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Let $K$ be a field of characteristic $0$ and $m,n\in\mathbb Z$, $m,n\ge 1$. Prove that $$K[X,Y]/(X^2-Y^m)\simeq K[X,Y]/(X^2-Y^n)$$ if and only if $m=n$. (Related to Isomorphism between quotient rings of $\mathbb{Z}[x,y]$.)

Remarks. 1) The isomorphism can be considered identity on $K$.

2) I didn't assume that $K$ is algebraically closed (although it's more or less obvious that we can) since I expect the eventual solutions will give some explanations that one can do it. One can also assume $K=\mathbb C$ if this helps.

3) Geometric arguments are welcome, but I'm really looking for an elementary solution.

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  • Won't geometric arguments hit the following obstacle? If $n=2k$, then the zero locus of $x^2-y^{2k}$ has two components. The curve $C_1$ consisting of points with $x=y^k$, and the curve $C_2$ consisting of points with $x=-y^k$. The coordinate rings of both $C_1$ and $C_2$ are essentially $K[y]$. Irrespective of the value of $k$! I didn't see a way around this, so I didn't post in the linked thread. Nor did I see a way of lifting this to an isomorphism of the coordinate rings of the reducible algebraic sets. – Jyrki Lahtonen Jul 22 '13 at 08:33
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    I believe the standard approach -- at least in characteristic zero -- is (first to reduce to the algebraically closed case, which as you say is immediate) and then use Puiseux series about the (unique, on both curves) singular point. The singularity $\mathbb{C}[[x,y]]/(y^2-x^m)$ has two branches, each one with associated uniformizer $t^{\frac{m}{2}}$. In other words, the theory intrprets $\frac{m}{2}$ intrinsically as a valuation at the singularity, so it is an isomorphism invariant. – Pete L. Clark Jul 22 '13 at 09:00
  • Similarly with $m=2k+1$, the function field of the curve $x^2=y^m$ is just $K(u)$ with $u=x/y^k$ (so $y=u^2$ and $x=uy^k=u^m$, so all those curves are birationally equivalent to the projective line. We need a tool studying (non)-isomorphism of coordinate rings of plane curves with a singularity. – Jyrki Lahtonen Jul 22 '13 at 09:00
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    For $m,n$ odd, these two rings are integral domains, so if they are isomorphic then their integrally closure are isomorphic (to $K[t]$). This isomorphism is very restricted, so if $m\neq n$ then we can easily get a contradiction. – Yuchen Liu Jul 22 '13 at 14:16
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    I know that the question asks for something different, but there is an isomorphism between $k[[x,y]]/(x^2-y^2)$ and $k[[x,y]]/(x^2-y^4)$ if $k$ is of characteristic 2. The maps $x\rightarrow x+y$ and $y\rightarrow y+y^2$ give one morphism, the maps $x\rightarrow x+y+y^2+y^4+...$ and $y\rightarrow y+y^2+y^4+...$ gives the other one. – Marci Jul 22 '13 at 14:28

4 Answers4

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Proof for $m$ and $n$ odd: Let $A_m = k[x,y]/y^2-x^m$. Let $B_m$ be integral closure of $A_m$ in $\mathrm{Frac} A_m$. So $B_m \cong k[t]$, with $x = t^2$ and $y=t^m$. Then $\dim B_m/A_m = (m-1)/2$. (A basis for the quotient is $t$, $t^3$, ..., $t^{m-2}$.) Since the vector space $B/A$ is an intrinsic invariant of the ring $A$, this shows that, if $A_m \cong A_n$ with $m$ and $n$ odd, then $(m-1)/2 = (n-1)/2$.

Proof in the general case, with $\mathrm{Char}(K) \neq 2$ Note that $A_m$ is a domain for $m$ odd and not for $m$ even, so $A_m \cong A_n$ imples that $m \equiv n \bmod 2$. We did the case where $m$ and $n$ are odd above; we now do the case that $m$ and $n$ are even. We replace the fraction field by the total ring of fractions. When $\mathrm{Char}(K) \neq 2$, this is $k(t) \oplus k(u)$, with $x = (t,u)$ and $y= (t^{m/2}, -u^{m/2})$. This integral closure of $A_m$ in this ring $k[t] \oplus k[u]$. (Notice that $t^{m/2} = (1/2)(x^{m/2} + y)$ and $u^{m/2} = (1/2)(x^{m/2} - y)$, so $t$ and $u$ are integral over $A_m$; the details are left to the reader.)

