How can we compute the ideal $I \subseteq K[x, y]$ such that $K[x, y]/I \cong K[z^2, z^3 - z]$?

Question

Let $K$ be an algebraically closed field with $\operatorname{char}(K) \neq 2$.

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Algebraically, my question goes as follows:

Consider the map $$ \alpha: K[x, y] \to K[z], \\ x \mapsto z^2, \\ y \mapsto z^3 - z \\ $$ Equivalently, this is the map $$ \alpha: K[x, y] \to K[z], \\ p(x, y) \mapsto p(z^2, z^3 - z) $$ Define $$ I = \operatorname{ker}(\alpha) = \{p(x, y) \in K[x, y] \mid p(z^2, z^3 - z) = 0\} $$ Now $\alpha$ descends to an isomorphism between the "coimage" and "image" of $\alpha$, i.e. the middle map is an isomorphism: $$ \newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} K[x, y] \twoheadrightarrow \quotient{K[x, \ y]}{I} \overset{\sim}{\to} K[z^2, z^3 - z] \hookrightarrow K[z] $$ We have: $$ \begin{array}{rl} & \left[ (z^2)^3 - 2 (z^2)^2 + z^2 \right] - \left[ (z^3 - z)^2 \right] \\ = & \left[ z^6 - 2 z^4 + z^2 \right] - \left[ z^6 - 2 z^4 + z^2 \right] \\ = & 0 \\ \end{array} $$ This shows that $$ q(x, y) = x^3 - 2 x^2 + x - y^2 \in I $$ and hence $\langle q \rangle \subseteq I$. I believe that $q$ is in fact a generator of $I$, i.e. $\langle q \rangle = I$, but I can't show it.

Question: How can we compute $I$?

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Geometrically, my question goes as follows:

Consider the map $$ \alpha: \mathbb{A}_K^1 \to \mathbb{A}_K^2, \\ z \mapsto (z^2, z^3 - z) $$ $\alpha$ descends to an isomorphism between its "coimage" and "image", i.e. there is an affine algebraic set $X \subseteq \mathbb{A}_K^2$ and the following sequence: $$ \mathbb{A}_K^1 \twoheadrightarrow X \hookrightarrow \mathbb{A}_K^2 $$ Question: How can we compute a set of equations that describes $X$?

Answer 1

In a pinch, if you don’t want to refer to this machinery linked to in the comment, you can do things like this by hand using the division algorithm.

Suppose $f \in k[x,y]$ is a polynomial so that $f(z^2, z^3 - z) = 0$ in $k[z]$. Viewing $k[x,y] = k[x][y]$ as the polynomial ring in $y$, over the ring $k[x]$, we can divide $f$ by $q(x,y)$ to write $$f(x,y) = g(x,y)q(x,y) + R_1(x) y - R_2(x)$$ where $R_1(x), R_2(x) \in k[x]$ are polynomials in one variable, $x$. (Note that the $y$-degree of $q$ is $2$, with the highest degree term being $-y^2$.)

Now, plugging in $x = z^2$ and $y = z^3 - z$, we are left with an equation $$R_1(z^2)(z^3 - z) = R_2(z^2)$$ in $k[z]$. The right side only has terms in even degree, whereas the left only has terms in odd degrees. Hence, both sides must be zero, so $R_1(x)$ and $R_2(x)$ had to be zero to start with, so $f$ is divisible by $q$.

Answer 2

We need to prove that the homomorphism of $k$-algebras $$\varphi : k[x,y]/(x^3-2x^2 + x - y^2) \to k[z^2,z^3-z]$$ defined by $[x] \mapsto z^2$, $[y] \mapsto z^3-z$ is an isomorphism. It is clearly surjective. The left hand side is a free module over $k[x]$ of rank $2$. The right hand side is also a module over $\varphi(k[x]) = k[z^2]$, and it is also free of rank $2$: Otherwise, $z^3-z \in k[z^2]$, which is absurd. Hence, $\varphi$ must be an isomorphism. (Of course, this is just another way of phrasing Daniel's answer.)