# if $f$ has Newton Polygon consisting of one segment $(0,0)$ to $(n,m)$ with $m,n$ coprime, then $f$ cannot be factored

Let $$f(X)\in 1+ X\mathbb{Z}_p[X]$$ have Newton Polygon consisting of one segment joining $$(0,0)$$ to $$(n,m)$$ with $$m,n$$ coprime. I have to show that $$f(X)$$ cannot be factored as a product of two polynomials with coefficients in $$\mathbb{Z}_p$$.

I know that the slope of the Newton Polygon is $$m/n$$ and since $$m,n$$ coprime, this does not lie in $$\mathbb{Z}_p$$. I also know there exists a theorem which says something about the number of same slopes, but I do not fully understand this theorem. Can someone help me to understand this question and hopefully to understand the theorem better? Thanks!

If you understand that every root $$\rho$$ of $$f$$ satisfies $$v(\rho)=-m/n$$, then you see that this will happen for both of $$g$$ and $$h$$ if $$f=gh$$.
Now, what can the Newton polygon of $$g$$ be? It will be of width $$r$$ for some integer with $$0, since neither $$g$$ nor $$h$$ is constant. And the right-hand vertex? Since the slope is still $$m/n$$, the vertex will be at $$(r, \frac{rm}n)$$. But the $$y$$-coordinate has to be an integer if the polynomial has $$\Bbb Z_p$$-coefficients. Thus $$g\notin\Bbb Z_p[X]$$.