Let $φ:\mathbb{C[x,y,z]}\to \mathbb{C[t]}$ by $φ(f(x,y,z))=f(t,t^2,t^3)$. Find generators for ker($φ$)

Question

Let $φ:\mathbb{C[x,y,z]}\to \mathbb{C[t]}$ by $φ(f(x,y,z))=f(t,t^2,t^3)$. Find generators for ker($φ$).

This is a scan of my work so far, with my conjecture boxed at the top. I want to be able to write k(x,y,z) in the form $(x^2-y)f(x,y,z)$+$(x^3-z)g(x,y,z)$+$(y^3-z^2)h(x,y,z)$ for some polynomials f,g, and h. How can I do this? Or, if my conjecture is wrong, please lead me on the right path.

EDIT

I have been informed that the $(y^3-z^2)$ generator is redundant. (Thanks Daniel!) That being said, how can I express $k(x,y,z)$ in the form $(x^2-y)f(x,y,z)$+$(x^3-z)g(x,y,z)$? I'm looking for an explicit formula; I would prefer to avoid the division algorithm unless absolutely necessary, since the division algorithm I am familiar with is for single variable polynomials only.

Answer 1

You have the homomorphism $\phi:\mathbb{C}[x,y,z]\to \mathbb{C}[t]$ defined by $$\phi(p(x,y,z))=p(t,t^2,t^3)$$ Let $I$ be the kernel of $\phi$.

By definition, $I=\{p \in \mathbb{C}[x,y,z]\mid p(t,t^2,t^3) = 0\}$.

Let $J$ be the ideal of $\mathbb{C}[x,y,z]$ given by $$J=(x^2-y,x^3-z)$$ It's easy to verify that each of the generators of $J$ satisfies the condition for membership in $I$, hence $J \subseteq I$.

The goal is to show $J = I$.

Suppose instead that $J$ is a proper subset of $I$.

Let $p \in I\setminus J$.

Reducing $p\;\text{mod}\;(x^2 - y)$, we can replace all occurrences of the variable $y$ by $x^2$, and the new polynomial, $p'$ say, is still in $I\setminus J$. Note that $p'$ is in the subring $\mathbb{C}[x,z]$ of $\mathbb{C}[x,y,z]$ (i.e., the variable $y$ is not present in any of the terms of $p'$).

Reducing $p'\;\text{mod}\;(x^3 - z)$, we can replace all occurrences of the variable $z$ by $x^3$, and the new polynomial, $p''$ say, is still in $I\setminus J$. Note that $p''$ is in the subring $\mathbb{C}[x]$ of $\mathbb{C}[x,y,z]$ (i.e., the variables $y,z$ are not present in any of the terms of $p''$).

But then \begin{align*} &p''(x) \in I\\[4pt] \implies\;&p''(t) = 0\;\text{in}\;\mathbb{C}[t]\\[4pt] \implies\;&p''(x)= 0\;\text{in}\;\mathbb{C}[x]\\[4pt] \implies\;&p''= 0\;\text{in}\;\mathbb{C}[x,y,z]\\[4pt] \end{align*} contradiction, since $p'' \notin J$.

It follows that $J=I$, as was to be shown.

Answer 2

Lemma: Let $R$ be a commutative ring and $n$ be a fixed positive integer. Then for any polynomial $p(x_1, x_2) \in R[x_1, x_2]$, there's $q(x_1, x_2), r(x_1) \in R[x_1, x_2]$ such that $p(x_1,x_2) = (x_1^n - x_2)q(x_1,x_2) + r(x_1)$

Proof: Induct on the highest degree of $x_2$ in $p(x_1, x_2)$. Base case (where no term of $x_2$ appears is trivial). For the inductive step $d \rightarrow d+1$, write $\displaystyle p(x_1,x_2) = \sum_{0 \leq i \leq d+1} x_2^i p_i(x_1) = x_2^{d+1} p_{d+1}(x_1) + \sum_{0 \leq i \leq d} x_2^i p_i(x+1)$. By induction hypothesis, write $\displaystyle - p_{d+1}(x_1)x_1^n x_2^d + \sum_{0 \leq i \leq d} x_2^i p_i(x+1) = (x_1^n-x_2)q(x_1, x_2) + r(x_1)$, and then we get $p(x_1, x_2) = (x_1^n - x_2)(x_2^d p_{d+1}(x_1) + q(x_1,x_2)) + r(x_1)$, finishing the inductive step.

Returning to the problem, fix arbitrary $p(x,y,z) \in \mathbb{C}[x,y,z]$. Set $R = \mathbb{C}[y]$, then $p_y(x,z) \in R[x,z]$ (you treat the $y$ as fixed) and by lemma $p(x,y,z) = p_y(x,z) = (x^3 - z)r_1(x,z) + r_2(x)$. As $r_1, r_2 \in R = \mathbb{C}[y]$, we get $p(x,y,z) = (x^3-z)R_1(x,y,z)+R_2(x,y)$ with $R_i \in \mathbb{C}[x,y,z]$. Again we also get $R_2(x,y) = (x^2-y)R_3(x,y) + R_4(x)$, so all in all $$ p(x,y,z) = (x^3-z)R_1(x,y,z) + (x^2 - y)R_3(x,y) + R_4(x) $$.

Let $J = \langle x^3 -z, x^2-y \rangle$. It's clear if $R_4(x) = 0$ then $p \in J$. But $\phi(p) = 0 \Rightarrow R_4(t) = 0$. Now note the mapping of $\phi$ restricted to $\mathbb{C}[x]$ is a isomorphism, so taking the inverse map we get $R_4(x) = 0$, or $p \in J$ as desired.