Let $F$ be a field. Let $\varphi : F[x] \rightarrow F[x]$ be an isomorphism such that $\varphi(a)=a$ for every $a \in F$. Prove that $f \in F[x]$ is irreducible if and only if $\varphi(f)$ is.
How will I be able to start this proof? Any help will be greatly appreciated. Thanks in advance!
If $R \to S$ is any ring isomorphism, then it maps irreducible elements of $R$ to irreducible elements of $S$. This can be derived directly from the definitions. Note that this is - in some sense - trivial a priori since isomorphisms are required to preserve the whole ring structure and irreducibility only refers to this structure.
It doesn't make any sense to restrict to polynomial rings here. The proof doesn't simplify in this special case, and this restriction obscures what is really going on. Also the assumption that the isomorphism is the identity on constants is not necessary. Of course this special case is useful etc., but if you have this extra assumption you are somehow invited to use it in your proof, but then your proof will be either a) wrong, or b) too complicated.