,$ :

$$3xy^2 - xy^4-y^3+2x^3y\mapsto3xy^2-x(yx^2)-x^2+2x^3y=2x^3y-x^2+3xy^2$$

Now, if $\,n\neq m\,$ , say WLOG $\,n<m\,$ , we have

$$x^2-y^m\neq0\;\;\text{in}\;\;R_n$$

since otherwise we'd have a polynomial in $\;R_n\;$ with $\,y-$degree higher than $\,n\,$ ...

DonAntonio
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    Seeing oxeimon's answer, please do note that my answer refers to the rings $,\Bbb Z[x,y];,;;\Bbb Z[x,y]/(x^2-y^n);$ , etc. – DonAntonio Jun 24 '13 at 08:13
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    I don't understand the last half of this. Also, I don't see how you rule out homomorphism that do things like $x\rightarrow x^2$. – Potato Jun 24 '13 at 08:16
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    Clearly the map induced by $x\rightarrow x$ and $y\rightarrow y$ won't work. It's the others we need to be concerned about. – Potato Jun 24 '13 at 08:17
  • Well, I don't understand one quarter of what you're asking: I pointed out a characteristic of $,x^2-y^n,$ in $,R_n,$ (namely, being zero) that isn't fulfilled in $,R_m,$ . What "ruling out homomorphism $,x\mapsto x^2,$ " are you talking about? Please do note that I am not saying my answer can't be improved or even that it is not completely wrong, but rather that I don't understand your remarks. – DonAntonio Jun 24 '13 at 08:24
  • The way I'm interpreting this (and it's late, so it might just be me) is that you prove the canonical map $\mathbb Z [x,y] \rightarrow \mathbb Z[x,y]/(x^2-y^n)$ doesn't descend to a map $\mathbb Z [x,y]/(x^2-y^m) \rightarrow \mathbb Z[x,y]/(x^2-y^n)$. If this isn't the case, could you perhaps explain again what you are doing? The problem is certainly me, but please humor me. – Potato Jun 24 '13 at 08:46
  • Well, I did not mean that but you could as well see it that way: for a homomorphism to factor through something we need that something to fulfill the relations the map's kernel fulfills. I think this is, bottom line, what I showed. – DonAntonio Jun 24 '13 at 08:53
  • Yeah, I originally thought the same way, so this problem is kind of wield. Anyway, thank you very much. BTW, due to my mistake, I think I should give credit to oxeimon... – hxhxhx88 Jun 24 '13 at 15:21
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    @DonAntonio Yes, so, if I am understanding right, you don't rule out that that any other map descends. So, for example, consider the map $\mathbb Z[x,y] \rightarrow \mathbb Z[x,y]/(x^2-y^n)$ that sends $x\rightarrow x$ and $y\rightarrow x+y$. Of course, this doesn't work, but I don't see where in your proof you account for this. – Potato Jun 24 '13 at 16:49