I have to show $p=p(x-\lambda)$ if and only if they have the same zeros in $F$

Question

Suppose $F$ is a field, $|F|\geq n \geq 2$. Given $\lambda \in F$ I know $p,p(x-\lambda)\in F[x]$ are irreducible monic polynomials. I have to show $p=p(x-\lambda)$ if and only if they have the same zeros in $\bar{F}$ or equivalently for each zero $a \in \bar{F}$ of $p$ , $a-\lambda$ is also a zero of $p$

Answer 1

I give you a more general result:

If you have two irreducible polynomials $f,g \in F[x]$, then the following are equivalent:

1. $f=\lambda g$ for some scalar $\lambda \in F$

2. $f$ and $g$ have the same zeroes

3. $f$ and $g$ have a common zero

Clearly 1 implies 2 and 2 implies 3. On the other hand, let $\alpha$ be a common root of $f$ and $g$. Then $f$ and $g$ are multiples of the minimal polynomial of $\alpha$ (say $m$), and since they are irreducible, they must be of the form $\mbox{scalar} \cdot m$, and so we have 1.