Automorphism preserves irreducibility?

Question

There is a claim in my notes saying: If $D$ is an integral domain and $f:D[X]\to D[X]$ is an automorphism then $p(x)\in D[x]$ is irreducible iff $f(p(x))$ is irreducible. I have tried to prove this but I get stuck because I need to prove that $f$ sends constants to constants. Any help?

Answer 1

The question is unclear. By automorphism, do you mean $D$-algebra automorphism? If you do, then this is easy, because $f(1)=1$ and so for any $d \in D$, $f(d)=f(d \cdot 1)=d f(1)=d$, thus $f$ preserves constants.

If $f$ is just a ring automorphism, then this is false with the (non-standard) definition of irreducible polynomials you have provided in the comments. Let $k$ be a field and let $D=k[Y]$, then we have a $k$-linear automorphism $f:D[X] \to D[X]$ that sends $X \mapsto Y, Y \mapsto X$. Under this definition $X$ is irreducible, but $f(X)=Y$ is not irreducible, because it is constant. (If you define constants to be irreducible, then here's another example: $Y^2$ is irreducible, because it's constant, but $f(Y^2)=X^2$ is not irreducible.)