Mixing
$p$-adic supremum of cyclotomic polynomial
$begingroup$
Let $p$ be a prime number, and $Phi_{n}(T)$ be the $p^{n}$th cyclotomic polynomial, which we consider as a function on $mathbb{C}_{p}$. In Pollack's paper 'On the $p$-adic L-function of a modular form at a supersingular prime' it is claimed that
$$
text{sup}_{vert z vert < r}vertPhi_{n}(1 + z)vert_{p} = frac{r}{p^{n-1}(p - 1)}.
$$
I have failed at deriving this . My most recent attempt was as follows:
The polynomial $Phi_{n}(T)$ has $p^{n-1}(p-1)$ roots given by $zeta_{n} - 1$, where $zeta$ is a primitive $p^{n}$th root of unity. Each of these roots has valuation
$$
v_{p}(zeta_{n} - 1) = frac{1}{p^{n-1}(p-1)}.
$$
If we write $phi_{n}(1 + T) = sum_{i = 0}^{p^{n-1}(p-1)}a_{i}T^{i}$, then for $r geq 0$ we can consider the growth modulus
$$
M_{Phi}(r) = text{sup}_{i}vert a_{i} vert r^{i}.
$$
Since $Phi_{n}(1 + T)$ has all of its zeroes at the same radius $r_{Phi}$, this is its only critical radius (radius $r$ at which $vert Phi_{n}(1 + r)vert_{p} neq M_{Phi}(r)$). Let $nu = text{sup}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ and $mu = text{inf}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ then it is a theorem that the number of zeroes of $Phi_{n}(1 + T)$ in the closed disc $overline{D(0, r_{Phi})}$ (centre $0$, radius $r_{Phi}$) is equal to $nu$ and the number of zeroes in the open disc $D(0, r_{Phi})$ is equal to $mu$. Since all of the zeroes lie on $r_{Phi}$ this forces
begin{align*}
nu &= p^{n-1}(p-1) \
mu &= 0.
end{align*}
Thus, if we consider $r > r_{Phi}$ then $r$ is a regular radius and so
begin{align*}
vert Phi_{n}(1 + r) vert_{p} &= text{sup}vert a_{i} vert r^{i} \
&= M_{Phi}(r) \
&= vert a_{p^{n - 1}(p - 1)} vert r^{p^{n - 1}(p-1)} \
&= r^{p^{n - 1}(p - 1)}.
end{align*}
So for this choice of $r$ we should have
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = r^{p^{n-1}(p-1)}
$$
which is certainly not what Pollack gets. Where (and how wildly) have I gone wrong?
(For completion) For $r < r_{Phi}$ we get
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = vert a_{0} vert = vert p vert.
$$
number-theory p-adic-number-theory
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
$endgroup$
add a comment |
$begingroup$
Let $p$ be a prime number, and $Phi_{n}(T)$ be the $p^{n}$th cyclotomic polynomial, which we consider as a function on $mathbb{C}_{p}$. In Pollack's paper 'On the $p$-adic L-function of a modular form at a supersingular prime' it is claimed that
$$
text{sup}_{vert z vert < r}vertPhi_{n}(1 + z)vert_{p} = frac{r}{p^{n-1}(p - 1)}.
$$
I have failed at deriving this . My most recent attempt was as follows:
The polynomial $Phi_{n}(T)$ has $p^{n-1}(p-1)$ roots given by $zeta_{n} - 1$, where $zeta$ is a primitive $p^{n}$th root of unity. Each of these roots has valuation
$$
v_{p}(zeta_{n} - 1) = frac{1}{p^{n-1}(p-1)}.
$$
If we write $phi_{n}(1 + T) = sum_{i = 0}^{p^{n-1}(p-1)}a_{i}T^{i}$, then for $r geq 0$ we can consider the growth modulus
$$
M_{Phi}(r) = text{sup}_{i}vert a_{i} vert r^{i}.
