Isomorphism of a polynomial ring $F[x]$.

Question

Show that the map $\phi:F[x]\to F[x]$ given by $\phi(f(x))=f(x+1)$ is an isomorphism such that $\phi(a)=a$ for every $a\in F$.

So far I have that $\phi$ is a homomorphism because, Let $f,g\in F[x]$ then

$\phi(f(x)+g(x))=(f+g)(x+1)=f(x+1)+g(x+1)=\phi(f(x))+\phi(g(x))$.

$\phi(f(x))\phi(g(x))=f(x+1)g(x+1)=(fg)(x+1)=\phi(f(x)g(x))$.

However I'm having trouble showing this is an isomorphic. Here's my attempt,

$\phi(f(x))=\phi(g(x))$ then $f(x+1)=g(x+1)$ then $x+1=x+1$ since f and g isomorphic then $x=x$. Therefore injective.

I'm not sure how to show it is surjective.

Answer 1

Hint: The inverse map sends $g(x)$ to $g(x-1)$

Answer 2

As thedilated shows in his answer: you can just construct an inverse map to show something is a bijection (as Ben P. says in the comments, this is a pretty easy way to show bijectivity for this problem).

I will approach the problem from the way that you were trying to do it (i.e. show injectivity and surjectivity of $\phi$).

Injectivity: suppose $\phi(f(x))=\phi(g(x))$. Then we have $f(x+1)=g(x+1)$, and thus $f((x-1)+1)=g((x-1)+1)\implies f(x)=g(x)$.

Surjectivity: take $f(x)\in F[x]$. Then $\phi (f(x-1))=f(x)$.

Now, clearly if $f_a(x)=a$ for each $x$ then $\phi(a)=\phi (f_a(x))=f_a(x+1)=a$.