Isomorphism between quotient rings of $\mathbb{Z}$

I need to find the condition on $m,n\in\mathbb{Z}^+$ under which the following ring isomorphism holds:

My strategy is to first find a homomorphism
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?


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

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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.

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


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