$$
Since $Phi_{n}(1 + T)$ has all of its zeroes at the same radius $r_{Phi}$, this is its only critical radius (radius $r$ at which $vert Phi_{n}(1 + r)vert_{p} neq M_{Phi}(r)$). Let $nu = text{sup}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ and $mu = text{inf}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ then it is a theorem that the number of zeroes of $Phi_{n}(1 + T)$ in the closed disc $overline{D(0, r_{Phi})}$ (centre $0$, radius $r_{Phi}$) is equal to $nu$ and the number of zeroes in the open disc $D(0, r_{Phi})$ is equal to $mu$. Since all of the zeroes lie on $r_{Phi}$ this forces
begin{align*}
nu &= p^{n-1}(p-1) \
mu &= 0.
end{align*}
Thus, if we consider $r > r_{Phi}$ then $r$ is a regular radius and so
begin{align*}
vert Phi_{n}(1 + r) vert_{p} &= text{sup}vert a_{i} vert r^{i} \
&= M_{Phi}(r) \
&= vert a_{p^{n - 1}(p - 1)} vert r^{p^{n - 1}(p-1)} \
&= r^{p^{n - 1}(p - 1)}.
end{align*}
So for this choice of $r$ we should have
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = r^{p^{n-1}(p-1)}
$$
which is certainly not what Pollack gets. Where (and how wildly) have I gone wrong?
(For completion) For $r < r_{Phi}$ we get
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = vert a_{0} vert = vert p vert.
$$
number-theory p-adic-number-theory
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
$endgroup$
$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
$endgroup$
– Torsten Schoeneberg
Jan 15 at 19:43
1
$begingroup$
Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
$endgroup$
– Rob Rockwood
Jan 15 at 20:11
1
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
$endgroup$
– Rob Rockwood
Jan 15 at 21:01
add a comment |
$begingroup$
Let $p$ be a prime number, and $Phi_{n}(T)$ be the $p^{n}$th cyclotomic polynomial, which we consider as a function on $mathbb{C}_{p}$. In Pollack's paper 'On the $p$-adic L-function of a modular form at a supersingular prime' it is claimed that
$$
text{sup}_{vert z vert < r}vertPhi_{n}(1 + z)vert_{p} = frac{r}{p^{n-1}(p - 1)}.
$$
I have failed at deriving this . My most recent attempt was as follows:
The polynomial $Phi_{n}(T)$ has $p^{n-1}(p-1)$ roots given by $zeta_{n} - 1$, where $zeta$ is a primitive $p^{n}$th root of unity. Each of these roots has valuation
$$
v_{p}(zeta_{n} - 1) = frac{1}{p^{n-1}(p-1)}.
$$
If we write $phi_{n}(1 + T) = sum_{i = 0}^{p^{n-1}(p-1)}a_{i}T^{i}$, then for $r geq 0$ we can consider the growth modulus
$$
M_{Phi}(r) = text{sup}_{i}vert a_{i} vert r^{i}.
$$
Since $Phi_{n}(1 + T)$ has all of its zeroes at the same radius $r_{Phi}$, this is its only critical radius (radius $r$ at which $vert Phi_{n}(1 + r)vert_{p} neq M_{Phi}(r)$). Let $nu = text{sup}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ and $mu = text{inf}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ then it is a theorem that the number of zeroes of $Phi_{n}(1 + T)$ in the closed disc $overline{D(0, r_{Phi})}$ (centre $0$, radius $r_{Phi}$) is equal to $nu$ and the number of zeroes in the open disc $D(0, r_{Phi})$ is equal to $mu$. Since all of the zeroes lie on $r_{Phi}$ this forces
begin{align*}
nu &= p^{n-1}(p-1) \
mu &= 0.
end{align*}
Thus, if we consider $r > r_{Phi}$ then $r$ is a regular radius and so
begin{align*}
vert Phi_{n}(1 + r) vert_{p} &= text{sup}vert a_{i} vert r^{i} \
&= M_{Phi}(r) \
&= vert a_{p^{n - 1}(p - 1)} vert r^{p^{n - 1}(p-1)} \
&= r^{p^{n - 1}(p - 1)}.
end{align*}
So for this choice of $r$ we should have
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = r^{p^{n-1}(p-1)}
$$
which is certainly not what Pollack gets. Where (and how wildly) have I gone wrong?
