Describe all ring homomorphisms from $\mathbb{R}[T] \rightarrow \mathbb{R}[T]$ One of the problems in a problem set I was given as homework in my Algebra course proposes the next problem:


  Describe all ring homomorphisms $\mathbb{R}[T] \rightarrow \mathbb{R}[T]$. Which of them are isomorphisms?


I would like some suggestions towards the right direction, not the answer to the problem.

This is what I've got so far:

Given a ring homomorphism $f:\mathbb{R}[T] \rightarrow \mathbb{R}[T]$:

Given an arbitrary polinomial $p(T) = a_{0} + a_{1}T + \ldots + a_nT^n$ we have that $f(p) = f(a_0) + f(a_1)f(T) + \ldots + f(a_n)f(T)^n$. So we get that $f$ is completely determined by the values it assumes on $\mathbb{R}$ and $f(T)$.

So this problem may now be separated in two: 


Classifying all homomorphisms of the form $f:\mathbb{R} \rightarrow \mathbb{R}[T]$
Classifying all possible values $f(T)$


With respect to (1):

Conjecture
The only ring homomorphism is $f(x)=x$ (we put as a condition that f(1) = 1, on the definition the professor gave us, so that discards $f(x)=0$)

I've shown by induction that $f(n) = n, ~\forall n\in\mathbb{N}$, then $f(m) = m, ~\forall m\in \mathbb{Z}$, then $f(q) = q, ~\forall q\in \mathbb{Q}$ but I
am having problems showing that $f(\alpha) = \alpha, ~\forall \alpha \in \mathbb{Q}'$ because I don't really know if $f(\alpha) \in \mathbb{R}$.

If I knew that $f(\mathbb{R}) \subset \mathbb{R}$ then I could do something like $f(\alpha) = f(\sqrt{\alpha})^2 &gt; 0$ if $\alpha &gt; 0$. And this would help me prove that $f(\beta) = \beta$ for all irrationals too. The only problem is that, what happens if say, $f(\alpha) = T$. Then $&gt;$ would make no sense.

I think I solved this problem but I am not sure, maybe here is where you guys can help me a little.

If we suppose that $f(\alpha)$ is a polynomial with degree $n$, we can then compute $f(\alpha^\frac{1}{n+1}) = f(\alpha)^\frac{1}{n+1}$ and that would be in $\mathbb{R}[T]$ only if the degree, $n$, of $f(\alpha)$ were $0$ thus proving that indeed $f(\alpha) \in \mathbb{R}$.

Is there any mistake or an easier way? Or is there any usefull comment anyone wants to make that could help me out. Thanks in advance :)
The only Homomorphism on $\mathbb R$ that fixes $1$, is the identity map. For the proof you can see Ring homomorphisms $\mathbb R\to\mathbb R$ .
Since $f$ is a homomorphism that fixes $1$, image of every invertible elements in $\mathbb R[T]$ should be invertible, and we should have $f_{|\mathbb R}=Id_{\mathbb R}$ .  

$f$ completely determined by the image of $T$, and so    

$f(a_0+a_1T+...+a_nT)=a_0+a_1(p(T))+...+a_n(p(T))^n$ where $p(T)=f(T)$.  

At last $f$ is isomorphism iff $f(T)=aT+b$
(Big) hint: Show that a (unital) ring homomorphism from $\mathbb{R}[T]$ to any other unital ring $A$ is completely determined by the image of $T$.