if $p(x)$ is irreducible show $cp(x)$ also is irreducible

Question

If $p(x)$ is irreducible in $K[x]$ and $c \ne 0$, $c \in K$ $\Rightarrow cp(x)$ is irreducible in $K[x]$.

I would start saying that $p(x)=a(x)b(x)$ with $a(x)$ a unit $\lambda \in K$, without loss of generality,

$\Rightarrow p(x)=\lambda b(x)$ $\Rightarrow cp(x)=(c\lambda) b(x)$, because $c \lambda \in K \Rightarrow cp(x)$ is irreducible.

I'm not sure if it shows the statement.

Answer 1

Your proof doesn't show the claimed statement.

First of all your proof basically goes: We can write $p(x)=\lambda b(x)$ with $\lambda$ a unit in $K$, $b(x)=\newcommand{\inv}{^{-1}}\lambda\inv p(x)$. Then you write $cp(x)=(c\lambda) b(x)$ and claim that this implies that $cp(x)$ is irreducible. But this doesn't follow. It's essentially circular, you don't know that $b(x)$ or $(c\lambda) b(x)$ are irreducible without using this lemma in the first place.

Also as pointed out in the comments, depending on your definition of irreducible, this statement may be false.

Also the choice of $K$ to represent an integral domain is less than ideal. $K,L,M$ are usually reserved for fields. More common for integral domains are either $A,B,\ldots$ or $R,S,T,\ldots$.