0
$\begingroup$

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.

  • 1
    "how can I pick elements a,b in R[x,y] to be able to start on the prove?" You don't. You show that 2. and 3. is true for _all possible_ (also known as "arbitrary") $a, b$. – 2017-01-19
  • 3
    The fact that this assignment gives you a well defined homomorphism is basically the definition of the polynomial ring, namely its universal property. The main part of the exercise is the bijectivity. Here, you can pretty easily find the inverse. – 2017-01-19
  • 0
    For $a$ chose a finite number of elements $a_{i,j} \in R$ and put $a = a_{0,0}+a_{1,0}x + a_{0,1}y +a_{2,0}x^2 + a_{1,1}xy+a_{0,2} y^2 + \ldots$ and likewise for $b$ take $b = b_{0,0}+b_{1,0}x + b_{0,1}y +b_{2,0}x^2 + b_{1,1}xy+b_{0,2} y^2 + \ldots$ – 2017-01-21
  • 0
    So if every element in set A has an inverse in set B, and every element in set B has an inverse in set A, then they're bijective? – 2017-01-22

1 Answers 1

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 $$\phi(\psi(h)) = \psi(\;\underbrace{h(x-f(y),y)}_{=g(x,y)}\;) = g(x+f(y),y) = h(x,y) = h. $$