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I am trying to prove a series of propositions:

Given any homomorphism p from $\mathbb{R}$[X] to $\mathbb{R}$[X], show that it is equal to $\phi_g$ for a unique g in $\mathbb{R}$[X], with $\phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(\sum_{i=0}^n a_iX^i) =\sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = \sum_{i=0}^n p(a_i)p(X)^i = \sum_{i=0}^n a_ig(X)^i = \phi_g(f)$.

Show that if h,g $\in \mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $\in$ R$^x$ and b$\in \mathbb{R}$.

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  • $\begingroup$ As it stands both statements you ask to show are false. What makes you believe they are true? $\endgroup$ – Servaes Nov 30 '18 at 14:42
  • $\begingroup$ And Servaes meant the non-trivial automorphism of $\mathbb{Q}(\sqrt{3})$ extends to an automorphism of $\mathbb{R}$ (that we can't define without things like the axiom of choice) and $\mathbb{R}[X]$ $\endgroup$ – reuns Nov 30 '18 at 22:16
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For the claim to be true you need the additional hypothesis that $p$ is $\Bbb{R}$-linear.

An $\Bbb{R}$-linear ring homomorphism $$p: \Bbb{R}[X] \longrightarrow\ \Bbb{R}[X]$$ is determined by where it maps $X$. It follows immediately from the ring axioms that $p=\phi_{p(X)}$. Indeed, if $p$ is $\Bbb{R}$-linear then $p(r)=r$ for all $r\in\Bbb{R}$, and your algebraic manipulations show that then $$p(f)=f(p(X))=\phi_{p(X)}(f),$$ for all $f\in\Bbb{R}[X]$. To see that the $\Bbb{R}$-linear automorphisms are precisely the linear maps $p$ for which $p(X)$ is linear, note that $\deg p(f)=\deg p(X)\cdot\deg f$ for all $f\in\Bbb{R}[X]$, so for $p$ to be surjective we must have $\deg p(X)=1$. Check that $\phi_g$ is invertible for all linear $g$.

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