Prove that the map R[x,y]→R[x,y] defined by $x\mapsto x+f(y)$, $y\mapsto y$ is an automorphism of R[x,y].

Question

An automorphism of a ring $R$ is an isomorphism from $R$ to itself.

Let $R$ be a ring, and let $f(y)$ be a polynomial in one variable with coefficients in $R$. Prove that the map $R[x,y]\to R[x,y]$ defined by $x\mapsto x+f(y), y\mapsto y$ is an automorphism of $R[x,y]$.

I am a bit lost on that one. To show it's a homomorphism, we have to prove the following:

Let $a,b \in R[x,y]$ and call the map $\phi$. Then:

1. $\phi(1) = 1$,

2. $\phi(a+b) = \phi(a)+\phi(b)$,

3. $\phi(ab) = \phi(a)\phi(b)$.

Since it has to be an isomorphism, it needs to be one-to-one. How can I show that? Also how can I pick elements $a,b$ in $R[x,y]$ to be able to start on the prove?

Thank you very much.

Answer 1

We can provide an inverse to $\phi$. Namely, let $$\psi(h) = h(x-f(y),y).$$ We have $$\psi(\phi(g)) = \psi(\;\underbrace{g(x+f(y),y)}_{=h(x,y)}\;) = h(x-f(y),y) = g(x,y) = g $$ and $$\phi(\psi(h)) = \psi(\;\underbrace{h(x-f(y),y)}_{=g(x,y)}\;) = g(x+f(y),y) = h(x,y) = h. $$