Trying to understand a proof for the automorphisms of a polynomial ring

Question

Consider the following proof for finding all automorphisms of the ring $\mathbb{Z}[x]$ which I am trying to understand.

Source

I have two question regarding the proof

1) They set $d = \deg(\phi(f(x))$. Why does for all non-constant polynomials $f(x) \in \mathbb{Z}[x]$ apply: $\deg(\phi(f(x))) \geq d$? They argue that it applies because $f(x)$ is a linear combination of multiples of x, I don't understand that, and know linear combinations only in terms of vectors. Can you please explain this in a different way?

2) Knowing that $\deg(\phi(f(x))) \geq d$ applies, they follow that $d = 1$, what I don't understand as well. They take a non-constant $f(x) \in \mathbb{Z}[x]$, which exists because the automorphism is surjective. But why does $\phi(f(x)) = x$ apply for it? And how do they follow $d = 1$ from it?

Answer 1

I claim that the set of automorphisms of $\mathbb{Z}[x]$ are the ring homomorphisms $\phi$ that satisfy $\phi(x) = \phi_0+\phi_1 x$ where $\phi_0 \in \mathbb{Z}$ is arbitrary and $\phi_1 \in \{\pm 1\}$.

Suppose $\phi$ is an automorphism. As above, we have $\phi(n) = n$ for $n \in \mathbb{Z}$. Since $\phi$ is an automorphism, we can find an inverse for $x$. Then for some $p_k$ we have $\phi(\sum_{k=0}^n p_k x^k) = x$ (with $p_n \neq 0$). That is, $\sum_{k=0}^n p_k \phi(x)^k = x$. It follows that we must have $\partial \phi(x) \ge 1$, and since $p_n \neq 0$, we must have $\partial \phi(x) = 1$ (and also, $n=1$).

Hence $\phi(x)$ has the form $\phi(x) = \phi_0+\phi_1 x$ which gives $p_0 + p_1 \phi_0 + p_1 \phi_1 x = x$. Then $p_1 = \phi_1 \in \{\pm 1\}$ (and $p_0+p_1 \phi_0 = 0$ which gives $p_0 = -p_1 \phi_0 = - \phi_0 \phi_1$).

Now for the other direction. Suppose $\phi$ is a ring homomorphism and $\phi(x) = \phi_0+\phi_1 x$, where $ \phi_1 \in \{\pm 1\}$. (Since $\phi$ is a ring homomorphism we have $\phi(n) = n$ for $n \in \mathbb{Z}$.) We need to show that $\phi$ has an inverse. By explicit computation, we have $\phi(\phi_1^{-1}( x-\phi_0)) = x$. Pick some element $\sum_k a_k x^k \in \mathbb{Z}[x]$ and note that $\phi ( \sum_k a_k (\phi_1^{-1}( x-\phi_0))^k) = \sum_k a_k x^k$, hence $\phi$ is invertible and so is an automorphism.

Answer 2

If $f(x)=a_0+a_1x+\cdots+a_nx^n$, $a_n\ne0$, $n\ge1$, then $\phi(f(x))=a_0+a_1\phi(x)+\cdots+a_n\phi(x)^n$ (note that $a_i\in\mathbb Z$ and then $\phi(a_i)=a_i$) which has the degree equal to $\deg\phi(x)^n=dn\ge d$. Moreover, if $d\ge2$ then $\deg\phi(f(x))=dn\ge2n\ge 2$, so the image of $\phi$ consists of constants and of polynomials of degree $\ge 2$. This shows that the image of $\phi$ is different from $\mathbb Z[x]$, contradicting the surjectivity of $\phi$.

Then $d=1$ and thus $\phi(x)=bx+a$. (Here is where the notes err.) Since $\phi$ is surjective there is $f\in\mathbb Z[x]$ such that $\phi(f(x))=x$ and from $\deg\phi(f(x))=1$ we get $\deg f=1$, so $f(x)=a_0+a_1x$. From $\phi(f(x))=x$ we get $a_0+a_1(bx+a)=x$, hence $a_1b=1$, so $b=\pm1$. Thus you showed that the ring automorphisms of $\mathbb Z[x]$ have the form $\phi(x)=\pm x+a$. (Well, now you have to prove that these are indeed automorphisms, but this is easy enough.)