Mixing
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)$ ?
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
p-adic-number-theory valuation-theory
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
add a comment |
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
p-adic-number-theory valuation-theory
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
add a comment |
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
p-adic-number-theory valuation-theory
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
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
p-adic-number-theory valuation-theory
p-adic-number-theory valuation-theory
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
asked Nov 20 '18 at 7:27
M. A. SARKAR
2,1831619
2,1831619
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 $(p^2,p+1)$.
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
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
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
add a comment |
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}
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
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
1
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
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
add a comment |
Your Answer
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2 Answers
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2 Answers
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active
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votes
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 $(p^2,p+1)$.
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
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
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
add a comment |
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 $(p^2,p+1)$.
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
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
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
add a comment |
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 $(p^2,p+1)$.
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
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 $(p^2,p+1)$.
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
answered Nov 24 '18 at 21:16
Lubin
43.7k44585
43.7k44585
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
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
add a comment |
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
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
excellent explanation. I got it
– M. A. SARKAR
Nov 25 '18 at 4:15
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
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
2
2
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
No, $(1,1)$ is not a vertex of the Newton polygon.
– Lubin
Nov 26 '18 at 20:59
add a comment |
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}
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
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
1
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
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
add a comment |
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}
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
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
1
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
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
add a comment |
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}
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
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}
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
edited Nov 20 '18 at 8:43
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
answered Nov 20 '18 at 7:59
sigmatau
1,7551924
1,7551924
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
1
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
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
add a comment |
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
1
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
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
what is the Newton polygon?
– M. A. SARKAR
Nov 20 '18 at 8:09
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
@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
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
1
1
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
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
2
2
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
or I just leave it as a partial answer.
– sigmatau
Nov 20 '18 at 8:44
add a comment |
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