(For completion) For $r < r_{Phi}$ we get
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = vert a_{0} vert = vert p vert.
$$
number-theory p-adic-number-theory
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
$endgroup$
Let $p$ be a prime number, and $Phi_{n}(T)$ be the $p^{n}$th cyclotomic polynomial, which we consider as a function on $mathbb{C}_{p}$. In Pollack's paper 'On the $p$-adic L-function of a modular form at a supersingular prime' it is claimed that
$$
text{sup}_{vert z vert < r}vertPhi_{n}(1 + z)vert_{p} = frac{r}{p^{n-1}(p - 1)}.
$$
I have failed at deriving this . My most recent attempt was as follows:
The polynomial $Phi_{n}(T)$ has $p^{n-1}(p-1)$ roots given by $zeta_{n} - 1$, where $zeta$ is a primitive $p^{n}$th root of unity. Each of these roots has valuation
$$
v_{p}(zeta_{n} - 1) = frac{1}{p^{n-1}(p-1)}.
$$
If we write $phi_{n}(1 + T) = sum_{i = 0}^{p^{n-1}(p-1)}a_{i}T^{i}$, then for $r geq 0$ we can consider the growth modulus
$$
M_{Phi}(r) = text{sup}_{i}vert a_{i} vert r^{i}.
$$
Since $Phi_{n}(1 + T)$ has all of its zeroes at the same radius $r_{Phi}$, this is its only critical radius (radius $r$ at which $vert Phi_{n}(1 + r)vert_{p} neq M_{Phi}(r)$). Let $nu = text{sup}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ and $mu = text{inf}{i : vert a_{i} vert r_{Phi}^{i} = M_{Phi}(r_{Phi})}$ then it is a theorem that the number of zeroes of $Phi_{n}(1 + T)$ in the closed disc $overline{D(0, r_{Phi})}$ (centre $0$, radius $r_{Phi}$) is equal to $nu$ and the number of zeroes in the open disc $D(0, r_{Phi})$ is equal to $mu$. Since all of the zeroes lie on $r_{Phi}$ this forces
begin{align*}
nu &= p^{n-1}(p-1) \
mu &= 0.
end{align*}
Thus, if we consider $r > r_{Phi}$ then $r$ is a regular radius and so
begin{align*}
vert Phi_{n}(1 + r) vert_{p} &= text{sup}vert a_{i} vert r^{i} \
&= M_{Phi}(r) \
&= vert a_{p^{n - 1}(p - 1)} vert r^{p^{n - 1}(p-1)} \
&= r^{p^{n - 1}(p - 1)}.
end{align*}
So for this choice of $r$ we should have
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = r^{p^{n-1}(p-1)}
$$
which is certainly not what Pollack gets. Where (and how wildly) have I gone wrong?
(For completion) For $r < r_{Phi}$ we get
$$
text{sup}_{vert z vert < r}vertphi_{n}(1 + T)vert_{p} = vert a_{0} vert = vert p vert.
$$
number-theory p-adic-number-theory
number-theory p-adic-number-theory
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
edited Jan 15 at 21:14
Rob Rockwood
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
asked Jan 15 at 13:52
Rob RockwoodRob Rockwood
1548
1548
$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
$endgroup$
– Torsten Schoeneberg
Jan 15 at 19:43
1
$begingroup$
Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
$endgroup$
– Rob Rockwood
Jan 15 at 20:11
1
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
$endgroup$
– Rob Rockwood
Jan 15 at 21:01
add a comment |
$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
$endgroup$
– Torsten Schoeneberg
Jan 15 at 19:43
1
$begingroup$
Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
$endgroup$
– Rob Rockwood
Jan 15 at 20:11
1
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
$endgroup$
– Rob Rockwood
Jan 15 at 21:01
$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
$endgroup$
– Torsten Schoeneberg
Jan 15 at 19:43
$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
$endgroup$
– Torsten Schoeneberg
Jan 15 at 19:43
1
1
$begingroup$
Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
$endgroup$
– Rob Rockwood
Jan 15 at 20:11
$begingroup$
Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
$endgroup$
– Rob Rockwood
Jan 15 at 20:11
1
1
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
$endgroup$
– Rob Rockwood
Jan 15 at 21:01
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
$endgroup$
– Rob Rockwood
Jan 15 at 21:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right and Pollack is wrong.
