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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 $\ker(\phi)=\langle x^3-y^2\rangle$.

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    $\begingroup$ Do you want to assume $\gcd(a,b)=1$? $\endgroup$ – Daniel Schepler Mar 21 at 19:34
  • $\begingroup$ @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 $\gcd(a,b)\neq 1$. $\endgroup$ – Wenze 'Sylvester' Zhang Mar 21 at 19:45
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    $\begingroup$ In general I think the kernel would be $\langle x^{b/d} - y^{a/d} \rangle$ where $d = \gcd(a,b)$. $\endgroup$ – Daniel Schepler Mar 21 at 19:46
  • $\begingroup$ @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? $\endgroup$ – Wenze 'Sylvester' Zhang Mar 21 at 23:55
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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 $r = 0$.

So, all we have left to show is the easy purely number theoretic result:

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.)

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