# showing map is a ring isomorphism (injective)

Let $k$ be an algebraically closed field with $char(k)\neq 2$. I want that $k[x,y]$ is isomorphic to $k[u,v]$, where $u=x+iy$ and $v=x-iy$ for some $i\in k$ such that $i^2=-1$. I have that the isomorphism $\phi$ should just map $f(x,y)$ to $f(u,v)$, and have convinced myself that this is a surjective homomorphism. I can't figure out why it should be injective though, or alternatively why the kernel is trivial. Can anyone point this out?

You need the expressions for $x$ and $y$ in terms of $u$ and $v$. They are $x=(u+v)/2$ and $y=(u-v)/(2i)=-i(u-v)/2$. Try it on $f(x,y)=x^2+y^3$: you will get your element of the ring as an element $g(u,v)$ which is quite different in appearance from $f$.