But:
This is a problem that cries out for the Newton copolygon, which in my opinion makes sense only when you jettison absolute value and use instead the (additive) valuation $v_p$, which you may define as $v_p(z)=-log_p|z|_p$—but I prefer to think of it just as the divisibility by $p$ and its fractional powers. I guessed from the framing of your question that you were not familiar with copolygon-talk, so below I sketch out its definition and use.
You get the copolygon of a polynomial $f(T)=sum_na_nT^n$ as a convex body in $Bbb R^2$, say coordinatized by $(xi,eta)$, by drawing for every monomial $a_nT^n$ of $f$, the closed half-plane $etale v_(a_n)+ nxi$. You see that the boundary line, $eta=v(a_n)+nxi$ tells you exactly $v(a_nz^n)$ as a function of $v(z)$. You intersect all of these half-planes to get a convex body, the Newton copolygon of $f$. Its boundary, you see immediately, tells you $v(f(z))$ as a function of $v(z)$, except when $v(z)$ is the $xi$-coordinate of a vertex of the copolygon.
A few minutes’ calculation shows you not only that the slopes of the segments of the copolygon are the $x$-coordinates of the vertices of the Newton polygon of $f$, but, much more interestingly, the $xi$-coordinates of the vertices of the copolygon are the (negative of) the slopes of the segments of the polygon.
To apply all this to your problem, you don’t need to consider the coefficents of $Phi_n$, all you need to know is that it’s an Eisenstein polynomial of degree $p^{n-1}(p-1)$ and therefore has the simplest possible Newton polygon, with vertices at $(0,1)$ and $(p^{n-1}(p-1),0)$. One segment, with slope $frac{p^{-(n-1)}}{p-1}$.
You can use the polygon, plus what I told you above, or else the knowledge that a monomial matters to the copolygon if and only if it matters to the polygon, to see that the copolygon’s boundary is the polygonal function given by $eta=p^{n-1}(p-1)xi$ if $xilefrac1{p^{n-1}(p-1)}$ and $eta=1$ otherwise. These formulas tell you the precise valuation of $Phi_n(1+z)$ depending on $v(z)$ except when this last number is equal to $frac1{p^{n-1}(p-1)}$ when of course $v(z)$ can be anything at all greater than or equal to $1$.
When translated into the language of valuations, Pollack asks for the minimum possible value of $v(Phi_n(z))$ subject to the condition that $v(z)gerho=-log_p(r)$. He’s just asking the about lowest point on the graph not to the left of $xi=rho$, and of course that’s $(rho,p^{n-1}(p-1)rho)$ and the $eta$-coordinate here is a $v$-value corresponding to an absolute value of $p^{-p^{n-1}(p-1)rho}=r^{p^{n-1}(p-1)}$, just what you got.
answered Jan 15 at 21:48
LubinLubin
44.5k44586
$endgroup$
add a comment |
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$begingroup$
You are right and Pollack is wrong.
But:
This is a problem that cries out for the Newton copolygon, which in my opinion makes sense only when you jettison absolute value and use instead the (additive) valuation $v_p$, which you may define as $v_p(z)=-log_p|z|_p$—but I prefer to think of it just as the divisibility by $p$ and its fractional powers. I guessed from the framing of your question that you were not familiar with copolygon-talk, so below I sketch out its definition and use.
