# Why is $\phi: \ R[X] \rightarrow R[X]: \ \sum_{i=0}^n a_i X^i \mapsto \sum_{i=0}^n a_{n-i}X^i$ multiplicative?

I did the following exercise:

Define $\phi: \ R[X] \rightarrow R[X]: \ \sum_{i=0}^n a_i X^i \mapsto \sum_{i=0}^n a_{n-i}X^i$. Let $f = \sum_{i=0}^n a_i X^i$ with $a_0 \neq 0 \neq a_n$. Show the following:

$$f \ \text{ is irreducible} \iff \phi(f) \ \text{is irreducible}$$

It's enough to show the implication "$\implies$" since $\phi(\phi(f))=f$. I made it to show that $\phi$ is additive, but failed to show that $\phi$ is multiplicative though I know it's true by a previous exercise. It's clear that $\phi$ is bijective. We know that $\phi$ is unitary, so units are mapped to units:

Let $\phi: R \mapsto R'$ unitary and an isomorphism. Let $x \in R$ be a unit. Then $\exists x^* \in R, \ x \cdot x^*= 1$. Then $\phi(1) = \phi(x\cdot x^*) = \phi(x)\cdot \phi(x^*) = 1'$.

Moreover, irreducible units are mapped to irreducible units: Assume that $\phi(x) = y' \cdot z'$ for some elements $x',y'\in R'$. By surjectivity, $\exists y,z \in R, \ \phi(x) =x' \ \text{and} \ \phi(y) = y'$. Now $z'$ is a unit or $y'$ is a unit because one of the elements $z,y$ is a unit.

I'd say that this would end the proof. You could help me by checking and showing that f is multiplicative. Thanks.

For $f\ne 0$, we have $\phi(f)(X)=f(X^{-1})\cdot X^n$, so for $f,g\ne0$ we have $\phi(fg)(X)=(fg)(X^{-1})X^{\deg(fg)}=f(X^{-1})g(X^{-1})X^{\deg(f)+\deg(g)$}=f(X^{-1})X^{\deg f}g(X^{-1}X^{\deg g}=\phi(f)(X)\cdot \phi(g)(X)$. By the way,$\phi$is not additive.$\phi(X^2+X)\ne\phi(X^2)+\phi(X)$. • Sorry for the wrong statements. (I noticed that$\phi(\phi(f)) = f\$ is far not always true as well.) Can you say something about the rest of my answer as well? – Koenraad van Duin Aug 19 '13 at 18:54