A basis for the quotient is $(1,0)$, $(t,0)$, $(t^2,0)$, ..., $(t^{m/2-1},0)$ (using again that the characteristic is not $2$.) So $\dim B_m/A_m = m/2$ and we conclude as before.

In the comments above, marci points out that the condition $\mathrm{Char}(K) \neq 2$ is essential.

Generalities The quantity $\dim B/A$ is called the "defect" or the "$\delta$-invariant" of the singularity. See this MO question for more.

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David E Speyer
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I claim that it is enough to have a ring isomorphism to conclude that $n=m$. ($K$ is a field, or more generally a normal domain, with $2$ invertible in $K$). Let $A=K[x,y]/(y^2-x^m)$. Let us see how to recover $m$ from $A$.

If $m=2r$ is even, then $A$ has two minimal prime ideals $\mathfrak p_1, \mathfrak p_2$ (generated respectively by $y-x^r$ and $y+x^r$). We have $\mathfrak p_1+\mathfrak p_2=(x^r, y)$. Let $\mathfrak m=\sqrt{\mathfrak p_1+\mathfrak p_2}=(x,y)$. Then $r$ is the smallest positive integer such that $\mathfrak m^r\subseteq \mathfrak p_1+\mathfrak p_2$.

If $m=2r+1$ is odd, consider the integral closure $B$ of $A$ as in David's answer. Then $B=K[t]$ and $A=K+Kt^2+\cdots +Kt^{2r-2}+t^{2r}K[t]$. Now if $I=\mathrm{Ann}_A(B/A)$, we have $I=(x^{r+1}, y)$ and $r+1$ is the smallest positive integer such that $(\sqrt{I})^{r+1}\subseteq I$.

By the way, here is a geometric argument (need $K$-isomorphism). Consider $C$ the unique projective curve over $K$ obtained by adding regular points to Spec$(A)$ (one point if $m$ is odd and two points otherwise). Then the arithmetic genus of $C$ is $[(m-1)/2]$. This can be seen by considering $C$ as the limit of a continous family of hyperelliptic curves $y^2=x^m+ax+b$ ($a, b\in K$ such that $x^m+ax+b$ is a separable polynomial) and by using the fact that the arithmetic genus is constant.

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Suppose that $n$ is an odd integer, $K$ is algebraically closed. Then $$K[X,Y]/(X^2-Y^n) \simeq K[Y,Y^{n/2}] \simeq K[Z^2,Z^n].$$

Let $n<m$ be odd integers and suppose that $f: K[Z^2,Z^n] \rightarrow K[Z^2,Z^m]$ is an isomorphism of $K$-algebras. Since $(Z^2,Z^r) \subset K[Z^2,Z^r]$ is the only maximal ideal $I$ for which $K[Z^2,Z^r]_I$ is non-regular, $f$ must send the ideal $(Z^2,Z^n)$ in the ideal $(Z^2,Z^m)$ (geometrically, I mean that the origin is the only singular point of the curve $X^2=Y^r$).

Denote $P(Z):=f(Z^2)$ and $Q(Z):=f(Z^n)$. Then $$nv_Z(P) = 2v_Z(Q) \quad (*).$$ By the above $v_Z(P),v_Z(Q) ≥ 1$. Since $f$ is surjective, at least $v_Z(P)=2$ or $v_Z(Q)$, but by $(*)$ it can only be $v_Z(P)=2$. Hence $v_Z(Q)=n$. But no element in $K[Z^2,Z^m]$ has valuation $n$ (because $n<m$). This is a contradiction.

I don't know yet how deal when $n,m$ are even.

user10676
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You can easily compute the Hochschild homology of the $k$-algebra $A=k[X,Y]/(X^2-Y^m)$ and find that $HH_2(A)$ is a $k$-vector space of dimension $2m$.

This shows that $m$ is invariant under $k$-algebra isomorphisms.

Mariano Suárez-Álvarez
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