You get the copolygon of a polynomial $f(T)=sum_na_nT^n$ as a convex body in $Bbb R^2$, say coordinatized by $(xi,eta)$, by drawing for every monomial $a_nT^n$ of $f$, the closed half-plane $etale v_(a_n)+ nxi$. You see that the boundary line, $eta=v(a_n)+nxi$ tells you exactly $v(a_nz^n)$ as a function of $v(z)$. You intersect all of these half-planes to get a convex body, the Newton copolygon of $f$. Its boundary, you see immediately, tells you $v(f(z))$ as a function of $v(z)$, except when $v(z)$ is the $xi$-coordinate of a vertex of the copolygon.
A few minutes’ calculation shows you not only that the slopes of the segments of the copolygon are the $x$-coordinates of the vertices of the Newton polygon of $f$, but, much more interestingly, the $xi$-coordinates of the vertices of the copolygon are the (negative of) the slopes of the segments of the polygon.
To apply all this to your problem, you don’t need to consider the coefficents of $Phi_n$, all you need to know is that it’s an Eisenstein polynomial of degree $p^{n-1}(p-1)$ and therefore has the simplest possible Newton polygon, with vertices at $(0,1)$ and $(p^{n-1}(p-1),0)$. One segment, with slope $frac{p^{-(n-1)}}{p-1}$.
You can use the polygon, plus what I told you above, or else the knowledge that a monomial matters to the copolygon if and only if it matters to the polygon, to see that the copolygon’s boundary is the polygonal function given by $eta=p^{n-1}(p-1)xi$ if $xilefrac1{p^{n-1}(p-1)}$ and $eta=1$ otherwise. These formulas tell you the precise valuation of $Phi_n(1+z)$ depending on $v(z)$ except when this last number is equal to $frac1{p^{n-1}(p-1)}$ when of course $v(z)$ can be anything at all greater than or equal to $1$.
When translated into the language of valuations, Pollack asks for the minimum possible value of $v(Phi_n(z))$ subject to the condition that $v(z)gerho=-log_p(r)$. He’s just asking the about lowest point on the graph not to the left of $xi=rho$, and of course that’s $(rho,p^{n-1}(p-1)rho)$ and the $eta$-coordinate here is a $v$-value corresponding to an absolute value of $p^{-p^{n-1}(p-1)rho}=r^{p^{n-1}(p-1)}$, just what you got.
answered Jan 15 at 21:48
LubinLubin
44.5k44586
$endgroup$
add a comment |
$begingroup$
You are right and Pollack is wrong.
But:
This is a problem that cries out for the Newton copolygon, which in my opinion makes sense only when you jettison absolute value and use instead the (additive) valuation $v_p$, which you may define as $v_p(z)=-log_p|z|_p$—but I prefer to think of it just as the divisibility by $p$ and its fractional powers. I guessed from the framing of your question that you were not familiar with copolygon-talk, so below I sketch out its definition and use.
You get the copolygon of a polynomial $f(T)=sum_na_nT^n$ as a convex body in $Bbb R^2$, say coordinatized by $(xi,eta)$, by drawing for every monomial $a_nT^n$ of $f$, the closed half-plane $etale v_(a_n)+ nxi$. You see that the boundary line, $eta=v(a_n)+nxi$ tells you exactly $v(a_nz^n)$ as a function of $v(z)$. You intersect all of these half-planes to get a convex body, the Newton copolygon of $f$. Its boundary, you see immediately, tells you $v(f(z))$ as a function of $v(z)$, except when $v(z)$ is the $xi$-coordinate of a vertex of the copolygon.
A few minutes’ calculation shows you not only that the slopes of the segments of the copolygon are the $x$-coordinates of the vertices of the Newton polygon of $f$, but, much more interestingly, the $xi$-coordinates of the vertices of the copolygon are the (negative of) the slopes of the segments of the polygon.
To apply all this to your problem, you don’t need to consider the coefficents of $Phi_n$, all you need to know is that it’s an Eisenstein polynomial of degree $p^{n-1}(p-1)$ and therefore has the simplest possible Newton polygon, with vertices at $(0,1)$ and $(p^{n-1}(p-1),0)$. One segment, with slope $frac{p^{-(n-1)}}{p-1}$.
