Prove that $Ker (\phi ) = (X^2 - Y^3, Y^2-Z^3)$ where $\phi : k[X,Y,Z] \rightarrow k[T]$ s.t. $X \mapsto t^9, \; Y \mapsto t^6, \; Z \mapsto t^4$.

Question

I want to prove that for a ring homomorphism $\phi : k[X,Y,Z] \rightarrow k[T]$ s.t. $X \mapsto t^9, \; Y \mapsto t^6, \; Z \mapsto t^4$, we have $Ker (\phi ) = (X^2 - Y^3, Y^2-Z^3)$.

My attempt: It is easy to prove that $ I = (X^2 - Y^3, Y^2-Z^3) \subset Ker (\phi ) $.

To prove $Ker (\phi) \subset I$,

Let $F(X,Y,Z) \in k[X,Y,Z]$ where $k$ is a field. We know that we can write $F$ around any point $(a_1,a_2,a_3) \in \mathbb{A}^3(k)$ as $$ F = \sum \lambda_{(i)} (X-a_1)^{i_1} (Y-a_2)^{i_2}(Z-a_3)^{i_3} $$

and if $F(a_1,a_2,a_3)=0$, then $F = \sum G_{i} (X_i-a_i) $ where $X_i$s are $X,Y,Z$.

We write $F$ around any general point $(t^9,t^6,t^4)$ as $F = G_1(X-t^9)+G_2(Y-t^6)+G_3(Z-t^4)$. How can I show that $F $ can also be written as $F = H_1 (X^2-Y^3) + H_2 (Y^2-Z^3)$?

Edit 1 : Additional question

Is the ring homomorphism $\phi$ surjective? (so that we can comment $k[X,Y,Z]/I \cong k[T]$)?

It doesn't seem surjective as I can't see how what will be mapped to $T \in k[T]$. But as I have verified, I am wrong.

Answer 1

Let's denote $I=(X^2-Y^3,Y^2-Z^3)$. It is, indeed, true that $\operatorname{Ker}(\phi)=I$.

One way of seeing this is that reducing any power $X^n, n\ge2$, modulo $X^2-Y^3$, and after that reducing any power $Y^m, m\ge2$, modulo $Y^2-Z^3$ shows that the quotient ring $k[X,Y,Z]/I$ is a free $k[Z]$ module of rank four with basis $1,X,Y,XY$. After all, all the appearances of $X^2,Y^2$ or higher have been replaced with other monomials.

This means that $k[X,Y,Z]/I$ has a basis (as a vector space over $k$) of monomials of one of the four types $Z^i$, $XZ^i$, $YZ^i$, and $XYZ^i$, $i\in\Bbb{N}$. The homomorphism $\phi$ sends these monomials to $t^{4i}$, $t^{4i+9}$, $t^{4i+6}$ and $t^{4i+15}$ respectively. The integers $0,9,6,15$ are pairwise non-congruent modulo $4$, so the images of these basic monomials are linearly independent over $k$. The claim follows.

---

We also see that the $k$-space $k[t]/\operatorname{Im}(\phi)$ is spanned by the cosets of the monomials $t,t^2,t^3,t^5,t^7,t^{11}$. Again, it is simplest to see this by looking at the exponents modulo $4$. The highest missing powers in each residue class are $t^5,t^2,t^{11}$ simply because $t^0,t^9,t^6$ and $t^{15}$ are in there. The problems of this type fall under the umbrella of numerical semigroups. The numerical semigroup generated by $4,6,9$ governs the image of $\phi$, and this is the complement. Whenever the generating set of natural numbers has no common factors, the complement will be finite.

Answer 2

I could not easily understand how "the claim follows" argument in @Jyrki solution. So, I complete the rest of the argument.

From @JyrkiLahtonen argument, we can infer the following two results.

(i) Every element in $k[X,Y,Z]/I$ is the $I$ residue of elements of form $F_1+F_2X+F_3Y+F_4XY$ for some $F_1,F_2,F_3,F_4 \in k[Z]$, and

(ii) In the above ring homomorphism $\phi : k[X,Y,Z] \rightarrow k[T] $, if we choose any function of the form $G = F_1+F_2X+F_3Y+F_4XY$ and s.t. $\phi (G)=0$, then $G=0$ and since all the members of the form $G$ are in correspondence with $k[X,Y,Z]/I$. So, the map $k[X,Y,Z]/I \hookrightarrow k[T] $ is an injection map (as kernel of the map is zero). Now, use the result that subring of any integral domain is also an integral domain and the result then follows.

More formally, in (ii), we can prove that $\psi : k[X,Y,Z]/I \rightarrow k[T]$ s.t. $F_1+F_2X+F_3Y+F_4XY +I \mapsto F_1+F_2\phi(X)+F_3\phi(Y)+F_4\phi(X)\phi(Y)$ and prove from there.