Division of a polynomial in three variables I would like to demonstrate the following
Let $f(x,y,z) \in K[x,y,z]$ a polynomial such that $f(a^n,b^n,ab)=0$; then f is divisible by $xy-z^n$.
Hint: use a polynomial division algorithm

EDIT
At this link Showing that $\mathbb{C}[x,y]^{\mu_n}$ and $\mathbb{C}[x,y,z]/(xy-z^n)$ are isomorphic as rings
I found the same problem. The answer suggest:
$$ $$
If $f(x,y,z)=\sum_{i,j,k} a_{i,j,k}x^iy^jz^k$ so
$$0=f(u^n,v^n,uv)=\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$$
The equation $\sum_{i,j,k} a_{i,j,k}u^{ni+k}v^{nj+k}$ give some dependence relation amongst the $a_{i,j,k}$ (since the exponent of $u$ and $v$ are the same for some $i,j,k$ and since the element $u^av^b$ are indipendent over $K[x,y]$.

Can anyone tell me what are these relations between coefficients
$a_{i,j,k}$ they are talking of? How do they obtain? How do these
allow me to prove that $f(x,y,z)=g(x,y,z)(xy-z^n)$? TO PLEASE SOMEONE
CAN HELP ME?

 A: I'm going to assume that $K$ is a field and $K[x, y, z]$ is the associated ring of polynomials. Unfortunately, I don't see a polynomial division solution to this...
The problem is equivalent to showing that $f(x, y, z)$ is an element of the $0$ coset of the quotient ring $K[x, y, z]/ (xy - z^n )$. Let's first consider $K[a^n, b^n, ab]$. Observe that there is a surjective homomorphism $\varphi : K[x, y, z] \rightarrow K[a^n, b^n, ab]$ where $(x, y, z) \mapsto (a^n, b^n, ab).$ Moreover, $\ker (\varphi) = (xy - z^n)$ (show why). Then by the First Isomorphism Theorem for rings, it follows that
$$K[x, y, z] / (xy - z^n) \cong K[a^n, b^n, ab].$$
(can you give the isomorphism explicitly?) The conclusion that $f(x, y, z) \in 0 + (xy - z^n)$ follows.
This form shows up in some fairly interesting results.
A: Here's how the polynomial division approach would work.
Lemma: Any polynomial $f\in k[x,y,z]$ can be written uniquely as $q(x,y,z)(xy-z^n) + \sum_{i=0}^{n-1} p_i(x,y)z^i$.
Proof: Induction on the highest power of $z$ appearing in $f$. Clearly any polynomial of degree $n-1$ or less in $z$ satisfies the statement, so suppose $f$ has degree $d\geq n$ in $z$. Write $f=\sum_{i=0}^d p_i(x,y)z^i$: then $$f= p_d(x,y)z^d + \sum_{i=0}^{d-1} p_i(x,y)z^i =$$ $$= p_d(x,y)xyz^{n-d}-p_d(x,y)z^{d-n}(xy-z^n)+\sum_{i=0}^{d-1} p_i(x,y)z^i$$ and so $f=-p_d(x,y)z^{d-n}(xy-z^n) + f'$, where $f'=p_d(x,y)xyz^{n-d}+\sum_{i=0}^{d-1} p_i(x,y)z^i$ has degree at most $d-1$ in $z$, so by the induction hypothesis we are done. $\blacksquare$
Now suppose $f$ is in the kernel of $\varphi:k[x,y,z]\to k[a,b]$ by $\varphi(x)=a^n$, $\varphi(y)=b^n$, $\varphi(z)=ab$. Writing it as above, we see that $\varphi(q(x,y,z)(xy-z^n))=\varphi(q(x,y,z))(a^nb^n-(ab)^n)=0$, so $f$ is in the kernel iff $$\varphi\left(\sum_{i=0}^{n-1} p_i(x,y)z^i\right)=0.$$
Distributing $\varphi$, we see that we should have $\sum_{i=0}^{n-1} p_i(a^n,b^n)(ab)^i=0$. Now I claim that no monomial $a^rb^s$ appears with nonzero coefficient in $p_i(a^n,b^n)(ab)^i$ and $p_j(a^n,b^n)(ab)^j$ for $i\neq j$: look at the residue class of the exponent of $a$ modulo $n$. If $a^rb^s$ appeared in both of those expressions with nonzero coefficient, we should have $i\equiv r \mod n$ and $j\equiv r \mod n$, or $i\equiv j\mod n$. But $0\leq i,j<n$, so $i\equiv j\mod n$ implies $i=j$. Hence the coefficient of $a^rb^s$ in $\sum_{i=0}^{n-1} p_i(a^n,b^n)(ab)^i$ comes from only one $p_i(a^n,b^n)(ab)^i$, so if all the coefficients of $a^rb^s$ are zero, then all the coefficients of $p_i$ are zero for all $i$.
But this exactly means that $f=q(x,y,z)(xy-z^n)$ and we're done.
