Consider the finite matrix group $C_{4} \subset$ GL$(2,\mathbb{C})$ generated by
$$A = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \in \text{GL}(2,\mathbb{C}).$$
(a)Prove that $C_{4}$ is cyclic of order 4.
(b)Use the method of Example 13 to determine $\mathbb{C}[x,y]^{C_{4}}$.
(c)Is there an algebraic relation between the invariants you found in part (b)? Can you give an example to show how uniqueness fails?
(d)Use the method of Exercise 13 to show that the relation found in part (c) is the only relation between the invariants.
I am comfortable with my results for parts (a) through (c) and in particular I found that the ring of invariants, $\mathbb{C}[x,y]^{C_{4}}$ = $[x^4, y^4, xy].$
I identified the relation $x^4 \cdot y^4 = (xy)^4$ so that uniqueness fails.
As mentioned above I was instructed to
(d) Use the method of exercise 13 to show that the relation found is the only relation between invariants.
Where exercise 13 states
We discovered the algebraic relation $x^2 \cdot y^2 = (xy)^2$ between the invariants $x^2, y^2, xy$. We want to show that this is essentially the only relation. More precisely, suppose that we have a polynomial $g(u,v,w) \in k[u,v,w]$ such that $g(x^2,y^2,xy) = 0$. We want to prove that $g(u,v,w)$ is a multiple (in $k[u,v,w]$) of $uv-w^2$ (which is the polynomial corresponding to the relation).
(a) If we divide $uv-w^2$ using lex order with u>v>w, show that the remainder can be written in the form $uA(u,w) + vB(v,w) + C(w)$.
(b) Show that a polynomial $r =uA(u,w) + vB(v,w) + C(w)$ satisfies $r(x^2, y^2, xy) = 0$ if and only if $r = 0$.
But I didn't quite understand that method, so what I said instead (switching back to my ring of invariants) was
Let $g(u,v,w) \in \mathbb{C}[u,v,w]$ such that $g(x^4,y^4,xy)=0$. Let $f_{1} = uv-w^4 \in \mathbb{C}[u,v,w]$ be the polynomial that corresponds to my identified relation. Dividing $g$ by $f_{1}$ with respect to lex order and $u>v>w$ we have $$g = a_{1}(uv-w^4) + r,$$ where $r$ = 0 or $r$ is a linear combination of monomials with coef in $\mathbb{C}$, where no monomial divisible by LT$(f_{1}) = uv$. If $r$ = 0 we are done, since that implies $g$ is a multiple of $f_{1}$, so suppose $r \neq 0$. Then $r$ must be a polynomial in $w$ alone and we can say $$g = a_{1}(uv-w^4) + r(w).$$ But by our hypothesis, $g(x^4,y^4,xy)=0$, so we have $$0 = g(x^4,y^4,xy) = a_{1}(x^4 \cdot y^4 - (xy)^4) + r(w).$$ Which implies $$0 = g(x^4,y^4,xy) = 0 + r(xy),$$ But since $\mathbb{C}$ is an infinite field we know $r$ must be the zero polynomial, and $g$ is a multiple of $uv-w^4$ as desired.
I feel convinced by my argument but I am a little hesitant since they clearly told me to use exercise 13... Please let me know if my argument fails, and if so maybe enlighten me as to why the argument recommended works. In part (a) it does not make sense to me since it seems to contradict that no term of $r$ is divisible by LT$(f)$ = $uv$?.
Any help greatly appreciated! Let me know if there is any more information I can provide or if anything I said was unclear.
Your proof uses the following claim (paraphrasing slightly):
This is not true. For a counterexample, consider $r(u,v,w)=2u+v^3+vw$. This is not a polynomial in $w$ alone, but none of its monomials are divisible by $uv$. Note that each monomial is divisible by either $u$ or $v$, but never both.
Here is a corrected version (it's the same as your exercise 13(a)):
To see why, note that no monomial of $r$ can contain both $u$ and $v$, so each belongs to one of the following three groups:
Monomials in the first group sum to a polynomial of the form $uA(u,w)$. Similarly the second group sums to $vB(v,w)$ and the third to $C(w)$.