Isomorphism between quotient rings of $K[X,Y]$

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.