What does $φ(a) = a$ mean in this statement? Let $F$ be a field and let $φ:F[x] \to F[x]$ be an isomorphism such that $φ(a)=a $ for every $a$ in $F$. Prove that $f(x)$ is irreducible in $F[x]$ if and only if $φ(f(x))$ is. [Hint: First prove that for any $g(x)$ in $F[x]$, $g(x)$ is nonconstant if and only if $φ(g(x))$ is nonconstant.]
I found the solution online and everywhere I find it, its basically translating this question to:
Prove that $f(x)$ is irreducible if and only if $f(x+1)$ is. I dont understand how this problem translates to this. 
When I tried proving this I thought the transformation meant $φ(f(x))= f(x)$.
So I guess the question Im asking is what does φ(a) = a mean in this situation.
 A: $F[x]$ is just the ring of polynomials with coefficients from the field $F$. So if $\phi(a)=a$ for all $a \in F$, it is just saying that all the constants in $F[x]$ get mapped to themselves under the mapping $\phi$ i.e. polynomials of the form $f(x)=a$ for $a \in F$ get mapped to themselves.
To prove the statement, first suppose that $f(x)$ is irreducible in $F[x]$. Then suppose $\phi(f(x)) = g(x)h(x)$ for some $g(x),h(x) \in F[x]$. Now,
$$\phi(f(x)) = g(x)h(x) \implies f(x)= \phi^{-1}(g(x)h(x)) = \phi^{-1}(g(x))\phi^{-1}(h(x))$$
But $f(x)$ is irreducible so either $\phi^{-1}(g(x))=a$ or $\phi^{-1}(h(x))=a$ for some $a \in F[x]$ and hence either $g(x)$ or $h(x)$ is a unit so $f(x)$ is irreducible.
For the other direction, suppose $\phi(f(x))$ is irreducible. Then if $f(x)$ were reducible, we could write $f(x)=g(x)h(x)$ for some non-constant polynomials in $g(x),h(x) \in F[x]$. But then we would have $$\phi(f(x))=\phi(g(x)h(x))= \phi(g(x)) \phi(h(x))$$
Using the hint you are given, $\phi(g(x))$ and $\phi(h(x))$ are non-constant since $g(x)$ and $h(x)$ are assumed to be non-constant and hence $\phi(f(x))$ is reducible which is a contradiction. Thus $f(x)$ must be irreducible.
