# How to find the Newton polygon of the polynomial product $\ \prod_{i=1}^{p^2} (1-iX)$

How to find the Newton polygon of the polynomial product $$\ \prod_{i=1}^{p^2} (1-iX)$$ ?

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

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 $$(p^2,p+1)$$.

• excellent explanation. I got it – M. A. SARKAR Nov 25 '18 at 4:15
• Does this result hold for $p=2$ ? Because for $p=2$, we have $f(X)=(1-X)(1-2X)(1-3X)(1-4X)=1-10X+35X^2-50X^3+24X^4$. Thus the vertices are $$(0,0), (1,1), (2,0),(3,1), (4,3)$$. The vertex $(1,1)$ makes disturbance . Would you please do little bit more? – M. A. SARKAR Nov 26 '18 at 14:50
• No, $(1,1)$ is not a vertex of the Newton polygon. – Lubin Nov 26 '18 at 20:59

Partial Answer: regarding the coefficients of the polynomial:

Fix one term in the brackets, say $$Y=(1-5X)$$. In order for the coefficient $$5$$ to contribute to $$a_j$$, we have to multiply $$Y$$ with $$j-1$$ other brackets, since this is the only way of getting a power of $$j$$ for $$X$$. This corresponds to choosing a subset $$S \in \{1,2,\ldots,p^{2}\}$$ of size $$j-1$$ since each term in the product has a unique coefficient for $$X$$ that is in $$\{1,2,\ldots,p^{2}\}$$. This leads to

$$$$a_j=(-1)^{j} \underset{ S \subset \{1,2, \ldots, p^{2} \}, \ |S|=j}{\sum} \prod \limits_{s \in S} s \ .$$$$

• what is the Newton polygon? – M. A. SARKAR Nov 20 '18 at 8:09
• @M.A. SARKAR What do you mean by $ord_p(a_j)$? I dont know these kind of polynomials but thought an expression for the coefficnets might help – sigmatau Nov 20 '18 at 8:13
• This is from discrete valuation field like p-adic field and $ord_p$ is a valuation function. If $a_j=\frac{a}{b}p^n$, where $a,b$ are coprime then $ord_p(a_j)=n$. – M. A. SARKAR Nov 20 '18 at 8:21
• I see, should I add my answer as a comment, sicne I don't see how to find $ord_{p}(a_j)$ right now. – sigmatau Nov 20 '18 at 8:42
• or I just leave it as a partial answer. – sigmatau Nov 20 '18 at 8:44