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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

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

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  • $\begingroup$ excellent explanation. I got it $\endgroup$ – M. A. SARKAR Nov 25 '18 at 4:15
  • $\begingroup$ 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? $\endgroup$ – M. A. SARKAR Nov 26 '18 at 14:50
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    $\begingroup$ No, $(1,1)$ is not a vertex of the Newton polygon. $\endgroup$ – Lubin Nov 26 '18 at 20:59
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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

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

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  • $\begingroup$ what is the Newton polygon? $\endgroup$ – M. A. SARKAR Nov 20 '18 at 8:09
  • $\begingroup$ @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 $\endgroup$ – sigmatau Nov 20 '18 at 8:13
  • $\begingroup$ 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$. $\endgroup$ – M. A. SARKAR Nov 20 '18 at 8:21
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    $\begingroup$ I see, should I add my answer as a comment, sicne I don't see how to find $ord_{p}(a_j)$ right now. $\endgroup$ – sigmatau Nov 20 '18 at 8:42
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    $\begingroup$ or I just leave it as a partial answer. $\endgroup$ – sigmatau Nov 20 '18 at 8:44

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