Why this is an Automorphism of Q[x]

Question

I got to use the Eisenstein Criterion but instead of using $f(x)$ using $f(x+1),$ I know that if I have an automorphism $\phi :\mathbb{Q}[x] \to \mathbb{Q}[x]$ being irreductible does not change. but i don't understand why $\phi$ is an automorphism. Thanks if you can help me with this.

Answer 1

I'm not quite sure if I've interpreted your question properly, but here's my answer:

Let $\phi:\mathbb{Q}[x]\rightarrow\mathbb{Q}[x]$ be given by $\phi(f(x)) = f(x+1)$. First note that if $\varphi:\mathbb{Q}\rightarrow\mathbb{Q}$ is given by $\varphi(x) = x+1$, we can re-write $\phi$ as $\phi(f) = f\circ\varphi$.

We want to see if $\phi$ is an automorphism.

Firstly, we check that $\phi$ is well defined. Clearly $\varphi$ is a polynomial, and the composition of polynomials is a polynomial, so $\phi$ is well defined.

Now we want to check that it preserves the structure. Let $f, g \in \mathbb{Q}[x]$. Then $\phi((f+g)(x)) = (f+g)(x+1) = f(x+1) + g(x+1) = \phi(f(x))+\phi(g(x))$.

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

Finally it suffices to show that $\phi$ is bijective.

First we show $\phi$ is injective. Let $f, g \in \mathbb{Q}[x]$ and suppose $\phi(f)=\phi(g)$. Then $f\circ \varphi = g\circ\varphi$, so $f = f\circ \varphi \circ \varphi^{-1}= g\circ\varphi\circ\varphi^{-1} = g$ (as we can easily see that $\varphi$ is invertible.) Similarly you can show that $\phi$ is surjective, and we are done :)