Invariants over reflections in 2 dimensions

Question

This is part of Exercise 1.7 of Einsenbud's commutative algebra book. Let $k$ be a field. It is clear that the ring of invariant functions under the reflection $(x,y)\mapsto -(x,y)$ in $k^2$ is $k[x^2,y^2,xy]$. Eisenbud asks us to understand this as the quotient of a ring of polynomial functions $k[u,v,w]/(uw-v^2)$. To show the isomorphism I though of using the first isomorphism theorem on the ring homomorphism $\varphi:k[u,v,w]\rightarrow k[x^2,y^2,xy]$ defined by $\varphi(u)=x^2$, $\varphi(v)=xy$, and $\varphi(w)=y^2$. It is clear that it is surjective and that $(uw-v^2)\subseteq\ker\varphi$. I however found trouble when trying to proof $\ker\varphi\subseteq(uw-v^2)$. By taking some general $f=\sum_{r,s,t=0}^\infty a_{rst}u^rv^sw^t\in\ker\varphi$, I was able to obtain some relations between its coefficients. I am sure that using them one can show that the polynomial is divisible by $uw-v^2$. However, this doesn't seem very enlightening and I was wondering if there is a better way of proving this.

Answer 1

I was able to find a nice answer after reading Example 11.3.16 in Artin's algebra book. I'll post it in case anyone is interested in the future. The idea is to note that $uw-v^2$ is monic as a polynomial on $v$ (since $k$ is a field this doesn't really matter). Therefore for any $f\in\ker \varphi$ there exists $q,r\in k[u,v,w]$, with $r$ at most linear in $v$ such that $f=q(uw-v^2)+r$. Since $f,uw-v^2\in\ker\varphi$, we conclude that $r\in\ker\varphi$. Now, let us write $r=g_0+g_1 v$ for $g_0,g_1\in k[u,w]$. We then have that $0=\varphi(r)=\varphi(g_0)+\varphi(g_1)xy$. All monomials $\varphi(g_0)$ and $\varphi(g_1)$ are of even degree in both $x$ and $y$. In particular, the monomials in $\varphi(g_1)xy$ are of odd degree in $x$ and $y$. We conclude that $g_0=0$ and $g_1=0$, i.e. $r=0$. therefore $f\in(uw-v^2)$.