1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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. $\phi(X^2+X)\ne\phi(X^2)+\phi(X)$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Koenraad van Duin Aug 19 '13 at 18:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.