Show that every homomorphism $\mathbb{R}$[X] $\rightarrow$ $\mathbb{R}$[X] can is equal to $φ_g$ for a unique g $\in$ $\mathbb{R}$[X], given by $φ_g(f)$ = $f(g(X))$

My guess for any homomorphism $h$, $g = h(X)$ but I'm not sure how to proceed from there.

  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – José Carlos Santos Nov 30 '18 at 8:40
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    $\begingroup$ Your guess is right! Now you can directly calculate $h(f) = \varphi_g(f)$ for any $f$ explicitly, using that $h$ is a homomorphism. $\endgroup$ – Gnampfissimo Nov 30 '18 at 9:00
  • $\begingroup$ Every homomorphism of $\mathbb R$-algebras, that is, fixing $\mathbb R$. $\endgroup$ – lhf Nov 30 '18 at 10:07

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