Are any of these rings isomorphic?

Question

As part of my ongoing struggle to understand the complex conics, I've reached the following problem:

Let $Q_1 = x^2 + y^2$, $Q_2 = x^2 - 1$, and $Q_3 = x^2$ be polynomials in $\mathbb{C}[x,y]$. Are any of the rings $\mathbb{C}[x,y]/(Q_i)$ isomorphic to one another?

I expect the answer to be no, but I might be wrong, and in any case I don't know how to prove it. Here's what I've tried:

For $Q_3$, we get $x^2 = 0$, so $x$ is a nilpotent. If there was an isomorphism, $x$ would have to map to a nilpotent, but the other rings have no non-trivial nilpotents. So $x$ maps to $0$. I'm hoping to be able to use this to prove some sort of contradiction.

For the other two rings, I'm not sure.

Any tips? In general, how can I better understand these rings? If the $Q_i$ had been irreducible curves, I would try to find a parametrization of that curve. But that doesn't seem possible here.

Answer 1

$\mathbb{C}[x,y]/(Q_3)$ is local (and is not reduced, i.e., has non-zero nilpotents), $\mathbb{C}[x,y]/(Q_2)\simeq\mathbb C\times\mathbb C$ is not local, is reduced and has Krull dimension $0$ (it has only two maximal ideals), $\mathbb{C}[x,y]/(Q_1)\simeq\mathbb C[U,V]/(UV)$ is not local, is reduced and has Krull dimension $1$ (it has infinitely many maximal ideals).

Answer 2

You can think of $\Bbb{C}[x,y]/(Q_2)$ and $\Bbb{C}[x,y]/(Q_3)$ as ordered pairs of polynomials in $y$ $$\Bbb{C}[x,y]/(Q_1) :=\{p_1(y),p_2(y)\}_1$$ $$\Bbb{C}[x,y]/(Q_2) :=\{p_1(y),p_2(y)\}_2$$ The operation of multiplication by $x$ does very different things in these rings. In the first case it swaps these polynomials, and in the latter case it performs $p_2 \leftarrow p_1$ but annihilates $p_2$ altogether. Looking at the properties of any ring isomorphism, you can show that $x \mapsto 0$, and therefore the would-be isomorphism falls apart.

As I commented it is even easier to show that $\Bbb{C}[x,y]/(Q_1)$ with its relation on $y$ cannot be isomorphic to either of the other.