You can use the polygon, plus what I told you above, or else the knowledge that a monomial matters to the copolygon if and only if it matters to the polygon, to see that the copolygon’s boundary is the polygonal function given by $eta=p^{n-1}(p-1)xi$ if $xilefrac1{p^{n-1}(p-1)}$ and $eta=1$ otherwise. These formulas tell you the precise valuation of $Phi_n(1+z)$ depending on $v(z)$ except when this last number is equal to $frac1{p^{n-1}(p-1)}$ when of course $v(z)$ can be anything at all greater than or equal to $1$.
When translated into the language of valuations, Pollack asks for the minimum possible value of $v(Phi_n(z))$ subject to the condition that $v(z)gerho=-log_p(r)$. He’s just asking the about lowest point on the graph not to the left of $xi=rho$, and of course that’s $(rho,p^{n-1}(p-1)rho)$ and the $eta$-coordinate here is a $v$-value corresponding to an absolute value of $p^{-p^{n-1}(p-1)rho}=r^{p^{n-1}(p-1)}$, just what you got.
answered Jan 15 at 21:48
LubinLubin
44.5k44586
$endgroup$
add a comment |
$begingroup$
You are right and Pollack is wrong.
But:
This is a problem that cries out for the Newton copolygon, which in my opinion makes sense only when you jettison absolute value and use instead the (additive) valuation $v_p$, which you may define as $v_p(z)=-log_p|z|_p$—but I prefer to think of it just as the divisibility by $p$ and its fractional powers. I guessed from the framing of your question that you were not familiar with copolygon-talk, so below I sketch out its definition and use.
You get the copolygon of a polynomial $f(T)=sum_na_nT^n$ as a convex body in $Bbb R^2$, say coordinatized by $(xi,eta)$, by drawing for every monomial $a_nT^n$ of $f$, the closed half-plane $etale v_(a_n)+ nxi$. You see that the boundary line, $eta=v(a_n)+nxi$ tells you exactly $v(a_nz^n)$ as a function of $v(z)$. You intersect all of these half-planes to get a convex body, the Newton copolygon of $f$. Its boundary, you see immediately, tells you $v(f(z))$ as a function of $v(z)$, except when $v(z)$ is the $xi$-coordinate of a vertex of the copolygon.
A few minutes’ calculation shows you not only that the slopes of the segments of the copolygon are the $x$-coordinates of the vertices of the Newton polygon of $f$, but, much more interestingly, the $xi$-coordinates of the vertices of the copolygon are the (negative of) the slopes of the segments of the polygon.
To apply all this to your problem, you don’t need to consider the coefficents of $Phi_n$, all you need to know is that it’s an Eisenstein polynomial of degree $p^{n-1}(p-1)$ and therefore has the simplest possible Newton polygon, with vertices at $(0,1)$ and $(p^{n-1}(p-1),0)$. One segment, with slope $frac{p^{-(n-1)}}{p-1}$.
You can use the polygon, plus what I told you above, or else the knowledge that a monomial matters to the copolygon if and only if it matters to the polygon, to see that the copolygon’s boundary is the polygonal function given by $eta=p^{n-1}(p-1)xi$ if $xilefrac1{p^{n-1}(p-1)}$ and $eta=1$ otherwise. These formulas tell you the precise valuation of $Phi_n(1+z)$ depending on $v(z)$ except when this last number is equal to $frac1{p^{n-1}(p-1)}$ when of course $v(z)$ can be anything at all greater than or equal to $1$.
When translated into the language of valuations, Pollack asks for the minimum possible value of $v(Phi_n(z))$ subject to the condition that $v(z)gerho=-log_p(r)$. He’s just asking the about lowest point on the graph not to the left of $xi=rho$, and of course that’s $(rho,p^{n-1}(p-1)rho)$ and the $eta$-coordinate here is a $v$-value corresponding to an absolute value of $p^{-p^{n-1}(p-1)rho}=r^{p^{n-1}(p-1)}$, just what you got.
answered Jan 15 at 21:48
LubinLubin
44.5k44586
$endgroup$
You are right and Pollack is wrong.
