# Kernel of a polynomial ring homomorphism

Let $$\mathbb{F}$$ be a field, and define a homomorphism $$\phi:\mathbb{F}[x,y]\rightarrow \mathbb{F}[z]$$ by: $$f(x,y)\mapsto f(z^a,z^b)$$ where $$a\neq b\in \mathbb{N}$$.

My question is: For what $$\mathbb{F},a$$ and $$b$$, we have: $$\ker(\phi)=\langle x^b-y^a\rangle \ ?$$

Or rather this is a statement true for any polynomial ring; otherwise, is there any general technique to find the kernel of a map like this?

For example, let $$\phi:\mathbb{C}[x,y]\rightarrow\mathbb{C}[z]$$ defined by $$f(x,y)\mapsto f(z^2,z^3)$$, then $$\ker(\phi)=\langle x^3-y^2\rangle$$.

• Do you want to assume $\gcd(a,b)=1$? – Daniel Schepler Mar 21 at 19:34
• @DanielSchepler No, I am assuming nothing but $a\neq b$. However, I am curious about whether $\gcd(a,b)=1$ is sufficient for the statement to be true, and what happens when $\gcd(a,b)\neq 1$. – Wenze 'Sylvester' Zhang Mar 21 at 19:45
• In general I think the kernel would be $\langle x^{b/d} - y^{a/d} \rangle$ where $d = \gcd(a,b)$. – Daniel Schepler Mar 21 at 19:46
• @DanielSchepler I agree with you, but I found it hard to prove $\ker(\phi)\subseteq \langle x^{b/d}-y^{a/d}\rangle$. Do you have a proof in mind? – Wenze 'Sylvester' Zhang Mar 21 at 23:55

Observe that in general, if $$d = \gcd(a,b)$$, then $$x^{b/d} - y^{a/d} \in \ker(\phi)$$, so $$\langle x^{b/d} - y^{a/d} \rangle \subseteq \ker(\phi)$$.
To show the reverse inclusion, notice that by treating $$\mathbb{F}[x,y]$$ as $$\mathbb{F}[y][x]$$, using the division algorithm we can write any $$f \in \mathbb{F}[x,y]$$ as $$q(x,y) (x^{b/d} - y^{a/d}) + r(x,y)$$ where each term in the remainder $$r(x,y)$$ has $$x$$-degree less than $$b/d$$. Now, $$\phi(f) = \phi(r)$$. On the other hand, each possible monomial $$x^m y^n$$ with $$0 \le m < b/d$$ maps to a distinct monomial $$z^{am + bn}$$ (see below for a proof), so if $$f \in \ker(\phi)$$ then $$\phi(r) = 0$$ implies $$r = 0$$.
The map $$\mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}_0, (m, n) \mapsto am + bn$$, is injective when restricted to $$[0, b/d) \times \mathbb{N}_0$$.
To show this, suppose $$am + bn = am' + bn'$$; then $$(a/d) (m-m') = (b/d) (n'-n)$$. Since $$\gcd(a/d, b/d) = 1$$, this implies that $$m \equiv m' \pmod{b/d}$$; but due to the restriction on the range of possible values of $$m,m'$$, that further implies that $$m=m'$$ and therefore $$n=n'$$ also.
(Note that this argument assumes that $$0 \notin \mathbb{N}$$ in your convention, so that $$a \ne 0$$ and $$b \ne 0$$. If either or both of $$a,b=0$$, then those special cases should be easier to work with.)