How to find the Newton polygon of the polynomial product $ \ \prod_{i=1}^{p^2} (1-iX)$ ?
Answer:
Let $ \ f(X)=\prod_{i=1}^{p^2} (1-iX)=(1-X)(1-2X) \cdots (1-pX) \cdots (1-p^2X).$
If I multiply , then we will get a polynomial of degree $p^2$.
But it is complicated to express it as a polynomial form.
So it is complicated to calculate the vertices $ (0, ord_p(a_0)), \ (1, ord_p(a_1)), \ (2, ord_p(a_2)), \ \cdots \cdots$
of the above product.
Help me doing this
Best Answer
It’s really quite simple. There are $p^2-p$ roots $\rho$ with $v(\rho)=0$, $p-1$ roots with $v(\rho)=-1$, and one root with $v(\rho)=-2$. Consequently, there is one segment of the polygon with slope $0$ and width $p^2-p$, one segment with slope $1$ and width $p-1$, and one segment with slope $2$ and width $1$.
Thus, the vertices are $(0,0)$, $(p^2-p,0)$, $(p^2-1,p-1)$, and $(p^2,p+1)$.