But:
This is a problem that cries out for the Newton copolygon, which in my opinion makes sense only when you jettison absolute value and use instead the (additive) valuation $v_p$, which you may define as $v_p(z)=-log_p|z|_p$—but I prefer to think of it just as the divisibility by $p$ and its fractional powers. I guessed from the framing of your question that you were not familiar with copolygon-talk, so below I sketch out its definition and use.
You get the copolygon of a polynomial $f(T)=sum_na_nT^n$ as a convex body in $Bbb R^2$, say coordinatized by $(xi,eta)$, by drawing for every monomial $a_nT^n$ of $f$, the closed half-plane $etale v_(a_n)+ nxi$. You see that the boundary line, $eta=v(a_n)+nxi$ tells you exactly $v(a_nz^n)$ as a function of $v(z)$. You intersect all of these half-planes to get a convex body, the Newton copolygon of $f$. Its boundary, you see immediately, tells you $v(f(z))$ as a function of $v(z)$, except when $v(z)$ is the $xi$-coordinate of a vertex of the copolygon.
A few minutes’ calculation shows you not only that the slopes of the segments of the copolygon are the $x$-coordinates of the vertices of the Newton polygon of $f$, but, much more interestingly, the $xi$-coordinates of the vertices of the copolygon are the (negative of) the slopes of the segments of the polygon.
To apply all this to your problem, you don’t need to consider the coefficents of $Phi_n$, all you need to know is that it’s an Eisenstein polynomial of degree $p^{n-1}(p-1)$ and therefore has the simplest possible Newton polygon, with vertices at $(0,1)$ and $(p^{n-1}(p-1),0)$. One segment, with slope $frac{p^{-(n-1)}}{p-1}$.
You can use the polygon, plus what I told you above, or else the knowledge that a monomial matters to the copolygon if and only if it matters to the polygon, to see that the copolygon’s boundary is the polygonal function given by $eta=p^{n-1}(p-1)xi$ if $xilefrac1{p^{n-1}(p-1)}$ and $eta=1$ otherwise. These formulas tell you the precise valuation of $Phi_n(1+z)$ depending on $v(z)$ except when this last number is equal to $frac1{p^{n-1}(p-1)}$ when of course $v(z)$ can be anything at all greater than or equal to $1$.
When translated into the language of valuations, Pollack asks for the minimum possible value of $v(Phi_n(z))$ subject to the condition that $v(z)gerho=-log_p(r)$. He’s just asking the about lowest point on the graph not to the left of $xi=rho$, and of course that’s $(rho,p^{n-1}(p-1)rho)$ and the $eta$-coordinate here is a $v$-value corresponding to an absolute value of $p^{-p^{n-1}(p-1)rho}=r^{p^{n-1}(p-1)}$, just what you got.
answered Jan 15 at 21:48
LubinLubin
44.5k44586
answered Jan 15 at 21:48
LubinLubin
44.5k44586
answered Jan 15 at 21:48
LubinLubin
44.5k44586
answered Jan 15 at 21:48
LubinLubin
44.5k44586
44.5k44586
add a comment |
add a comment |
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$begingroup$
The first case where your result and the one you quote differ is $p=3, n=1$, where $Phi(1+z) = z^2+3z+3$; which I would think makes that supremum equal to $max(r^2, 1/3)$, which at least partially matches yours, but certainly not $r/2$ (in general, I think we have the maximum of your result and $p^{-n}$ as answer). --- However, I do not even find that claim in the paper (which I assume is this: pdfs.semanticscholar.org/c13a/…). Where is it?
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– Torsten Schoeneberg
Jan 15 at 19:43
1
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Should be in the proof of lemma 4.5 (I chose the case j = 0 to simplify things)
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– Rob Rockwood
Jan 15 at 20:11
1
$begingroup$
I didn't include the analysis for radii below the critical radius, but that would give that the supremum on those radii is the absolute value of the constant term which fits with your example.
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– Rob Rockwood
Jan 15 at 21:01