Showing two rings are not isomorphic

Question

Consider $\mathbb{C}[x,y]/(y^2-x^3)$ and $\mathbb{C}[x,y]/(y^2-x^3-x^2)$. Someone told me that these two rings are not isomorphic, but I don't know how to prove. Please help me.

Answer 1

As mentioned by Stahl in the comments, if these rings were isomorphic, then $\mathbb{C}[[x, y]] / (y^2 - x^3)$ and $\mathbb{C}[[x, y]]/(y^2 - x^3 - x^2)$ would be isomorphic.

But the former ring is an integral domain, and the latter is not.

Proof that $\mathbb{C}[[x, y]] / (y^2 - x^3)$ is an integral domain: map $\mathbb{C}[[x, y]] \to \mathbb{C}[[t]]$ by $x \mapsto t^2$ and $y \mapsto t^3$. Let's check that the kernel is exactly $(y^2 - x^3)$. Any power series $$ f(x, y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^2 + a_{11}xy + a_{02}y^2 + a_{30} x^3 + \ldots $$ in the kernel of this map satisfies $$ a_{00} + a_{10}t^2 + a_{01}t^3 + a_{20}t^4 + a_{11}t^5 + (a_{02} + a_{30})t^6 + \ldots = 0, $$ so in particular all the coefficients below $a_{02}$ are zero, and for each $n$, $$ \sum_{2i + 3j = n} a_{ij} = 0. $$ We are reduced to showing that any polynomial of the form $$ \sum_{2i + 3j = n} a_{ij}x^iy^j $$ is divisible by $y^2 - x^3$ when the sum of its coefficients is zero. We can replace every copy of $y^2$ with $x^3$ in the polynomial to check this, so (depending on the parity of $n$) our expression is either $y \sum a_{ij} x^{(n-1)/2}$ or $\sum a_{ij} x^{n/2}$, which is obviously zero either way.

Proof that $\mathbb{C}[[x, y]]/(y^2 - x^3 - x^2)$ is not an integral domain: we have $x^3 + x^2 = x^2(1+x)$. Now $(1+x)$ is a square in $\mathbb{C}[[x, y]]$, using the Taylor expansion of the square root function, that is, there exists $z$ with $x^3 + x^2 = z^2x^2$. Then $(y+zx)(y-zx) = y^2 - x^3 - x^2 = 0$ in the quotient, but neither $y + zx$ nor $y - zx$ are (they are degree $1$ in $y$ and $zx$ doesn't have any $y$-terms).

(Technically this only proves there is no $\mathbb{C}$-algebra isomorphism; I am not sure if there is an exotic ring isomorphism that is non-linear on the $\mathbb{C}$ part, since you have a bunch of transcendental garbage in $\mathbb{C}$ to work with in that case.)

(EDIT: oh, it's fine. The first ring has the property that for any maximal ideal, the completion is a domain. The latter doesn't have this property. So they can't be isomorphic.)