# Homomorphisms and automorphisms on polynomial rings

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$$\in \mathbb{R}$$.

• As it stands both statements you ask to show are false. What makes you believe they are true? – Servaes Nov 30 '18 at 14:42
• 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 $\mathbb{R}[X]$ – reuns Nov 30 '18 at 22:16

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 $$g$$.